Beginning Algebra

2 CHAPTER 1 REVIEW OF REAL NUMBERS

1.1 T I P S F O R S U C C E S S I N M AT H E M AT I C S

O b j e c t i v e s

1 Get ready for this course.

2 Understand some general tips for success.

3 Understand how to use this text.

4 Get help as soon as you need it.

5 Learn how to prepare for and take an exam.

6 Develop good time management.

Before reading this section, remember that your instructor is your best source forinformation. Please see your instructor for any additional help or information.

1 Getting Ready for This CourseNow that you have decided to take this course, remember that a positive atti-tude will make all the difference in the world.Your belief that you can succeedis just as important as your commitment to this course. Make sure that you areready for this course by having the time and positive attitude that it takes tosucceed.

Next, make sure that you have scheduled your math course at a time thatwill give you the best chance for success. For example, if you are also working,you may want to check with your employer to make sure that your work hourswill not conflict with your course schedule. Also, schedule your class during atime of day when you are more attentive and do your best work.

On the day of your first class period, double-check your schedule and allowyourself extra time to arrive in case of traffic problems or difficulty locating yourclassroom. Make sure that you bring at least your textbook, paper, and a writinginstrument. Are you required to have a lab manual, graph paper, calculator, orsome other supply besides this text? If so, also bring this material with you.

2 General Tips for SuccessBelow are some general tips that will increase your chance for success in amathematics class. Many of these tips will also help you in other courses youmay be taking.

Exchange names and phone numbers with at least one other person inclass.This contact person can be a great help if you miss an assignment or wantto discuss math concepts or exercises that you find difficult.

Choose to attend all class periods and be on time. If possible, sit nearthe front of the classroom. This way, you will see and hear the presentationbetter. It may also be easier for you to participate in classroom activities.

Do your homework. You’ve probably heard the phrase “practice makesperfect” in relation to music and sports. It also applies to mathematics.You willfind that the more time you spend solving mathematics problems, the easierthe process becomes. Be sure to schedule enough time to complete your as-signments before the next class period.

Check your work. Review the steps you made while working a problem.Learn to check your answers to the original problems. You may also compareyour answers with the answers to selected exercises section in the back of the

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TIPS FOR SUCCESS IN MATHEMATICS SECTION 1.1 3

book. If you have made a mistake, try to figure out what went wrong.Then cor-rect your mistake. If you can’t find what went wrong, don’t erase your work orthrow it away. Bring your work to your instructor, a tutor in a math lab, or aclassmate. It is easier for someone to find where you had trouble if they look atyour original work.

Learn from your mistakes and be patient with yourself. Everyone, evenyour instructor, makes mistakes. (That definitely includes me—Elayn Martin-Gay.) Use your errors to learn and to become a better math student.The key isfinding and understanding your errors.

Was your mistake a careless one, or did you make it because you can’tread your own math writing? If so, try to work more slowly or write more neatlyand make a conscious effort to carefully check your work.

Did you make a mistake because you don’t understand a concept? Takethe time to review the concept or ask questions to better understand it.

Did you skip too many steps? Skipping steps or trying to do too manysteps mentally may lead to preventable mistakes.

Know how to get help if you need it. It’s all right to ask for help. In fact,it’s a good idea to ask for help whenever there is something that you don’t un-derstand. Make sure you know when your instructor has office hours and howto find his or her office. Find out whether math tutoring services are availableon your campus. Check out the hours, location, and requirements of the tutor-ing service. Videotapes and software are available with this text. Learn how toaccess these resources.

Organize your class materials, including homework assignments, gradedquizzes and tests, and notes from your class or lab. All of these items will bevaluable references throughout your course especially when studying for up-coming tests and the final exam. Make sure that you can locate these materialswhen you need them.

Read your textbook before class. Reading a mathematics textbook is un-like leisure reading such as reading a novel or newspaper. Your pace will bemuch slower. It is helpful to have paper and a pencil with you when you read.Try to work out examples on your own as you encounter them in your text.You should also write down any questions that you want to ask in class. Whenyou read a mathematics textbook, some of the information in a section may beunclear. But when you hear a lecture or watch a videotape on that section,you will understand it much more easily than if you had not read your textbeforehand.

Don’t be afraid to ask questions. Instructors are not mind readers. Manytimes we do not know a concept is unclear until a student asks a question. Youare not the only person in class with questions. Other students are normallygrateful that someone has spoken up.

Hand in assignments on time. This way you can be sure that you will notlose points for being late. Show every step of a problem and be neat and or-ganized. Also be sure that you understand which problems are assigned forhomework. You can always double-check this assignment with another stu-dent in your class.

3 Using This TextThere are many helpful resources that are available to you in this text. It is im-portant that you become familiar with and use these resources.They should in-crease your chances for success in this course.



• The main section of exercises in each exercise set is referenced by an exam-ple(s). Use this referencing if you have trouble completing an assignmentfrom the exercise set.

• If you need extra help in a particular section, look at the beginning of thesection to see what videotapes and software are available.

• Make sure that you understand the meaning of the icons that are besidemany exercises. The video icon tells you that the corresponding exercisemay be viewed on the videotape that corresponds to that section. The pen-cil icon tells you that this exercise is a writing exercise in which youshould answer in complete sentences.The icon tells you that the exerciseinvolves geometry.

• Integrated Reviews in each chapter offer you a chance to practice—in oneplace—the many concepts that you have learned separately over severalsections.

• There are many opportunities at the end of each chapter to help you under-stand the concepts of the chapter.

Chapter Highlights contain chapter summaries and examples.

Chapter Reviews contain review problems organized by section.

Chapter Tests are sample tests to help you prepare for an exam.

Cumulative Reviews are reviews consisting of material from the begin-ning of the book to the end of that particular chapter.

See the preface at the beginning of this text for a more thorough explanationof the features of this text.

4 Getting HelpIf you have trouble completing assignments or understanding the mathemat-ics, get help as soon as you need it! This tip is presented as an objective on itsown because it is so important. In mathematics, usually the material presentedin one section builds on your understanding of the previous section. Thismeans that if you don’t understand the concepts covered during a class period,there is a good chance that you will not understand the concepts covered dur-ing the next class period. If this happens to you, get help as soon as you can.

Where can you get help? Many suggestions have been made in the sectionon where to get help, and now it is up to you to do it.Try your instructor, a tutor-ing center, or a math lab, or you may want to form a study group with fellowclassmates. If you do decide to see your instructor or go to a tutoring center,make sure that you have a neat notebook and are ready with your questions.

5 Preparing for and Taking an ExamMake sure that you allow yourself plenty of time to prepare for a test. If youthink that you are a little “math anxious,” it may be that you are not preparingfor a test in a way that will ensure success. The way that you prepare for a testin mathematics is important. To prepare for a test,

1. Review your previous homework assignments. You may also want to reworksome of them.

2. Review any notes from class and section-level quizzes you have taken. (Ifthis is a final exam, also review chapter tests you have taken.)

3. Review concepts and definitions by reading the Highlights at the end ofeach chapter.

4. Practice working out exercises by completing the Chapter Review found atthe end of each chapter. (If this is a final exam, go through a Cumulative


TIPS FOR SUCCESS IN MATHEMATICS SECTION 1.1 5

Review. There is one found at the end of each chapter except Chapter 1.Choose the review found at the end of the latest chapter that you have cov-ered in your course.) Don’t stop here!

5. It is important that you place yourself in conditions similar to test condi-tions to find out how you will perform. In other words, as soon as you feelthat you know the material, get a few blank sheets of paper and take a sam-ple test.There is a Chapter Test available at the end of each chapter. Duringthis sample test, do not use your notes or your textbook. Once you com-plete the Chapter Test, check your answers in the back of the book. If anyanswer is incorrect, there is a CD available with each exercise of each chap-ter test worked. Use this CD or your instructor to correct your sample test.Your instructor may also provide you with a review sheet. If you are notsatisfied with the results, study the areas that you are weak in and try again.

6. Get a good night’s sleep before the exam.

7. On the day of the actual test, allow yourself plenty of time to arrive atwhere you will be taking your exam.

When taking your test,

1. Read the directions on the test carefully.

2. Read each problem carefully as you take the test. Make sure that you an-swer the question asked.

3. Pace yourself by first completing the problems you are most confidentwith. Then work toward the problems you are least confident with. Watchyour time so you do not spend too much time on one particular problem.

4. If you have time, check your work and answers.

5. Do not turn your test in early. If you have extra time, spend it double-checking your work.

6 Managing Your Time

As a college student, you know the demands that classes, homework, work,and family place on your time. Some days you probably wonder how you’llever get everything done. One key to managing your time is developing aschedule. Here are some hints for making a schedule:

1. Make a list of all of your weekly commitments for the term. Include classes,work, regular meetings, extracurricular activities, etc. You may also find ithelpful to list such things as laundry, regular workouts, grocery shopping,etc.

2. Next, estimate the time needed for each item on the list. Also make a noteof how often you will need to do each item. Don’t forget to include time es-timates for reading, studying, and homework you do outside of your classes.You may want to ask your instructor for help estimating the time needed.

3. In the following exercise set, you are asked to block out a typical week onthe schedule grid given. Start with items with fixed time slots like classesand work.

4. Next, include the items on your list with flexible time slots. Think carefullyabout how best to schedule some items such as study time.

5. Don’t fill up every time slot on the schedule. Remember that you need toallow time for eating, sleeping, and relaxing! You should also allow a littleextra time in case some items take longer than planned.



6. If you find that your weekly schedule is too full for you to handle, you mayneed to make some changes in your workload, classload, or in other areasof your life. You may want to talk to your advisor, manager or supervisor atwork, or someone in your college’s academic counseling center for helpwith such decisions.

Note: In this chapter, we begin a feature called Study Skills Reminder. Thepurpose of this feature is to remind you of some of the information given inthis section and to further expand on some topics in this section.


SYMBOLS AND SETS OF NUMBERS SECTION 1.2 7

O b j e c t i v e s

1 Use a number line to order numbers.

2 Translate sentences into mathematical statements.

3 Identify natural numbers, whole numbers, integers, rational numbers, irrational num-bers, and real numbers.

4 Find the absolute value of a real number.

1 We begin with a review of the set of natural numbers and the set of whole num-bers and how we use symbols to compare these numbers. A set is a collection of ob-jects, each of which is called a member or element of the set. A pair of brace symbols

encloses the list of elements and is translated as “the set of” or “the set containing.”5 6

Natural NumbersThe set of natural numbers is 51, 2, 3, 4, 5, 6, Á 6.

1.2 S Y M B O L S A N D S E T S O F N U M B E R S

The three dots (an ellipsis) at the end of the list of elements of a set means that the listcontinues in the same manner indefinitely.

These numbers can be pictured on a number line. We will use number lines oftento help us visualize distance and relationships between numbers. Visualizing mathe-matical concepts is an important skill and tool, and later we will develop and exploreother visualizing tools.

To draw a number line, first draw a line. Choose a point on the line and label it 0.To the right of 0, label any other point 1. Being careful to use the same distance as from0 to 1, mark off equally spaced distances. Label these points 2, 3, 4, 5, and so on. Sincethe whole numbers continue indefinitely, it is not possible to show every whole numberon this number line. The arrow at the right end of the line indicates that the patterncontinues indefinitely.

Picturing whole numbers on a number line helps us to see the order of the num-bers. Symbols can be used to describe concisely in writing the order that we see.

These symbols may be used to form a mathematical statement.The statement might betrue or it might be false. The two statements below are both true.

If two numbers are not equal, then one number is larger than the other. Themeans “is greater than.” The “is less than.” For example,symbol 6 meanssymbol 7

2 Z 6 states that “two is not equal to six”

2 = 2 states that “two is equal to two”

The symbol Z means “is not equal to.”

The equal symbol = means “is equal to.”

0 4 52 31

Whole NumbersThe set of whole numbers is 50, 1, 2, 3, 4, Á 6.



On a number line, we see that a number to the right of another number is larger.Similarly, a number to the left of another number is smaller. For example, 3 is to the leftof 5 on a number line, which means that 3 is less than 5, or Similarly, 2 is to theright of 0 on a number line, which means 2 is greater than 0, or Since 0 is to theleft of 2, we can also say that 0 is less than 2, or

The symbols and are called inequality symbols.7Z , 6 ,0 6 2.

2 7 0.3 6 5.

3 6 5 states that “three is less than five”

2 7 0 states that “two is greater than zero”


TEACHING TIPIf students are having trouble withinequality symbols, remind them toread the Helpful Hint. If the sym-bols “point” to the smaller number,the inequality statement will becorrect. For example,

points to smaller numberq

7 7 5

E X A M P L E 1

H e l p f u l H i n tNotice that has exactly the same meaning as Switching the orderof the numbers and reversing the “direction of the inequality symbol” does notchange the meaning of the statement.

Also notice that, when the statement is true, the inequality arrow points to thesmaller number.

5 7 3 has the same meaning as 3 6 5.

0 6 2.2 7 0

Insert the space between each pair of numbers to make each statementtrue.

a. 2 3 b. 7 4 c. 72 27

6 , 7 , or = in

a. since 2 is to the left of 3 on the number line.

b. since 7 is to the right of 4 on the number line.

c. since 72 is to the right of 27 on the number line.

Two other symbols are used to compare numbers. The symbol means “is lessthan or equal to.” The symbol means “is greater than or equal to.” For example,

states that “seven is less than or equal to ten”

This statement is true since is true. If either or is true, thenis true.

states that “three is greater than or equal to three”

This statement is true since is true. If either or is true, then istrue.

The statement is false since neither nor is true. The sym-bols and are also called inequality symbols.Ú…

6 = 106 7 106 Ú 10

3 Ú 33 = 33 7 33 = 3

3 Ú 3

7 … 107 = 107 6 107 6 10

7 … 10

Ú…

72 7 27

7 7 4

2 6 3

0 4 52 31

3 5

Solut ion

CLASSROOM EXAMPLEInsert each pair ofnumbers.a. 9 20 b. 100 99answers: a. b. 76

6 , 7 , or = between

0 4 52 31

2 0 or 0 2


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Translate each sentence into a mathematical statement.

a. Nine is less than or equal to eleven.b. Eight is greater than one.c. Three is not equal to four.


Tell whether each statement is true or false.

a. b. c. d. 23 Ú 023 … 08 … 88 Ú 8Solut ion a. True, since is true. b. True, since is true.

c. False, since neither nor is true. d. True, since is true.

2 Now, let’s use the symbols discussed above to translate sentences into mathe-matical statements.

23 7 023 = 023 6 0

8 = 88 = 8

Solut ion

CLASSROOM EXAMPLETell whether each statement is true orfalse.a. b.c. d.answers:a. false b. true c. true d. true

0 … 521 Ú 21

21 … 218 6 6

E X A M P L E 2

E X A M P L E 3

CLASSROOM EXAMPLETranslate each sentence into a mathe-matical statement.a. Fourteen is greater than or equal to

two.b. Nine is not equal to ten.answers: a. b. 9 Z 1014 Ú 2

a. b.

c.

3 Whole numbers are not sufficient to describe many situations in the real world.For example, quantities smaller than zero must sometimes be represented, such as tem-peratures less than 0 degrees.

We can picture numbers less than zero on a number line as follows:

4Z3

fouris not

equal tothree

17811…9

oneis greater

thaneighteleven

is less thanor equal to

nine

0 44 55 22 33 11

Numbers less than 0 are to the left of 0 and are labeled and so on.A sign, such as the one in tells us that the number is to the left of 0 on a numberline. In words, is read “negative one.” A sign or no sign tells us that a number liesto the right of 0 on the number line. For example, 3 and both mean positive three.

The numbers we have pictured are called the set of integers. Integers to the left of0 are called negative integers; integers to the right of 0 are called positive integers. Theinteger 0 is neither positive nor negative.

+3+-1

-1,--1, -2, -3,

IntegersThe set of integers is 5Á , -3, -2, -1, 0, 1, 2, 3, Á 6.

0 44 55 22 33 11

negative integers positive integers


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The integer represents 72 degrees below zero.

A problem with integers in real-life settings arises when quantities are smallerthan some integer but greater than the next smallest integer. On a number line, thesequantities may be visualized by points between integers. Some of these quantities be-tween integers can be represented as a quotient of integers. For example,

The point on a number line halfway between 0 and 1 can be represented by aquotient of integers.

The point on a number line halfway between 0 and can be represented byOther quotients of integers and their graphs are shown.

The set numbers, each of which can be represented as a quotient of integers, iscalled the set of rational numbers. Notice that every integer is also a rational numbersince each integer can be expressed as a quotient of integers. For example, the integer5 is also a rational number since 5 = 5

1 .

- 12 .

-1

12 ,

-72


Use an integer to express the number in the following. “Pole of Inaccessibility, Antarc-tica, is the coldest location in the world, with an average annual temperature of 72 degreesbelow zero.” (Source: The Guinness Book of Records)

Solut ion

The number line also contains points that cannot be expressed as quotients of inte-gers. These numbers are called irrational numbers because they cannot be representedby rational numbers. For example, and are irrational numbers.p22

0 4 522 33 11

h q q # t

Rational NumbersThe set of rational numbers is the set of all numbers that can be expressed asa quotient of integers with denominator not zero.

1 unit

2 units

irrationalnumber

Irrational NumbersThe set of irrational numbers is the set of all numbers that correspond topoints on the number line but that are not rational numbers. That is, an irra-tional number is a number that cannot be expressed as a quotient of integers.

E X A M P L E 4

CLASSROOM EXAMPLEUse an integer to express the number inthe following. “The lowest altitude inNorth America is found in Death Valley,California. Its altitude is 282 feet belowsea level.” (Source: The World Almanac)answer: -282

Notice the ellipses (three dots) to the left and to the right of the list for the integers.This indicates that the positive integers and the negative integers continue indefinitely.

square

Rational numbers and irrational numbers can be written as decimal numbers. The dec-imal equivalent of a rational number will either terminate or repeat in a pattern. For


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example, upon dividing we find that

The decimal representation of an irrational number will neither terminate nor repeat.For example, the decimal representations of irrational numbers and are

(For further review of decimals, see the Appendix.)Combining the natural numbers with the irrational numbers gives the set of real

numbers. One and only one point on a number line corresponds to each real number.

1decimal number does not terminate or repeat in a pattern2 p = 3.141592653 Á1decimal number does not terminate or repeat in a pattern2 22 = 1.414213562 Á

p22

23

= 0.66666Á 1decimal number repeats in a pattern2 34

= 0.75 1decimal number terminates or ends2 and

Real NumbersThe set of real numbers is the set of all numbers each of which corresponds toa point on a number line.

On the following number line, we see that real numbers can be positive, negative,or 0. Numbers to the left of 0 are called negative numbers; numbers to the right of 0 arecalled positive numbers. Positive and negative numbers are also called signed numbers.

2 2 3 4 5045 113

Negative numbersZero

Positive numbers

Several different sets of numbers have been discussed in this section. The follow-ing diagram shows the relationships among these sets of real numbers.

Common Sets of Numbers

Natural Numbers orPositive Integers

Zero

0

Whole Numbers

0, 2, 56, 198

Negative Integers

20, 13, 1

Noninteger RationalNumbers

Real Numbers

18, q, 0, 2, p, 4710

Rational Numbers

35, √, 0, 5,

Irrational Numbers

p, 7

1, 16, 170

Integers

10, 0, 8

2711

3013, Ï,


E X A M P L E 6

E X A M P L E 5


STUDY SKILLS REMINDERDrawing a diagram, like the one shown on the previous

page, for yourself in your study notes will help you understandand remember mathematical relationships.

Given the set list the numbers in this set that belongto the set of:

a. Natural numbers b. Whole numbers c. Integersd. Rational numbers e. Irrational numbers f. Real numbers

E -2, 0, 14 , -1.5, 112, -3, 11, 22F ,

Solut ion

Insert the appropriate space to make each statement true.

a. b. c. -5 -67 142-1 0

6 , 7 , or = in

Solut ion a. since is to the left of 0 on a number line.-1-1 6 0

a. The natural numbers are 11 and 112.

b. The whole numbers are 0, 11, and 112.

c. The integers are 0, 11, and 112.

d. Recall that integers are rational numbers also. The rational numbers are11, and 112.

e. The irrational number is

f. The real numbers are all numbers in the given set.

We can now extend the meaning and use of inequality symbols such as and to apply to all real numbers.7

6

22.

-3, -2, -1.5, 0, 14 ,

-3, -2,

Order Property for Real NumbersGiven any two real numbers a and b, if a is to the left of b on a numberline. Similarly, if a is to the right of b on a number line.a 7 b

a 6 b

a b

a b

CLASSROOM EXAMPLEGiven the set

list the numbers in this set that belong to the set of:a. Natural numbersb. Whole numbersc. Integersd. Rational numberse. Irrational numbersf. Real numbersanswers:a. 6,913 b. 0, 6, 913

c. 0, 6, 913d. 0, 6, 913 e.f. all numbers in the given set

p-100, - 25,

-100,

-100, - 25, 0, p, 6, 913,

2 20 11

1 0

7 35 46

5 6

b. since simplifies to 7.

c. since is to the right of on the number line.-6-5-5 7 -6

1427 = 14

2

b a

a b

CLASSROOM EXAMPLEInsert in the appropriatespace to make each statement true.a.b.

answers: a. b. 77-9 -190 -7

6 , 7 , or =


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E X A M P L E 8


4 A number line not only gives us a picture of the real numbers, it also helps usvisualize the distance between numbers.The distance between a real number a and 0 isgiven a special name called the absolute value of a. “The absolute value of a” is writtenin symbols as ƒ a ƒ .

For example, and since both 3 and are a distance of 3 units from 0on a number line.

-3ƒ -3 ƒ = 3ƒ 3 ƒ = 3

Absolute ValueThe absolute value of a real number a, denoted by is the distance betweena and 0 on a number line.

ƒ a ƒ ,

0 22 33 1

3 units 3 units

1

H e l p f u l H i n tSince is a distance, is always either positive or 0, never negative. That is,for any real number a, ƒ a ƒ Ú 0.

ƒ a ƒƒ a ƒ

E X A M P L E 7Find the absolute value of each number.

a. b. c. d. e. ƒ 5.6 ƒƒ - 12 ƒƒ 0 ƒƒ -5 ƒƒ 4 ƒ

Solut ion a. since 4 is 4 units from 0 on a number line.

b. since is 5 units from 0 on a number line.

c. since 0 is 0 units from 0 on a number line.

d. since is units from 0 on a number line.

e. since 5.6 is 5.6 units from 0 on a number line.ƒ 5.6 ƒ = 5.6

12-

12ƒ -

12 ƒ = 1

2

ƒ 0 ƒ = 0

-5ƒ -5 ƒ = 5

ƒ 4 ƒ = 4

CLASSROOM EXAMPLEFind the absolute value of each number.a. b.c.

answers: a. 7 b. 8 c. 23

ƒ - 23 ƒ

ƒ -8 ƒƒ 7 ƒ

Insert the appropriate space to make each statement true.

a. b. c. d. e. ƒ -7 ƒ ƒ 6 ƒƒ 5 ƒ ƒ 6 ƒƒ -3 ƒ ƒ -2 ƒƒ -5 ƒ 5ƒ 0 ƒ 2

6 , 7 , or = in

Solut ion a. since and

b. since

c. since

d. since

e. since 7 7 6.ƒ -7 ƒ 7 ƒ 6 ƒ5 6 6.ƒ 5 ƒ 6 ƒ 6 ƒ

3 7 2.ƒ -3 ƒ 7 ƒ -2 ƒ5 = 5.ƒ -5 ƒ = 5

0 6 2.ƒ 0 ƒ = 0ƒ 0 ƒ 6 2

CLASSROOM EXAMPLEInsert the appropriatespace to make each statement true.

a. b.c.answers: a. b. c. 76=

ƒ -2.7 ƒ ƒ -2 ƒ-3 ƒ 0 ƒƒ -4 ƒ 4

6 , 7 , or = in


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Suppose you are a quality con-trol engineering technician in afactory that makes machinescrews. You have just helped toinstall programmable machin-ery on the production line thatmeasures the length of each

screw. If a screw’s length is greater than 4.05 centimeters or less thanor equal to 3.98 centimeters, the machinery is programmed to discardthe screw. To check that the machinery works properly, you test sixscrews with known lengths.The results of the test are displayed. Is thenew machinery working properly? Explain.

TEST RESULTS

Test Screw Actual Length Machineof Test Action on

Screw (cm) Test Screw

A 4.03 AcceptB 3.96 RejectC 4.05 AcceptD 4.08 RejectE 3.98 RejectF 4.01 Accept


1.3 F R AC T I O N S


O b j e c t i v e s

1 Write fractions in simplest form.

2 Multiply and divide fractions.

3 Add and subtract fractions.

1 A quotient of two numbers such as is called a fraction. In the fraction the topnumber, 2, is called the numerator and the bottom number, 9, is called the denominator.

A fraction may be used to refer to part of a whole. For example, of the circlebelow is shaded. The denominator 9 tells us how many equal parts the whole circle isdivided into and the numerator 2 tells us how many equal parts are shaded.To simplify fractions, we can factor the numerator and the denominator. In the state-ment 3 and 5 are called factors and 15 is the product. (The raised dot symbolindicates multiplication.)

To factor 15 means to write it as a product. The number 15 can be factored as or as

A fraction is said to be simplified or in lowest terms when the numerator and thedenominator have no factors in common other than 1. For example, the fraction is inlowest terms since 5 and 11 have no common factors other than 1.

To help us simplify fractions, we write the numerator and the denominator as aproduct of prime numbers.

511

1 # 15.3 # 5

3 # 5 = 15q q q

factor factor product

3 # 5 = 15,

29

29 ,2

9

fl of the circleis shaded.

Prime NumberA prime number is a whole number, other than 1, whose only factors are 1and itself. The first few prime numbers are

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, and so on.

A natural number, other than 1, that is not a prime number is called a compositenumber. Every composite number can be written as a product of prime numbers. Wecall this product of prime numbers the prime factorization of the composite number.



E X A M P L E 1

FRACTIONS SECTION 1.3 17

Write each of the following numbers as a product of primes.

a. 40 b. 63

CLASSROOM EXAMPLEWrite 60 as a product of primes.answer: 2 # 2 # 3 # 5

Solut ion a. First, write 40 as the product of any two whole numbers, other than 1.

Next, factor each of these numbers. Continue this process until all of the factors areprime numbers.

All the factors are now prime numbers. Then 40 written as a product of primes is

b. 63 = 9 # 7 øΩ

= 3 # 3 # 7

40 = 2 # 2 # 2 # 5

40 = 4 # 10øΩ øΩ

= 2 # 2 #2 # 5

40 = 4 # 10

TEACHING TIPHelp students understand that itmakes no difference which two fac-tors they start with. The resultingprime factorization is the same. Forexample:

40 = 5 # 8 T TΩ

= 5 # 2 # 4

T T T Ω= 5 # 2 # 2 # 2

To use prime factors to write a fraction in lowest terms, apply the fundamentalprinciple of fractions.

To understand why this is true, we use the fact that since c is not zero, then

We will call this process dividing out the common factor of c.

a # c

b # c=

a

b# cc

=a

b# 1 =

a

b

cc

= 1.

Fundamental Principle of FractionsIf is a fraction and c is a nonzero real number, then

a # c

b # c=

a

b

ab

E X A M P L E 2Write each fraction in lowest terms.

a. b. c.8820

1127

4249

Solut ion

CLASSROOM EXAMPLE

Write in lowest terms.

answer:47

2035

a. Write the numerator and the denominator as products of primes; then apply thefundamental principle to the common factor 7.

b.

There are no common factors other than 1, so is already in lowest terms.

c.8820

=2 # 2 # 2 # 11

2 # 2 # 5=

225

1127

1127

=11

3 # 3 # 3

4249

=2 # 3 # 7

7 # 7=

2 # 37

=67


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E X A M P L E 3

Multiply and Write the product in lowest terms.513

.2

15

Solut ion Multiply numerators.Multiply denominators.

Next, simplify the product by dividing the numerator and the denominator by anycommon factors.

Before dividing fractions, we first define reciprocals.Two fractions are reciprocalsof each other if their product is 1. For example and are reciprocals since Also, the reciprocal of 5 is since

To divide fractions, multiply the first fraction by the reciprocal of the secondfraction.

5 # 15 = 5

1# 1

5 = 1.15

23# 3

2 = 1.32

23

=2

39

=2 # 5

3 # 5 # 13

215

# 513

=2 # 5

15 # 13

Multiplying Fractionsab# cd

= a # cb # d, if b Z 0 and d Z 0

CLASSROOM EXAMPLE

Multiply . Write the product in low-

est terms.answer:

335

27

# 310

Concept Check Answer:answers may vary

E X A M P L E 4

Divide. Write all quotients in lowest terms.

a. b. c.38

,310

710

, 1445

,516

Dividing Fractionsa

b,

c

d=

a

b# d

c, if b Z 0, d Z 0, and c Z 0

CONCEPT CHECKExplain the error in the following steps.

a. b.

2 To multiply two fractions, multiply numerator times numerator to obtain thenumerator of the product; multiply denominator times denominator to obtain the de-nominator of the product.

67

=5 + 15 + 2

=12

1555

=1 55 5

=15


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E X A M P L E 5


Solut ion

CLASSROOM EXAMPLE

Divide . Write the quotient in

lowest terms.

answer:8

27

29

,34

Add or subtract as indicated. Write each result in lowest terms.

a. b. c. d.53

-13

97

-27

310

+210

27

+47

Solut ion

Adding and Subtracting Fractions with the Same Denominator

a

b-

c

b=

a - c

b, if b Z 0

a

b+

c

b=

a + c

b, if b Z 0

CLASSROOM EXAMPLEAdd or subtract as indicated. Then simplify if possible.

a. b.

answers:

a. b. 17

11

1310

-3

10211

+5

11

1216

Whole

!

a.

b.

c.

3 To add or subtract fractions with the same denominator, combine numeratorsand place the sum or difference over the common denominator.

38

,310

=38

# 103

=3 # 2 # 5

2 # 2 # 2 # 3=

54

710

, 14 =710

,141

=7

10# 114

=7 # 1

2 # 5 # 2 # 7=

120

.

45

,516

=45

# 165

=4 # 165 # 5

=6425

a. b.

c. d.

To add or subtract fractions without the same denominator, first write the frac-tions as equivalent fractions with a common denominator. Equivalent fractions arefractions that represent the same quantity. For example, and are equivalent frac-tions since they represent the same portion of a whole, as the diagram shows. Count thelarger squares and the shaded portion is Count the smaller squares and the shadedportion is Thus,

We can write equivalent fractions by multiplying a given fraction by 1, as shown inthe next example. Multiplying a fraction by 1 does not change the value of the fraction.

34 = 12

16 .1216 .

34 .

1216

34

53

-13

=5 - 1

3=

43

97

-27

=9 - 2

7=

77

= 1

310

+210

=3 + 2

10=

510

=5

2 # 5=

12

27

+47

=2 + 4

7=

67

E X A M P L E 6

Write as an equivalent fraction with a denominator of 20.25

Solut ionCLASSROOM EXAMPLE

Write as an equivalent fraction with a

denominator of 42.

answer:1842

37

Since multiply the fraction by Multiplying by does not change the

value of the fraction.Multiply by or 1.4

425

=25

# 44

=2 # 45 # 4

=820

44

= 144

.5 # 4 = 20,


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E X A M P L E 7Add or subtract as indicated. Write each answer in lowest terms.

a. b. c. 3 16

- 1 1112

12

+1722

-211

25

+14

Solut ion a. Fractions must have a common denominator before they can be added or subtracted.Since 20 is the smallest number that both 5 and 4 divide into evenly, 20 is the leastcommon denominator. Write both fractions as equivalent fractions with denomina-tors of 20. Since

then

b. The least common denominator for denominators 2, 22, and 11 is 22. First, writeeach fraction as an equivalent fraction with a denominator of 22. Then add or sub-tract from left to right.

Then

c. To find lets use a vertical format.

2 1412

#

-1 1112

1 312

or 1

14

3 212

#

-1 1112q

Need toborrow

=

=

3 16

-1 1112

=

=

ƒ— 2+1 212 —

T

3 16 - 1

1112 ,

12

+1722

-211

=1122

+1722

-422

=2422

=1211

12

=12

# 1111

=1122

, 1722

=1722

, and 211

=211

# 22

=422

25

+14

=820

+520

=1320

25

# 44

=2 # 45 # 4

=820 and 1

4# 55

=1 # 54 # 5

=520

CLASSROOM EXAMPLEAdd or subtract as indicated.

a. b.

answers: a. b. 11 7

12

1740

18 14 - 6

23

38

+1

20

Suppose you are fishing on a freshwater lake in Canada. You catch a whitefishweighing pounds. According to the International Game Fish Association,the world’s record for largest lake whitefish ever caught is pounds. Did youset a new world’s record? Explain. By how much did you beat or miss the exist-ing world record?

14 38

14 532

TEACHING TIPOnce you have reviewed all fouroperations on fractions separately,ask students how they will performeach operation:

12

# 13

12

,13

12

+13

12

-13


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MENTAL MATHRepresent the shaded part of each geometric figure by a fraction.

1. 2. 3. 4.

For Exercises 5 and 6, fill in the blank.

5. In the fraction 3 is called the and 5 is called the

6. The reciprocal of is___117

.711

denominator.numerator35

,

25

57

14

38

TEACHING TIPA Group Activity for this section is available in the Instructor’s Resource Manual.

1. 2. 3. 4. 5. 6. 7. 8. 9. 3 3 52 2 2 2 23 5 52 2 2 72 2 53 3 32 7 72 2 3 53 11


O b j e c t i v e s

1 Define and use exponents and the order of operations.

2 Evaluate algebraic expressions, given replacement values for variables.

3 Determine whether a number is a solution of a given equation.

4 Translate phrases into expressions and sentences into equations.

1 Frequently in algebra, products occur that contain repeated multiplication ofthe same factor. For example, the volume of a cube whose sides each measure 2 cen-timeters is cubic centimeters. We may use exponential notation to write suchproducts in a more compact form. For example,

2 # 2 # 2 may be written as 23.

12 # 2 # 22


1.4 I N T R O D U C T I O N TO VA R I A B L E E X P R E S S I O N S A N D E Q UAT I O N S




2cm

Volume is (2 . 2 . 2)cubic centimeters.

Using symbols for mathematical operations is a great convenience. However, the moreoperation symbols presented in an expression, the more careful we must be when per-forming the indicated operation. For example, in the expression do we addfirst or multiply first? To eliminate confusion, grouping symbols are used. Examples ofgrouping symbols are parentheses ( ), brackets [ ], braces and the fraction bar. Ifwe wish to be simplified by adding first, we enclose in parentheses.

If we wish to multiply first, may be enclosed in parentheses.

To eliminate confusion when no grouping symbols are present, use the followingagreed upon order of operations.

2 + 13 # 72 = 2 + 21 = 23

3 # 7

12 + 32 # 7 = 5 # 7 = 35

2 + 32 + 3 # 75 6,

2 + 3 # 7,

Order of OperationsSimplify expressions using the order below. If grouping symbols such asparentheses are present, simplify expressions within those first, starting withthe innermost set. If fraction bars are present, simplify the numerator and thedenominator separately.1. Evaluate exponential expressions.2. Perform multiplications or divisions in order from left to right.

3. Perform additions or subtractions in order from left to right.

TEACHING TIPOrder of operations will be usedthroughout this algebra course. Tellstudents to make sure that theymaster this order now.

Evaluate the following:

a. [read as “3 squared” or as “3 to second power”]b. [read as “5 cubed” or as “5 to the third power”]c. [read as “2 to the fourth power”]

d. e. a37b2

71

245332

E X A M P L E 1

Solut ion a. b.

c. d.

e. a37b2

= a37b a3

7b =

949

71 = 724 = 2 # 2 # 2 # 2 = 16

53 = 5 # 5 # 5 = 12532 = 3 # 3 = 9

H e l p f u l H i n tsince indicates repeated multiplication of the same factor.

23 = 2 # 2 # 2 = 8, whereas 2 # 3 = 6.

2323 Z 2 # 3

CLASSROOM EXAMPLEEvaluate.

a. b. c.answers:a. 16 b. 81 c.

425

a 25b2

3442

TEACHING TIPWarn students that a common mis-take when working with exponentsis to multiply base exponent in-stead of using the exponent to tellthem how many times the base is afactor.

The 2 in is called the base; it is the repeated factor.The 3 in is called the exp-onent and is the number of times the base is used as a factor.The expression is calledan exponential expression.

—— exponent

cbase —— 2 is a factor 3 times23 = 2 # 2 # 2 = 8T

232323


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INTRODUCTION TO VARIABLE EXPRESSIONS AND EQUATIONS SECTION 1.4 25

Now simplify There are no grouping symbols and no exponents, so we multi-ply and then add.

Multiply.Add. = 23

2 + 3 # 7 = 2 + 21

2 + 3 # 7.

Simplify each expression.

a. b. c. d. e.32

# 12

-12

3 # 423 # 10 - 7 , 72112 + 32

ƒ -15 ƒ6 , 3 + 52

E X A M P L E 2

Solut ion a. Evaluate first.

Next divide, then add.

Divide.Add. = 27

= 2 + 25

6 , 3 + 52 = 6 , 3 + 25

52

b. First, simplify the numerator and the denominator separately.

Simplify numerator and denominator separately.

Simplify.

c. Multiply and divide from left to right. Then subtract.

Subtract.

d. In this example, only the 4 is squared. The factor of 3 is not part of the base be-cause no grouping symbol includes it as part of the base.

Evaluate the exponential expression.Multiply.

e. The order of operations applies to operations with fractions in exactly the same wayas it applies to operations with whole numbers.

Multiply.

The least common denominator is 4.

Subtract. =14

=34

-24

32

# 12

-12

=34

-12

= 48 3 # 42 = 3 # 16

= 29 3 # 10 - 7 , 7 = 30 - 1

= 2

=3015

2112 + 32

ƒ -15 ƒ=

2115215

CLASSROOM EXAMPLESimplify.

a. b.

c.

answers:

a. 35 b. c. 72415

8[216 + 32 - 9]

95

# 13

-13

3 + 2 # 42

H e l p f u l H i n tBe careful when evaluating an exponential expression. In the exponent 2applies only to the base 4. In we multiply first because of parentheses,so the exponent 2 applies to the product

13 # 422 = 11222 = 1443 # 42 = 3 # 16 = 48

3 # 4.13 # 422, 3 # 42,



CLASSROOM EXAMPLE

Simplify:

answer:133

1 + ƒ 7 - 4 ƒ + 32

8 - 5

Solut ion

STUDY SKILLS REMINDERSMake a practice of neatly writing down enough steps so that youare comfortable with your computations.

E X A M P L E 4Simplify 3[4 + 2110 - 12].

Solut ion Notice that both parentheses and brackets are used as grouping symbols. Start with theinnermost set of grouping symbols.

Simplify the expression in parentheses.

Multiply.

Add.

Multiply. = 66

= 3[22]

= 3[4 + 18]

3[4 + 2110 - 12] = 3[4 + 2192]

CLASSROOM EXAMPLESimplify: answer: 106

2[3 + 5114 - 42]

H e l p f u l H i n tBe sure to follow orderof operations and resistthe temptation to incor-rectly add 4 and 2 first.

Expressions that include many grouping symbols can be confusing. When simpli-fying these expressions, keep in mind that grouping symbols separate the expressioninto distinct parts. Each is then simplified separately.

Simplify 3 + ƒ 4 - 3 ƒ + 22

6 - 3.

E X A M P L E 3

The fraction bar serves as a grouping symbol and separates the numerator and denom-inator. Simplify each separately. Also, the absolute value bars here serve as a groupingsymbol. We begin in the numerator by simplifying within the absolute value bars.

Simplify the expression inside the absolute valuebars.

Find the absolute value and simplify thedenominator.

Evaluate the exponential expression.

Simplify the numerator. =83

=3 + 1 + 4

3

=3 + 1 + 22

3

3 + ƒ 4 - 3 ƒ + 22

6 - 3=

3 + ƒ 1 ƒ + 22

6 - 3


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E X A M P L E 5

Simplify 8 + 2 # 3

22 - 1.

Solut ion

2 In algebra, we use symbols, usually letters such as x, y, or z, to represent un-known numbers. A symbol that is used to represent a number is called a variable. Analgebraic expression is a collection of numbers, variables, operation symbols, andgrouping symbols. For example,

are algebraic expressions. The expression 2x means Also, meansand means If we give a specific value to a variable, we can

evaluate an algebraic expression. To evaluate an algebraic expression means to find itsnumerical value once we know the values of the variables.

Algebraic expressions often occur during problem solving. For example, the ex-pression

gives the distance in feet (neglecting air resistance) that an object will fall in t seconds.(See Exercise 63 in this section.)

16t2

3 # y2.3y25 # 1p2 + 12 51p2 + 122 # x.

2x, -3, 2x + 10, 51p2 + 12, and 3y2 - 6y + 1

5

8 + 2 # 3

22 - 1=

8 + 64 - 1

=143

CLASSROOM EXAMPLE

Simplify .

answer:12

6 , 3 # 2

32 - 1

E X A M P L E 6

Evaluate each expression if and

a. b. c. d. x2 - y2xy

+y

23x

2y2x - y

y = 2.x = 3

CLASSROOM EXAMPLEEvaluate each expression when and

a. b. c.

answers:a. 7 b. c. 153

2

y2 - x2x

y+

5y

2y - x

y = 4.x = 1

Solut ion a. Replace x with 3 and y with 2.

Let and Multiply.Subtract.

b. Let and

c. Replace x with 3 and y with 2. Then simplify.

d. Replace x with 3 and y with 2.

x2 - y2 = 32 - 22 = 9 - 4 = 5

xy

+y

2=

32

+22

=52

y 2.x 33x

2y=

3 # 32 # 2

=94

= 4 = 6 - 2

y 2.x 3 2x - y = 2132 - 2

TEACHING TIPIf your students are having difficul-ty replacing variables with num-bers, try advising them to placeparentheses about the replacementnumbers in order to avoid confu-sion such as and A -3 B2.-32

3 Many times a problem-solving situation is modeled by an equation. Anequation is a mathematical statement that two expressions have equal value.The equalsymbol is used to equate the two expressions. For example,

and are all equations.I = PRT21x - 12

3= 0,

3 + 2 = 5, 7x = 35,“=”

TEACHING TIPRemind students of the differencebetween an equation and an ex-pression throughout their algebracourse.


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CONCEPT CHECKWhich of the following are equations? Which are expressions?

a. b. c. d. 12y = 3x12y + 3x5x - 85x = 8

When an equation contains a variable, deciding which values of the variable makean equation a true statement is called solving an equation for the variable.A solution ofan equation is a value for the variable that makes the equation true. For example, 3 is asolution of the equation because if x is replaced with 3 the statement is true.

Replace with 3.

True.

Similarly, 1 is not a solution of the equation because is not a truestatement.

1 + 4 = 7x + 4 = 7,

7 = 7

x 3 + 4 = 7

T x + 4 = 7

x + 4 = 7,

E X A M P L E 7

Decide whether 2 is a solution of 3x + 10 = 8x.

Solut ion

CLASSROOM EXAMPLEDecide whether 3 is a solution of

answer: yes5x - 10 = x + 2.

Concept Check Answers:equations: a, d; expressions: b, c.

TEACHING TIPThis is certainly an incomplete listof key words and phrases. Also,warn students that if a key wordappears in a sentence, they mustdecide whether it translates to anoperation. To see this, discuss theword “of” in these two examples.The sum of and 5. Find of 5.2

323

Replace x with 2 and see if a true statement results.

Original equation

Replace with 2.

Simplify each side.

True.

Since we arrived at a true statement after replacing x with 2 and simplifying both sidesof the equation, 2 is a solution of the equation.

4 Now that we know how to represent an unknown number by a variable, let’spractice translating phrases into algebraic expressions and sentences into equations.Oftentimes solving problems requires the ability to translate word phrases and sen-tences into symbols. Below is a list of some key words and phrases to help us translate.

16 = 16

6 + 10 16

x 3122 + 10 8122 3x + 10 = 8x

ADDITION SUBTRACTION MULTIPLICATION DIVISION EQUALITY(=)

Sum Difference of Product Quotient EqualsPlus Minus Times Divide GivesAdded to Subtracted from Multiply Into Is/was/

should beMore than Less than Twice Ratio YieldsIncreased by Decreased by Of Divided by Amounts toTotal Less Represents

Is the same as

121 # 21212

H e l p f u l H i n tAn equation contains the equal symbol An algebraic expression does not.“=”.


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H e l p f u l H i n tMake sure you understand the difference when translating phrases containing“decreased by,” “subtracted from,” and “less than.”

Phrase TranslationA number decreased by 10A number subtracted from 10

Notice the order.10 less than a numberA number less 10 x - 10

x - 1010 - xx - 10

Write an algebraic expression that represents each phrase. Let the variable x representthe unknown number.

a. The sum of a number and 3b. The product of 3 and a numberc. Twice a numberd. 10 decreased by a numbere. 5 times a number, increased by 7


TEACHING TIPTell students to be careful whentranslating phrases that include sub-traction or division.With these op-erations, order makes a difference.

E X A M P L E 8

Solut ion a. since “sum” means to add

b. and 3x are both ways to denote the product of 3 and x

c. or 2x

d. because “decreased by” means to subtract

e.

5 times a number()*

5x + 7

10 - x

2 # x

3 # x

x + 3

Now let’s practice translating sentences into equations.

CLASSROOM EXAMPLEWrite an algebraic expression that repre-sents each phrase. Let the variable x rep-resent the unknown number.a. The product of a number and 5b. A number added to 7c. Three times a numberd. A number subtracted from 8e. Twice a number, plus 1answers:a. 5x b. c. 3xd. e. 2x + 18 - x

7 + x

Write each sentence as an equation. Let x represent the unknown number.

a. The quotient of 15 and a number is 4.b. Three subtracted from 12 is a number.c. Four times a number, added to 17, is 21.

a. In words:

T T T

Translate:

b. In words:T T

Translate:

Care must be taken when the operation is subtraction.The expression wouldbe incorrect. Notice that 3 - 12 Z 12 - 3.

3 - 12

x=12 - 3T

a numberisthree subtracted from 12

4=15x

4isthe quotient of 15

and a number

ME X A M P L E 9

CLASSROOM EXAMPLEWrite each sentence as an equation. Letx represent the unknown number.a. The difference of 10 and a number is

18.b. Twice a number decreased by 1 is 99.answers:a. b. 2x - 1 = 9910 - x = 18

Solut ion


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Exponents

To evaluate exponential expressions on a scientific calculator, find the key marked or To evaluate, for example, press the following keys:

or

The display should read or

Order of OperationsSome calculators follow the order of operations, and others do not. To see whetheror not your calculator has the order of operations built in, use your calculator tofind To do this, press the following sequence of keys:

The correct answer is 14 because the order of operations is to multiply beforewe add. If the calculator displays then it has the order of operations builtin.

Even if the order of operations is built in, parentheses must sometimes be in-

serted. For example, to simplify press the keys

The display should read or

Use a calculator to evaluate each expression.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.956 - 45289 - 86

4623 + 12936 - 34

99 + 1401 + 9622241862 - 4552 + 893114 - 72 + 212120 - 5286957454

5/112 - 721 1

5 , 1 1 2 - 7 2 = . r or ENTER

512 - 7

,

14 ,

2 + 3 * 4 = . r or ENTER

2 + 3 # 4.

3 ¿ 5243

243

3 ¿ 5 = . r or ENTER

3 yx 5 = 35, ¿ . yx

Calculator Explorations

c. In words:

Translate: 21=17+4x

TTTTT

21is17added tofour timesa number

TEACHING TIPThroughout this course, studentsneed to know the difference be-tween an expression and an equa-tion. Start reminding them nowthat an equation contains an equalsign while an expression does not.



Suppose you are a local area network (LAN) administrator for a small collegeand you are configuring a new LAN for the mathematics department. The de-partment would like a network of 20 computers so that each user can transmitdata over the network at a speed of 0.25 megabits per second. The collectivespeed for a LAN is given by the expression rn, where r is the data transmissionspeed needed by each of the n computers on the LAN.You know that the net-work will drastically lose its efficiency if the collective speed of the networkexceeds 8 megabits per second. Decide whether the LAN requested by themath department will operate efficiently. Explain your reasoning.

MENTAL MATHFill in the blank with add, subtract, multiply, or divide.

1. To simplify the expression first2. To simplify the expression first3. To simplify the expression first

4. To simplify the expression first divide .20 - 4 , 2,

subtract .120 - 42 # 2, add .11 + 32 # 6,

multiply .1 + 3 # 6,



O b j e c t i v e s

1 Add real numbers with the same sign.

2 Add real numbers with unlike signs.

3 Solve problems that involve addition of real numbers.

4 Find the opposite of a number.

1 Real numbers can be added, subtracted, multiplied, divided, and raised to pow-ers, just as whole numbers can. We use a number line to help picture the addition ofreal numbers.

1.5 A D D I N G R E A L N U M B E R S

Add: 3 + 2

Solut ion

E X A M P L E 1

Recall that 3 and 2 are called addends. We start at 0 on a number line, and draw anarrow representing the addend 3. This arrow is three units long and points to the rightsince 3 is positive. From the tip of this arrow, we draw another arrow representing theaddend 2. The number below the tip of this arrow is the sum, 5.

1 0 1 2 3 4 55 4 3 2

Start End3 2

3 2 5

CLASSROOM EXAMPLEAdd using a number line: answer:

4 + 1

0 1 2 43 51

14



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E X A M P L E 2Add: -1 + 1-22

Solut ion Here, and are addends. We start at 0 on a number line, and draw an arrow rep-resenting This arrow is one unit long and points to the left since is negative.From the tip of this arrow, we draw another arrow representing The number belowthe tip of this arrow is the sum, -3.

-2.-1-1.

-2-1

1 0 1 2 3 4 55 4 3 2

StartEnd2 1

1 (2) 3


-3 + 1-42

567 0 1 24 3 2 1

34

Thinking of signed numbers as money earned or lost might help make additionmore meaningful. Earnings can be thought of as positive numbers. If $1 is earned andlater another $3 is earned, the total amount earned is $4. In other words,

On the other hand, losses can be thought of as negative numbers. If $1 is lost andlater another $3 is lost, a total of $4 is lost. In other words,

Using a number line each time we add two numbers can be time consuming. In-stead, we can notice patterns in the previous examples and write rules for addingsigned numbers. When adding two numbers with the same sign, notice that the sign ofthe sum is the same as the sign of the addends.

1-12 + 1-32 = -4.

1 + 3 = 4.

Adding Two Numbers with the Same SignAdd their absolute values. Use their common sign as the sign of the sum.

E X A M P L E 3Add.

a. b. c. -2 + 1-102-1 + 1-202-3 + 1-72Solut ion Notice that each time, we are adding numbers with the same sign.

a. Add their absolute values: ________Use their common sign.

b. Add their absolute values:______ Common sign.

c. Add their absolute values.______ Common sign. q

;-2 + 1-102 = -12

q1 20 21.;-1 + 1-202 = -21

q3 7 10.;-3 + 1-72 = -10

CLASSROOM EXAMPLEAdd: answer: -14

-5 + 1-92

2 Adding numbers whose signs are not the same can also be pictured on a num-ber line.


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ADDING REAL NUMBERS SECTION 1.5 35

Add: -4 + 6

Solut ion

Using temperature as an example, if the thermometer registers 4 degrees below 0 degrees and then rises 6 degrees, the new temperature is 2 degrees above 0 degrees.Thus, it is reasonable that

Once again, we can observe a pattern: when adding two numbers with differentsigns, the sign of the sum is the same as the sign of the addend whose absolute value islarger.

-4 + 6 = 2.

Adding Two Numbers with Different SignsSubtract the smaller absolute value from the larger absolute value. Use thesign of the number whose absolute value is larger as the sign of the sum.

E X A M P L E 4

4

0

2Rises 6Degrees

1 0 1 2 3 45 4 3 2

StartEnd4

6

4 6 2


-5 + 3

56 0 1 24 3 2 1

35

E X A M P L E 5Add.

a. b. c. 0.2 + 1-0.52-2 + 103 + 1-72Solut ion Notice that each time, we are adding numbers with different signs.

a. Subtract their absolute values:c_____The negative number, has the larger absolute value so the sum is

negative.

b. Subtract their absolute values: _____ The positive number, 10, has the larger absolute value so the sum is

positive.

c. Subtract their absolute values:

_______ The negative number, has the larger absolute value so thesum is negative.

0.5, q0.5 0.2 0.3. ;0.2 + 1-0.52 = -0.3

q 10 2 8. ;-2 + 10 = 8

7, 7 3 4. ;3 + 1-72 = -4

CLASSROOM EXAMPLEAdd:

answer: 4

14 + 1-102

E X A M P L E 6Add.

a. b. c.

d. e. f. - 38

+25

11.4 + 1-4.72- 710

+ a - 110b

0.6 + 1-1.12-5 + 35-8 + 1-112CLASSROOM EXAMPLEAdd.

a. b.

c.

answers: a. b. c. 9.2-8- 911

12.8 + 1-3.62-12 + 4-

611

+ a - 311b


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Solut ion a. Same sign. Add absolute values and use the common sign.b. Different signs. Subtract absolute values and use the

sign of the number with the larger absolute value.

c. Different signs.

d. Same sign.

e.

f. - 38

+25

= - 1540

+1640

=140

11.4 + 1-4.72 = 6.7

- 710

+ a - 110b = -

810

= - 45

0.6 + 1-1.12 = -0.5

-5 + 35 = 30-8 + 1-112 = -19

H e l p f u l H i n tDon’t forget that a com-mon denominator isneeded when adding orsubtracting fractions.The common denomina-tor here is 40.

E X A M P L E 7Add.

a. b. [7 + 1-102] + [-2 + ƒ -4 ƒ ]3 + 1-72 + 1-82

Solut ion a. Perform the additions from left to right.

Adding numbers with different signs.

Adding numbers with like signs.

b. Simplify inside brackets first.

Add.

3 Positive and negative numbers are often used in everyday life. Stock market re-turns show gains and losses as positive and negative numbers.Temperatures in cold cli-mates often dip into the negative range, commonly referred to as “below zero”temperatures. Bank statements report deposits and withdrawals as positive and nega-tive numbers.

= -1

= [-3] + [2]

[7 + 1-102] + [-2 + ƒ -4 ƒ ] = [-3] + [-2 + 4]

= -12

3 + 1-72 + 1-82 = -4 + 1-82

CLASSROOM EXAMPLEAdd.

answer: -7 [3 + 1-132] + [-4 + ƒ -7 ƒ ]

H e l p f u l H i n tDon’t forget that brack-ets are grouping sym-bols. We simplify withinthem first.

E X A M P L E 8FINDING THE GAIN OR LOSS OF A STOCK

During a three-day period, a share of Lamplighter’s International stock recorded thefollowing gains and losses:

Monday Tuesday Wednesdaya gain of $2 a loss of $1 a loss of $3

Find the overall gain or loss for the stock for the three days.

Solut ion Gains can be represented by positive numbers. Losses can be represented by negativenumbers. The overall gain or loss is the sum of the gains and losses.

In words:

T T T T T

Translate:

The overall loss is $2.

1-32 = -2+1-12+2

losspluslossplusgain

CLASSROOM EXAMPLEDuring a four-day period, a share ofWalco stock recorded the following gainsand losses:

Tuesday Wednesdaya loss of $2 a loss of $1

Thursday Fridaya gain of $3 a gain of $3

Find the overall gain or loss for the stock for the four days.answer: A gain of $3.


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ADDING REAL NUMBERS SECTION 1.5 37

TEACHING TIPAfter covering adding numberswith the same sign and with differ-ent signs, take a moment to reviewthis with students. Have them workthese problems:

Give students the answers, and askthem if they knew how to apply therules.

-5 + 1-62 7 + 1-42-11 + 5

4 To help us subtract real numbers in the next section, we first review the concept ofopposites. The graphs of 4 and are shown on a number line below.-4

1 0 1 2 3 4 55 4 3 2

4 units4 units

Notice that 4 and lie on opposite sides of 0, and each is 4 units away from 0.This relationship between and is an important one. Such numbers are

known as opposites or additive inverses of each other.+4-4

-4

Opposites or Additive InversesTwo numbers that are the same distance from 0 but lie on opposite sides of 0are called opposites or additive inverses of each other.

Let’s discover another characteristic about opposites. Notice that the sum of anumber and its opposite is 0.

In general, we can write the following:

12

+ a - 12b = 0

-3 + 3 = 0

10 + 1-102 = 0

The sum of a number a and its opposite is 0.

a + 1-a2 = 0

-a

This is why opposites are also called additive inverses. Notice that this also means thatthe opposite of 0 is then 0 since 0 + 0 = 0.

E X A M P L E 9

Find the opposite or additive inverse of each number.

a. 5 b. c. d. -4.512

-6

Solut ion a. The opposite of 5 is Notice that 5 and are on opposite sides of 0 when plot-ted on a number line and are equal distances away.

b. The opposite of is 6.

c. The opposite of is

d. The opposite of is 4.5.-4.5

- 12

.12

-6

-5-5.

CLASSROOM EXAMPLEFind the opposite of each number.

a. b. 1.9 c.answers:a. 35 b. c.

311

-1.9

- 311

-35


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a. b. c.

d. Since then q

- ƒ -6 ƒ = -6.ƒ -6 ƒ = 6,

-1-2x2 = 2x- a - 12b =

12

-1-102 = 10

TEACHING TIPThis would be a good time to re-view the definition of reciprocal.Students often confuse “opposite”with “reciprocal.”

TEACHING TIPHelp students understand that

can be positive or negative.-a

We use the symbol to represent the phrase “the opposite of” or “the additiveinverse of.” In general, if a is a number, we write the opposite or additive inverse of a as

We know that the opposite of is 3. Notice that this translates as

T T T T

This is true in general.

3=1-32-

3is-3the opposite of

-3-a.

“-”

If a is a number, then -1-a2 = a.

E X A M P L E 1 0Simplify each expression.

a. b. c. d. - ƒ -6 ƒ-1-2x2- a - 12b-1-102

Solut ion


a. b.

c.answers: a. 22 b. c. 5x-14

-1-5x2- ƒ -14 ƒ-1-222

MENTAL MATHTell whether the sum is a positive number, a negative number, or 0. Do not actually find the sum.

1. negative 2. positive 3. 0

4. negative 5. negative 6. 0- 23

+23

-3.68 + 0.27-1.26 + 1-8.32-162 + 162-162 + 164-80 + 1-1272

TEACHING TIPA Group Activity for this section is available in the Instructor’sResource Manual.



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1.6 S U B T R AC T I N G R E A L N U M B E R S

O b j e c t i v e s

1 Subtract real numbers.

2 Add and subtract real numbers.

3 Evaluate algebraic expressions using real numbers.

4 Solve problems that involve subtraction of real numbers.

1 Now that addition of signed numbers has been discussed, we can explore sub-traction. We know that Notice that also. This means that

Notice that the difference of 9 and 7 is the same as the sum of 9 and the opposite of 7.In general, we have the following.

9 - 7 = 9 + 1-729 + 1-72 = 2,9 - 7 = 2.



SUBTRACTING REAL NUMBERS SECTION 1.6 41

Subtracting Two Real NumbersIf a and b are real numbers, then a - b = a + 1-b2.

In other words, to find the difference of two numbers, add the first number to the op-posite of the second number.

E X A M P L E 1

Subtract.

a. b. c. d. -1 - 1-723 - 65 - 1-62-13 - 4

Solut ion

CLASSROOM EXAMPLESubtract.a. b.answers: a. b. -10-8

9 - 19-7 - 1

add

a. Add to the opposite of which is

opposite

add

b. Add 5 to the opposite of which is 6.

opposite

c. Add 3 to the opposite of 6, which is

d. -1 - 1-72 = -1 + 172 = 6

= -3

6. 3 - 6 = 3 + 1-62 = 11

qq 6, 5 - 1-62 = 5 + 162TT

= -17

qq 4.4,13 -13 - 4 = -13 + 1-42TT

TEACHING TIPBefore Example 3, you may wantto ask your students to mentally“subtract 8 from 10.” Then askthem to write this subtractionproblem in symbols and notice theorder of the numbers.

TEACHING TIPAfter covering addition and sub-traction of numbers, ask studentsto work the following exercises.

Classify each problem as you re-view the results in class.

5 + 1-12 5 - 1-12

-7 - 3 -8 + 1-42 -7 + 2 -8 - 1-42

H e l p f u l H i n tStudy the patterns indicated.

No Change to addition.change Change to opposite.

7 - 1-12 = 7 + 112 = 8

-3 - 4 = -3 + 1-42 = -7

5 - 11 = 5 + 1-112 = -6T

ƒT∂

E X A M P L E 2Subtract.

a. b. c. - 23

- a - 45b-

310

-510

5.3 - 1-4.62Solut ion a.

b.

c. The common denominator is 15.- 23

- a - 45b = -

23

+ a45b = -

1015

+1215

=2

15

- 310

-510

= - 310

+ a - 510b = -

810

= - 45

5.3 - 1-4.62 = 5.3 + 14.62 = 9.9

CLASSROOM EXAMPLESubtract.

a. b.

answers: a. b. 0.912

-1.8 - 1-2.7238

- a - 18b


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Subtract 8 from -4.

Solut ion Be careful when interpreting this: The order of numbers in subtraction is important. 8is to be subtracted from

2 If an expression contains additions and subtractions, just write the subtractionsas equivalent additions. Then simplify from left to right.

-4 - 8 = -4 + 1-82 = -12

-4.CLASSROOM EXAMPLESubtract from 10.answer: 19

-9


a. b. 1.6 - 1-10.32 + 1-5.62-14 - 8 + 10 - 1-62Solut ion a.

b.

When an expression contains parentheses and brackets, remember the order ofoperations. Start with the innermost set of parentheses or brackets and work your wayoutward.

= 6.3

1.6 - 1-10.32 + 1-5.62 = 1.6 + 10.3 + 1-5.62 = -6

-14 - 8 + 10 - 1-62 = -14 + 1-82 + 10 + 6

CLASSROOM EXAMPLESimplify.answer: -5

-6 + 8 - 1-42 - 11


a. b. 23 - ƒ 10 ƒ + [-6 - 1-52]-3 + [1-2 - 52 - 2]


E X A M P L E 3

E X A M P L E 4

E X A M P L E 5

Solut ion a. Start with the innermost sets of parentheses. Rewrite as a sum.

Add:

Write as a sum.

Add.

Add.

b. Start simplifying the expression inside the brackets by writing as a sum.

Add.

Evaluate and

Write as a sum.

Add.

Add.

3 Knowing how to evaluate expressions for given replacement values is helpfulwhen checking solutions of equations and when solving problems whose unknownssatisfy given expressions. The next example illustrates this.

= -3

= -2 + 1-128 10 = 8 + 1-102 + 1-12

10 .23 = 8 - 10 + 1-12 = 23 - ƒ 10 ƒ + [-1]

23 - ƒ 10 ƒ + [-6 - 1-52] = 23 - ƒ 10 ƒ + [-6 + 5]

-6 - 1-52 = -12

= -3 + [-9]

7 2 = -3 + [-7 + 1-22]2 152. = -3 + [1-72 - 2]

-3 + [1-2 - 52 - 2] = -3 + [1-2 + 1-522 - 2]

-2 - 5

CLASSROOM EXAMPLESimplify.answer: -3

52 - 20 + [-11 - 1-32]


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Find the value of each expression when and

a. b. x2 - yx - y

12 + x

y = -5.x = 2

Solut ion a. Replace x with 2 and y with Be sure to put parentheses around to separatesigns. Then simplify the resulting expression.

b. Replace the x with 2 and y with and simplify.

4 One use of positive and negative numbers is in recording altitudes above andbelow sea level, as shown in the next example.

= 9 = 4 + 5 = 4 - 1-52

x2 - y = 22 - 1-52-5

=7

14=

12

=2 + 5

14

x - y

12 + x=

2 - 1-5212 + 2

-5-5.

CLASSROOM EXAMPLEFind the value of each expression when

and

a. b.

answers:

a. b. 513

x2 - yx - y

14 + x

y = -4. x = 1

F I N D I N G T H E D I F F E R E N C E I N E L E V A T I O N SThe lowest point on the surface of the Earth is the Dead Sea, at an elevation of 1349feet below sea level. The highest point is Mt. Everest, at an elevation of 29,035 feet.How much of a variation in elevation is there between these two world extremes?(Source: National Geographic Society)

Solut ion To find the variation in elevation between the two heights, find the difference of thehigh point and the low point.

In words:

T T TTranslate:

Thus, the variation in elevation is 30,384 feet.

= 30,384 feet

= 29,035 + 13491-13492- 29,035

low pointminushigh pointCLASSROOM EXAMPLEAt 6:00 P.M., the temperature at the Win-ter Olympics was 14°; by morning thetemperature dropped to Find theoverall change in temperature.answer: -37°

-23°.

Mt. Everest

Dead Sea

29,035 feet above sea level

(29,035)

1349 feetbelow sea level

(1349)

Sea level (0)

E X A M P L E 6

E X A M P L E 7


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Suppose you are a dental hygienist. As part of a new patient assessment, youmeasure the depth of the gum tissue pocket around the patient’s teeth with adental probe and record the results. If these pockets deepen over time, thiscould indicate a problem with gum health or be an indication of gum disease.Now, a year later, you measure the patient’s gum tissue pocket depth again tocompare to the intial measurement. Based on these findings, would you alertthe dentist to a problem with the health of the patien’s gums? Explain.

Dental Chart

Gum Tissue Pocket Depth (millimeters)

Tooth: 22 23 24 25 26 27Initial 2 3 3 2 4 2Current 2 2 4 5 6 5

A knowledge of geometric concepts is needed by many professionals, such as doctors, car-penters, electronic technicians, gardeners, machinists, and pilots, just to name a few. Withthis in mind, we review the geometric concepts of complementary and supplementary angles.

Complementary and Supplementary Angles

Two angles are complementary if their sum is 90°.

Two angles are supplementary if their sum is 180°.

x y 90

xy

x y 180

x y



Find each unknown complementary or supplementary angle.

a. b.

38x

62 y

Solut ion a. These angles are complementary, so their sum is 90°. This means that x is

b. These angles are supplementary, so their sum is 180°. This means that y is

y = 180° - 62° = 118°

180° - 62°.

x = 90° - 38° = 52°

90° - 38°.

CLASSROOM EXAMPLEFind each unknown complementary orsupplementary angle.

a.

b.

answers: a. 102° b. 9°

E X A M P L E 8



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O b j e c t i v e s

1 Multiply and divide real numbers.

2 Evaluate algebraic expressions using real numbers.

1 In this section, we discover patterns for multiplying and dividing real numbers.To discover sign rules for multiplication, recall that multiplication is repeated addition.Thus means that 2 is an addend 3 times. That is,

which equals 6. Similarly, means is an addend 3 times. That is,

Since then This suggests that the productof a positive number and a negative number is a negative number.

What about the product of two negative numbers? To find out, consider the fol-lowing pattern.

Factor decreases by 1 each time

This pattern continues as

Factor decreases by 1 each time

This suggests that the product of two negative numbers is a positive number.

-3 # -1 = 3-3 # -2 = 6

f Product increases by 3 each time.

T

-3 # 2 = -6-3 # 1 = -3-3 # 0 = 0

s Product increases by 3 each time.

T

3 # 1-22 = -6.1-22 + 1-22 + 1-22 = -6,

1-22 + 1-22 + 1-22 = 3 # 1-22-23 # 1-22

2 + 2 + 2 = 3 # 2

3 # 2

1.7 M U LT I P LY I N G A N D D I V I D I N G R E A L N U M B E R S




Multiply.

a. b. c. 1-521-10221-121-62142Solut ion

Multiplying Real Numbers1. The product of two numbers with the same sign is a positive number.

2. The product of two numbers with different signs is a negative number.

E X A M P L E 1

a. b. c.

We know that every whole number multiplied by zero equals zero. This remainstrue for real numbers.

1-521-102 = 5021-12 = -21-62142 = -24

Zero as a FactorIf b is a real number, then Also, 0 # b = 0.b # 0 = 0.

CLASSROOM EXAMPLEMultiply.a. b.answers: a. 72 b. -8

(-4)122(-8)1-92

E X A M P L E 2

Perform the indicated operations.

a. b. c. d. 1-421-112 - 1521-221-121521-921-221-321-421721021-62Solut ion a. By the order of operations, we multiply from left to right. Notice that, because one

of the factors is 0, the product is 0.

b. Multiply two factors at a time, from left to right.

Multiply

c. Multiply from left to right.

Multiply

d. Follow the rules for order of operation.

Find each product.

Add 44 to the opposite of

Add.= 54

10.= 44 + 10

1-421-112 - 1521-22 = 44 - 1-102

= 45

(1)(5). 1-121521-92 = 1-521-92

= -24

(2)(3). 1-221-321-42 = 1621-42

1721021-62 = 01-62 = 0

H e l p f u l H i n tYou may have noticed from the example that if we multiply:

• an even number of negative numbers, the product is positive.

• an odd number of negative numbers, the product is negative.

Multiplying signed decimals or fractions is carried out exactly the same way asmultiplying by integers.

CLASSROOM EXAMPLEPerform the indicated operations.a.

b.answers: a. 64 b. 43

-31-92 - 41-421-221421-82


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MULTIPLYING AND DIV IDING REAL NUMBERS SECTION 1.7 49

Multiply.

a. b. c. a - 45b1-2022

3# a -

710b1-1.2210.052

Solut ion

Evaluate.

a. b. c. d. -321-322-231-223

E X A M P L E 3

CLASSROOM EXAMPLEMultiply.

a. b.

answers: a. b. 13.8- 524

-61-2.32- 56

# 14

TEACHING TIPSpend a lot of class time reviewingthe difference between and

for example.-52,1-522

E X A M P L E 4

a. The product of two numbers with different signs is negative.

b.

c.

Now that we know how to multiply positive and negative numbers, let’s see howwe find the values of and for example.Although these two expressions looksimilar, the difference between the two is the parentheses. In the parenthesestell us that the base, or repeated factor, is In only 4 is the base. Thus,

The base is

The base is 4. -42 = -14 # 42 = -16

4. 1-422 = 1-421-42 = 16

-42,-4.1-422,-42,1-422

a - 45b1-202 =

4 # 205 # 1

=4 # 4 # 5

5 # 1=

161

or 16

23

# a - 710b = -

2 # 73 # 10

= - 2 # 7

3 # 2 # 5= -

715

= -0.06

1-1.2210.052 = -[11.2210.052]

Solut ion a. The base is

b. The base is 2.

c. The base is

d. The base is 3.-32 = -13 # 32 = -9

3.1-322 = 1-321-32 = 9

-23 = -12 # 2 # 22 = -8

2.1-223 = 1-221-221-22 = -8

CLASSROOM EXAMPLEEvaluate.a. b.c. d.answers:a. b.c. 25 d. -25

-27-27

-521-522-331-323

H e l p f u l H i n tBe careful when identifying the base of an exponential expression.

-32

Base is 3-32 = -13 # 32 = -9

1-322Base is -3

1-322 = 1-321-32 = 9

Just as every difference of two numbers can be written as the sumso too every quotient of two numbers can be written as a product. For ex-

ample, the quotient can be written as Recall that the pair of numbers 3 andhas a special relationship.Their product is 1 and they are called reciprocals or multip-

licative inverses of each other.

13

6 # 13 .6 , 3

a + 1-b2,a - b


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Notice that 0 has no multiplicative inverse since 0 multiplied by any number is never 1but always 0.

E X A M P L E 5Find the reciprocal of each number.

a. 22 b. c. d. - 913

-103

16

Solut ion a. The reciprocal of 22 is since

b. The reciprocal of is since

c. The reciprocal of is

d. The reciprocal of is

We may now write a quotient as an equivalent product.

- 139 .-

913

- 110 .-10

316

# 163 = 1.16

33

16

22 # 122 = 1.1

22

Quotient of Two Real NumbersIf a and b are real numbers and b is not 0, then

a , b =a

b= a # 1

b

CLASSROOM EXAMPLEFind the reciprocal.

a. b.

answers:

a. b. - 15

157

-57

15

Reciprocals or Multiplicative InversesTwo numbers whose product is 1 are called reciprocals or multiplicative in-verses of each other.

In other words, the quotient of two real numbers is the product of the first number andthe multiplicative inverse or reciprocal of the second number.

Use the definition of the quotient of two numbers to divide.

a. b. c.20-4

-14-2

-18 , 3

Solut ion a. b.

c.

Since the quotient can be written as the product it follows that signpatterns for dividing two real numbers are the same as sign patterns for multiplyingtwo real numbers.

a # 1b ,a , b

20-4

= 20 # - 14

= -5

-14-2

= -14 # -12

= 7-18 , 3 = -18 # 13

= -6

CLASSROOM EXAMPLEDivide.a. b.answers:a. b. 8-3

-80-10

-12 , 4

Multiplying and Dividing Real Numbers1. The product or quotient of two numbers with the same sign is a positive

number.

2. The product or quotient of two numbers with different signs is a negativenumber.

E X A M P L E 6


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Divide.

a. b. c. d. - 32

, 923

, a - 54b-36

3-24-4

Solut ion

CLASSROOM EXAMPLEDivide.

a. b.

answers:a. b.

107

-25

- 27

, a - 15b50

-2

E X A M P L E 7

a. b. c.

d.

The definition of the quotient of two real numbers does not allow for division by0 because 0 does not have a multiplicative inverse.There is no number we can multiply0 by to get 1. How then do we interpret We say that division by 0 is not allowed ornot defined and that does not represent a real number.The denominator of a fractioncan never be 0.

Can the numerator of a fraction be 0? Can we divide 0 by a number? Yes. Forexample,

In general, the quotient of 0 and any nonzero number is 0.

03

= 0 # 13

= 0

30

30 ?

- 32

, 9 = - 32

# 19

= - 3 # 12 # 9

= - 3 # 1

2 # 3 # 3= -

16

23

, a - 54b =

23

# a - 45b = -

815

-363

= -12-24-4

= 6

Zero as a Divisor or Dividend

1. The quotient of any nonzero real number and 0 is undefined. In symbols, if

is undefined.

2. The quotient of 0 and any real number except 0 is 0. In symbols, if

a Z 0, 0a

= 0.

a Z 0, a

0


a. b. c.01-82

20

-310

E X A M P L E 8

Solut ion a. is undefined b. c.01-82

2=

02

= 00

-3= 0

10

CLASSROOM EXAMPLEPerform the indicated operations.

a. b.

answers: a. undefined b. 0

01-22-3

50

Notice that and This means that

In words, a single negative sign in a fraction can be written in the denominator, inthe numerator, or in front of the fraction without changing the value of the fraction.Thus,

1-7

=-17

= - 17

12-2

= - 122

=-12

2

-122

= -6.12-2

= -6, - 122

= -6,

TEACHING TIPMake sure that students under-stand the difference between and 0

- 4.

- 40


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In general, if a and b are real numbers, b Z 0, a

-b=

-a

b= -

a

b.

Examples combining basic arithmetic operations along with the principles oforder of operations help us to review these concepts.

E X A M P L E 9


a. b.21-322 - 20

-5 + 4

1-1221-32 + 3

-7 - 1-22

TEACHING TIPRemind students that

39-5

=-39

5= -

395

.

CLASSROOM EXAMPLE

Simplify.

answer: -4

51-223 + 52

-4 + 1

Solut ion a. First, simplify the numerator and denominator separately, then divide.

=39-5

or - 395

1-1221-32 + 3

-7 - 1-22 =36 + 3-7 + 2

If and evaluate each expression.

a. b. c.3x

2yx4 - y25x - y

y = -4,x = -2

CLASSROOM EXAMPLEIf and evaluate

answer: 26

x2 - y

2.

y = -3,x = -7

Solut ion a. Replace x with and y with and simplify.

b. Replace x with and y with

Substitute the given values for the variables.

Evaluate exponential expressions.

Subtract.

c. Replace x with and y with and simplify.

3x

2y=

31-2221-42 =

-6-8

=34

-4-2

= 0

= 16 - 1162 x4 - y2 = 1-224 - 1-422

-4.-2

5x - y = 51-22 - 1-42 = -10 - 1-42 = -10 + 4 = -6

-4-2

b. Simplify the numerator and denominator separately, then divide.

2 Using what we have learned about multiplying and dividing real numbers, wecontinue to practice evaluating algebraic expressions.

21-322 - 20

-5 + 4=

2 # 9 - 20-5 + 4

=18 - 20-5 + 4

=-2-1

= 2

E X A M P L E 1 0


Entering Negative Numbers on a Scientific Calculator

To enter a negative number on a scientific calculator, find a key marked (On

some calculators, this key is marked for “change sign.”) To enter for ex-

ample, press the keys The display will read (continued) -8 . 8 + /- .-8, CHS

+ /- .


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Entering Negative Numbers on a Graphing Calculator

To enter a negative number on a graphing calculator, find a key marked

Do not confuse this key with the key which is used for subtraction. To enter

for example, press the keys The display will read

Operations with Real Numbers

To evaluate on a calculator, press the keys

or

The display will read or .

Use a calculator to simplify each expression.

1. 2.

3. 4.

5. 6.

7. 8.

9. (Be careful.) 10. (Be careful.)-12521-1252258 - 625995 - 4550

-444 - 444.8-181 - 324

-5012942175 - 265

451322 - 812182134 + 25168 - 912-591-82 + 1726-38126 - 272

-217 - 92 - 20-16

-16

1-2 2 1 7 - 9 2 - 2 0 ENTER . 2 + /- * 1 7 - 9 2 - 2 0 = ,

-217 - 92 - 20

-8 . 1-2 8 .-8,

- , 1-2 .

MENTAL MATHAnswer the following with positive or negative.

1. The product of two negative numbers is a number.2. The quotient of two negative numbers is a number.3. The quotient of a positive number and a negative number is a number.4. The product of a positive number and a negative number is a number.5. The reciprocal of a positive number is a number.6. The opposite of a positive number is a number. negative

positive

negative

negative

positive

positive



O b j e c t i v e s

1 Use the commutative and associative properties.

2 Use the distributive property.

3 Use the identity and inverse properties.

1 In this section we give names to properties of real numbers with which weare already familiar. Throughout this section, the variables a, b, and c represent realnumbers.

We know that order does not matter when adding numbers. For example, weknow that is the same as This property is given a special name—thecommutative property of addition. We also know that order does not matter whenmultiplying numbers. For example, we know that This propertymeans that multiplication is commutative also and is called the commutative propertyof multiplication.

-5162 = 61-52.5 + 7.7 + 5

These properties state that the order in which any two real numbers are added ormultiplied does not change their sum or product. For example, if we let and

then the commutative properties guarantee that

3 + 5 = 5 + 3 and 3 # 5 = 5 # 3

b = 5,a = 3

Commutative PropertiesAddition:

Multiplication: a # b = b # a

a + b = b + a

H e l p f u l H i n tIs subtraction also commutative? Try an example. Is No! Theleft side of this statement equals 1; the right side equals There is no com-mutative property of subtraction. Similarly, there is no commutative propertyfor division. For example, does not equal 2 , 10.10 , 2

-1.3 - 2 = 2 - 3?

STUDY SKILLS REMINDERYou may want to start a list in your notes describing the situations inwhich you should take extra caution, such as

1. cannot divide by 0

2. subtraction is not commutative—order matters

1.8 P R O P E R T I E S O F R E A L N U M B E R S



PROPERTIES OF REAL NUMBERS SECTION 1.8 57

Use a commutative property to complete each statement.

a. b. 3 # x = x + 5 = Solut ion

CLASSROOM EXAMPLEUse a commutative property to completeeach statement.a. b.answers: a. b. x + 4y # 7

4 + x = 7 # y =

These properties state that the way in which three numbers are grouped does notchange their sum or their product.

a. By the commutative property of addition

b. By the commutative property of multiplication

CONCEPT CHECKWhich of the following pairs of actions are commutative?

a. “raking the leaves” and “bagging the leaves”b. “putting on your left glove” and “putting on your right glove”c. “putting on your coat” and “putting on your shirt”d. “reading a novel” and “reading a newspaper”

Let’s now discuss grouping numbers. We know that when we add three numbers,the way in which they are grouped or associated does not change their sum. For exam-ple, we know that This result is the same if we group thenumbers differently. In other words, also.Thus,

This property is called the associative property of addition.We also know that changing the grouping of numbers when multiplying does not

change their product. For example, (check it). This is theassociative property of multiplication.

2 # 13 # 42 = 12 # 32 # 4

= 12 + 32 + 4.2 + 13 + 4212 + 32 + 4 = 5 + 4 = 9,

2 + 13 + 42 = 2 + 7 = 9.

3 # x = x # 3

x + 5 = 5 + x

Concept Check Answer:b, d

Use an associative property to complete each statement.

a. b. 1-1 # 22 # 5 = 5 + 14 + 62 = Solut ion

Associative PropertiesAddition:

Multiplication: 1a # b2 # c = a # 1b # c2 1a + b2 + c = a + 1b + c2

a. By the associative property of addition

b. By the associative property of multiplication1-1 # 22 # 5 = -1 # 12 # 525 + 14 + 62 = 15 + 42 + 6

E X A M P L E 1

E X A M P L E 2

H e l p f u l H i n tRemember the difference between the commutative properties and the associa-tive properties. The commutative properties have to do with the order of num-bers, and the associative properties have to do with the grouping of numbers.

Let’s now illustrate how these properties can help us simplify expressions.

CLASSROOM EXAMPLEUse an associative property to completeeach statement.

a.b.answers:

a. b. -2 + 1-7 + 3215 # -32 # 6

1-2 - 72 + 3 = 5 # 1-3 # 62 =


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a. b. -317x210 + 1x + 122Solut ion

Since multiplication is commutative, this property can also be written as

The distributive property can also be extended to more than two numbers insidethe parentheses. For example,

Since we define subtraction in terms of addition, the distributive property is also truefor subtraction. For example

= 2x - 2y

21x - y2 = 21x2 - 21y2

= 3x + 3y + 3z

31x + y + z2 = 31x2 + 31y2 + 31z2

1b + c2a = ba + ca

a. By the commutative property of addition

By the associative property of addition

Add.

b. By the associative property of multiplication

Multiply.

2 The distributive property of multiplication over addition is used repeatedlythroughout algebra. It is useful because it allows us to write a product as a sum or asum as a product.

We know that Compare that with Since both original expressions equal 42, they must equal each other, or

This is an example of the distributive property.The product on the left side of the equalsign is equal to the sum on the right side. We can think of the 7 as being distributed toeach number inside the parentheses.

712 + 42 = 7122 + 7142= 42.

7122 + 7142 = 14 + 28712 + 42 = 7162 = 42.

= -21x

-317x2 = 1-3 # 72x = 22 + x

= 110 + 122 + x

10 + 1x + 122 = 10 + 112 + x2CLASSROOM EXAMPLESimplify each expression.a. b. 4(5x)answers: a. b. 20x14 + x1-3 + x2 + 17

Distributive Property of Multiplication Over Addition

a1b + c2 = ab + ac

Use the distributive property to write each expression without parentheses. Then sim-plify if possible.

a. b. c.d. e. f. 413x + 72 + 10-13 + x - w2-112 - y2

51x + 3y - z2-51-3 + 2z221x + y2

E X A M P L E 3

E X A M P L E 4


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PROPERTIES OF REAL NUMBERS SECTION 1.8 59

Solut ion

f. Apply the distributive property.

Multiply.

Add.

We can use the distributive property in reverse to write a sum as a product.

= 12x + 38

= 12x + 28 + 10

413x + 72 + 10 = 413x2 + 4172 + 10

a.

b.

c.

d.

e.

= -3 - x + w

= 1-12132 + 1-121x2 - 1-121w2 -13 + x - w2 = -113 + x - w2

= -2 + y

-112 - y2 = 1-12122 - 1-121y2 = 5x + 15y - 5z

51x + 3y - z2 = 51x2 + 513y2 - 51z2 = 15 - 10z

-51-3 + 2z2 = -51-32 + 1-5212z2 = 2x + 2y

21x + y2 = 2 # x + 2 # y

CLASSROOM EXAMPLEUse the distributive property to write eachexpression without parentheses. Thensimplify if possible.a. b.c.

answers:a. b.c. 18x + 45

-8 - a + b-6 - 21x

912x + 42 + 9

-18 + a - b2-312 + 7x2

Use the distributive property to write each sum as a product.

a. b. 7s + 7t8 # 2 + 8 # x

H e l p f u l H i n tNotice in part e that is first rewritten as -113 + x - w2.-13 + x - w2

TEACHING TIPWhen using the distributive prop-erty, a common mistake among students is to not distribute cor-rectly. Ask students to simplify thefollowing:

Review the difference betweenthese two expressions.

31x + y2 + 731x + y + 72

E X A M P L E 5

Solut ion a. b.

3 Next, we look at the identity properties.The number 0 is called the identity for addition because when 0 is added to any

real number, the result is the same real number. In other words, the identity of the realnumber is not changed.

The number 1 is called the identity for multiplication because when a real num-ber is multiplied by 1, the result is the same real number. In other words, the identity ofthe real number is not changed.

7s + 7t = 71s + t28 # 2 + 8 # x = 812 + x2CLASSROOM EXAMPLEUse the distributive property to write eachsum as a product.a. b.answers:a. b. 41x + y2913 + y2

4x + 4y9 # 3 + 9 # y

Identities for Addition and Multiplication0 is the identity element for addition.

1 is the identity element for multiplication.

a # 1 = a and 1 # a = a

a + 0 = a and 0 + a = a


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Notice that 0 is the only number that can be added to any real number with theresult that the sum is the same real number.Also, 1 is the only number that can be mul-tiplied by any real number with the result that the product is the same real number.

Additive inverses or opposites were introduced in Section 1.5. Two numbers arecalled additive inverses or opposites if their sum is 0. The additive inverse or oppositeof 6 is because The additive inverse or opposite of is 5 because

Reciprocals or multiplicative inverses were introduced in Section 1.3. Two nonzeronumbers are called reciprocals or multiplicative inverses if their product is 1.The reciprocal

or multiplicative inverse of is because Likewise, the reciprocal of is

because

CONCEPT CHECKWhich of the following is the

a. opposite of and which is the b. reciprocal of

1, - 103

, 3

10, 0,

103

, - 310

- 310

?- 310

,

-5a - 15b = 1.-

15

-523

# 32

= 1.32

23

-5 + 5 = 0.-56 + 1-62 = 0.-6

Additive or Multiplicative InversesThe numbers a and are additive inverses or opposites of each other be-cause their sum is 0; that is,

The numbers b and (for ) are reciprocals or multiplicative inverses of

each other because their product is 1; that is,

b # 1b

= 1

b Z 01b

a + 1-a2 = 0

-a

Name the property or properties illustrated by each true statement.

Solut ion a. Commutative property of multiplication (order changed)

b. Associative property of addition (grouping changed)

c. Identity element for addition

d. Commutative property of multiplication (order changed)

e. Multiplicative inverse property

f. Additive inverse property

g. Commutative and associative properties of multiplication (order and grouping changed)

-6 # 1y # 22 = 1-6 # 22 # y

-2 + 2 = 0

-2 # a - 12b = 1

0.2 # 1z # 52 = 0.2 # 15 # z21b + 02 + 3 = b + 3

1x + 72 + 9 = x + 17 + 923 # y = y # 3

Concept Check Answers:

a. b. - 103

310

E X A M P L E 6

CLASSROOM EXAMPLEName the property illustrated by eachtrue statement.a.b.c.

d.

answers:

a. additive inverse propertyb. associative property of multiplicationc. commutative property of additiond. multiplicative inverse property

3a13b = 1

6 + 1z + 22 = 6 + 12 + z2-4 # 16 # x2 = 1-4 # 62 # x

5 + 1-52 = 0


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O b j e c t i v e s

1 Read bar graphs.

2 Read line graphs.

In today’s world, where the exchange of information must be fast and entertaining,graphs are becoming increasingly popular. They provide a quick way of making com-parisons, drawing conclusions, and approximating quantities. Bar and line graphs arecommon in previous sections, but in this section, we do not label the heights of the barsor points.

1 A bar graph consists of a series of bars arranged vertically or horizontally. Thebar graph in Example 1 shows a comparison of the rates charged by selected electrici-ty companies. The names of the companies are listed horizontally and a bar is shownfor each company. Corresponding to the height of the bar for each company is a num-ber along a vertical axis. These vertical numbers are cents charged for each kilowatt-hour of electricity used.

The following bar graph shows the cents charged per kilowatt-hour for selected elec-tricity companies.

a. Which company charges the highest rate?b. Which company charges the lowest rate?c. Approximate the electricity rate charged by the first four companies listed.d. Approximate the difference in the rates charged by the companies in parts (a) and (b).

50

40

30

20

10

0AmericanElectricPower

(Kentucky)

East Miss.Electric

MontanaPower Co.

Cen

ts p

er K

ilow

att-

hour

Alaska VillageElectric Coop.

GreenMountain

Power(Vermont)

SouthernCal. Edison

Source: Electric Company Listed

1.9 R E A D I N G G R A P H S

E X A M P L E 1



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READING GRAPHS SECTION 1.9 63

Solut ion

The height of the bar is approximately halfway between the 0 and 10 marks. Wetherefore conclude that

American Electric Power charges approximately 5¢ per kilowatt-hour.Green Mountain Power charges approximately 11¢ per kilowatt-hour.Montana Power Co. charges approximately 6¢ per kilowatt-hour.Alaska Village Electric charges approximately 42¢ per kilowatt-hour.

d. The difference in rates for Alaska Village Electric Cooperative and American Elec-tric Power is approximately or 37¢.42¢ - 5¢

a. The tallest bar corresponds to the company that charges the highest rate. AlaskaVillage Electric Cooperative charges the highest rate.

b. The shortest bar corresponds to the company that charges the lowest rate.AmericanElectric Power in Kentucky charges the lowest rate.

c. To approximate the rate charged by American Electric Power, we go to the top ofthe bar that corresponds to this company. From the top of the bar, we move hori-zontally to the left until the vertical axis is reached.

50

40

30

20

10

0AmericanElectricPower

(Kentucky)

East Miss.Electric

MontanaPower Co.

Cen

ts p

er K

ilow

att-

hour

Alaska VillageElectric Coop.

GreenMountain

Power(Vermont)

SouthernCal. Edison

Source: Electric Company Listed

42

11

56

CLASSROOM EXAMPLEUse the bar graph from Example 1 toanswer.a. Approximate the rate charged by East

Mississippi Electric.b. Approximate the rate charged by

Southern California Edison.c. Find the difference in rates charged

by Southern California Edison andEast Mississippi Electric.

answers: a. 8¢ per kilowatt-hourb. 14¢ per kilowatt-hourc. 6¢

E X A M P L E 2

The following bar graph shows the estimated worldwide number of Internet users byregion as of 2002.

Source: Nua Internet Surveys

Internet Users (in millions)

Reg

ion

Africa/Middle East

Asia/Pacific

Europe

U.S./Canada

LatinAmerica

200 40 60 80 100 120 140 160 180 200

Worldwide Internet Users


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a. Find the region that has the most Internet users and approximate the number ofusers.

b. How many more users are in Europe than Latin America?

Solut ion

TEACHING TIPSome students may have troublefollowing the height of a bar to anaxis. If so, show students how touse a straightedge to help them.

a. Since these bars are arranged horizontally, we look for the longest bar, which is thebar representing Europe. To approximate the number associated with this region,we move from the right edge of this bar vertically downward to the Internet useraxis. This region has approximately 190 million Internet users.

b. Europe has approximately 190 million Internet users. Latin America has approxi-mately 33 million Internet users. To find how many more users are in Europe, wesubtract million more Internet users.

2 A line graph consists of a series of points connected by a line. The graph in Ex-ample 3 is a line graph.

190 - 33 = 157 or 157

Source: Nua Internet Surveys

Internet Users (in millions)

Reg

ion

Africa/Middle East

Asia/Pacific

Europe

U.S./Canada

LatinAmerica

200 40 60 80 100 120 140 160 180 200

Worldwide Internet Users

190

E X A M P L E 3

The line graph below shows the relationship between the distance driven in a 14-footU-Haul truck in one day and the total cost of renting this truck for that day. Notice thatthe horizontal axis is labeled Distance and the vertical axis is labeled Total Cost.

180

160

140

120

100

80

60

40

20

0

Tot

al C

ost (

dolla

rs)

Distance (miles)300250200150100500

One Day 14-foot Truck Rental

CLASSROOM EXAMPLEUse the bar graph from Example 2 to answer.Find the region that has the second-mostInternet users and approximate the number of users.answer: Asia/Pacific; 187 million


READING GRAPHS SECTION 1.9 65

a. Find the total cost of renting the truck if 100 miles are driven.b. Find the number of miles driven if the total cost of renting is $140.

Solut ion a. Find the number 100 on the horizontal scale and move vertically upward until theline is reached. From this point on the line, we move horizontally to the left until thevertical scale is reached. We find that the total cost of renting the truck if 100 milesare driven is approximately $80.

180

160

140

120

100

80

60

40

20

0

Tot

al C

ost (

dolla

rs)

Distance (miles)300250200150100500

One Day 14-foot Truck Rental

b. We find the number 140 on the vertical scale and move horizontally to the rightuntil the line is reached. From this point on the line, we move vertically downwarduntil the horizontal scale is reached. We find that the truck is driven approximately225 miles.

From the previous example, we can see that graphing provides a quick way to ap-proximate quantities. In Chapter 6 we show how we can use equations to find exact an-swers to the questions posed in Example 3.The next graph is another example of a linegraph. It is also sometimes called a broken line graph.

The line graph shows the relationship between time spent smoking a cigarette andpulse rate. Time is recorded along the horizontal axis in minutes, with 0 minutes beingthe moment a smoker lights a cigarette. Pulse is recorded along the vertical axis inheartbeats per minute.

E X A M P L E 4

100

90

80

70

60

50

40

30

20

10

0

Pul

se R

ate

(hea

rtbe

ats

per

min

ute)

10 20 30 35 40255 0 5 15Time (minutes)

a. What is the pulse rate 15 minutes after lighting a cigarette?b. When is the pulse rate the lowest?c. When does the pulse rate show the greatest change?

CLASSROOM EXAMPLEUse the graph from Example 3 to answerthe following.a. Find the total cost of renting the truck

if 50 miles are driven.b. Find the total number of miles driven

if the total cost of renting is $100.answers: a. $55 b. 135 miles


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Solut ion a. We locate the number 15 along the time axis and move vertically upward until theline is reached. From this point on the line, we move horizontally to the left until thepulse rate axis is reached. Reading the number of beats per minute, we find that thepulse rate is 80 beats per minute 15 minutes after lighting a cigarette.

100

90

80

70

60

50

40

30

20

10

0P

ulse

Rat

e(h

eart

beat

s pe

r m

inut

e)10 20 30 35 40255 0 5 15

Time (minutes)

b. We find the lowest point of the line graph, which represents the lowest pulse rate.From this point, we move vertically downward to the time axis.We find that the pulserate is the lowest at minutes, which means 5 minutes before lighting a cigarette.

c. The pulse rate shows the greatest change during the 5 minutes between 0 and 5. No-tice that the line graph is steepest between 0 and 5 minutes.

-5TEACHING TIPA Group Activity for this section isavailable in the Instructor’s Re-source Manual.

CLASSROOM EXAMPLEUse the graph from Example 4 to answerthe following.a. What is the pulse rate 40 minutes

after lighting a cigarette?b. What is the pulse rate when the ciga-

rette is being lit?c. When is the pulse rate the highest?answers: a. 70 b. 60

c. 5 minutes after lighting


SIMPLIFYING ALGEBRAIC EXPRESSIONS SECTION 2.1 81

Term Numerical Coefficient

3x 3

since means

0.7z 1

-5-5-1-y

0.7ab3c5

1

5# y

3y

3

5

15

y3

5

E X A M P L E 1Identify the numerical coefficient in each term.

a. b. c. y d. e.x

7-x22z4-3y

Solut ion a. The numerical coefficient of is b. The numerical coefficient of is 22.c. The numerical coefficient of y is 1, since y is 1y.d. The numerical coefficient of is since is

e. The numerical coefficient of is since means

Terms with the same variables raised to exactly the same powers are called liketerms. Terms that aren’t like terms are called unlike terms.

17

# x.x

717

,x

7

-1x.-x-1,-x

22z4-3.-3y

2.1 S I M P L I F Y I N G A L G E B R A I C E X P R E S S I O N S

TEACHING TIPRemind students that a variablewithout a coefficient has an under-stood coefficient of 1. You maywant to advise your students towrite in the coefficient of 1.

CLASSROOM EXAMPLEIdentify the numerical coefficient in eachterm.

a. b.

c. d.answers:a. b. 2

c. d.13

-1

-5

z

3-y

2x2-5x

O b j e c t i v e s

1 Identify terms, like terms, and unlike terms.

2 Combine like terms.

3 Use the distributive property to remove parentheses.

4 Write word phrases as algebraic expressions.

As we explore in this section, an expression such as is not as simple as possible,because—even without replacing x by a value—we can perform the indicated addition.

1 Before we practice simplifying expressions, some new language of algebra is pre-sented.A term is a number or the product of a number and variables raised to powers.

Terms

The numerical coefficient (sometimes also simply called the coefficient) of a termis the numerical factor. The numerical coefficient of 3x is 3. Recall that 3x means 3 # x.

-y, 2x3, -5, 3xz2,2y

, 0.8z

3x + 2x

H e l p f u l H i n tThe term means and thus has a numerical coefficient of The term zmeans 1z and thus has a numerical coefficient of 1.

-1.-1y-y



82 CHAPTER 2 EQUATIONS, INEQUALIT IES, AND PROBLEM SOLVING

Why? Same variable x, but different powers x and Why? Different variablesWhy? Different variables and different powers

x 2

a. Unlike terms, since the exponents on x are not the same.

b. Like terms, since each variable and its exponent match.

c. Like terms, since by the commutative property.

d. Like terms.

2 An algebraic expression containing the sum or difference of like terms can besimplified by applying the distributive property. For example, by the distributive prop-erty, we rewrite the sum of the like terms as

Also,

Simplifying the sum or difference of like terms is called combining like terms.

-y2 + 5y2 = 1-1 + 52y2 = 4y2

3x + 2x = 13 + 22x = 5x

3x + 2x

zy = yz

Like Terms Unlike Terms

3x, 2x

6abc3, 6ab22ab2c3, ac3b27y, 3z, 8x2-6x2y, 2x2y, 4x2y

5x, 5x2

E X A M P L E 2Determine whether the terms are like or unlike.

a. b. c. d. -x4, x4-2yz, -3zy4x2y, x2y, -2x2y2x, 3x2

CLASSROOM EXAMPLEDetermine whether the terms are like orunlike.

a. b.

c.answers:

a. like b. unlike c. like

-5ab, 3ba

3x

2y 2, -x

2y 2, 2xy7x, -6x

Solut ion

E X A M P L E 3Simplify each expression by combining like terms.

a. b. c. 8x2 + 2x - 3x10y2 + y27x - 3x

Solut ion a.

b.

c. 8x2 + 2x - 3x = 8x2 + 12 - 32x = 8x2 - x

10y2 + y2 = 10y2 + 1y2 = 110 + 12y2 = 11y2

7x - 3x = 17 - 32x = 4x

E X A M P L E 4Simplify each expression by combining like terms.

a. b. c.

d. e. - 12

b + b2.3x + 5x - 6

4y - 3y2-5a - 3 + a + 22x + 3x + 5 + 2

H e l p f u l H i n tIn like terms, each variable and its exponent must match exactly, but these fac-tors don’t need to be in the same order.

are like terms.2x2y and 3yx2

CLASSROOM EXAMPLESimplify each expression by combininglike terms.

a. b.

c.answers:

a. 5y b. c. 5y + 3x12x 2

5y - 3x + 6x

11x 2 + x

29y - 4y


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CLASSROOM EXAMPLESimplify by combining like terms.

a.

b.

c.

answers:

a. b.

c. - 43

x

8.6x 2 - 4.3x-8a + 3

23

x - 2x

8.6x 2 - 4.3x

-7a - 2 - a + 5

E X A M P L E 5

Solut ion Use the distributive property to combine the numerical coefficients of like terms.

a.

b.

c. These two terms cannot be combined because they are unlike terms.

d.

e.

The examples above suggest the following:

- 12

b + b = a - 12

+ 1bb =12

b

= 7.3x - 6

2.3x + 5x - 6 = 12.3 + 52x - 6

4y - 3y2

= -4a - 1

= 1-5 + 12a + 1-3 + 22 -5a - 3 + a + 2 = -5a + 1a + 1-3 + 22

= 5x + 7

2x + 3x + 5 + 2 = 12 + 32x + 15 + 22

Combining Like TermsTo combine like terms, add the numerical coefficients and multiply the resultby the common variable factors.

Find each product by using the distributive property to remove parentheses.

a. b. c. -1x + y - 2z + 62-21y + 0.3z - 1251x + 22

Solut ion

CLASSROOM EXAMPLEFind each product by using the distribu-tive property to remove parentheses.

a.

b.answers:

a.

b. -3x - 2y - z + 1

-3y - 18

-13x + 2y + z - 12-31y + 62

a. Apply the distributive property.Multiply.

b. Apply the distributive property.

Multiply.

c. Distribute over each term.

= -x - y + 2z - 6

= -11x2 - 11y2 - 11-2z2 - 11621 -1x + y - 2z + 62 = -11x + y - 2z + 62

= -2y - 0.6z + 2

-21y + 0.3z - 12 = -21y2 + 1-2210.3z2 + 1-221-12 = 5x + 10

51x + 22 = 5 # x + 5 # 2

3 Simplifying expressions makes frequent use of the distributive property to alsoremove parentheses.

H e l p f u l H i n tIf a sign precedes parentheses, the sign of each term inside the parenthesesis changed when the distributive property is applied to remove parentheses.

Examples:

= 3x + 4y + 1-1-3x - 4y - 12-1x - 2y2 = -x + 2y

= 5x - y + z-1-5x + y - z2-12x + 12 = -2x - 1

“-”

When simplifying an expression containing parentheses, we often use the distrib-utive property in both directions—first to remove parentheses and then again to com-bine any like terms.




a. In words:

T T TTranslate: 6+2x

6added

totwice anumber

a. Apply the distributive property.

Combine like terms.

b. Apply the distributive property.

Combine like terms.

c. Apply the distributive property.

Combine like terms. = -21 - 12x

9 - 314x + 102 = 9 - 12x - 30

= -11x - 13

-214x + 72 - 13x - 12 = -8x - 14 - 3x + 1

= 6x - 14

312x - 52 + 1 = 6x - 15 + 1

E X A M P L E 6


a.

b.answers:a. b. 30y - 7x + 4

8 + 516y - 32312x + 12 - 15x - 12

Simplify the following expressions.

a. b. c. 9 - 314x + 102-214x + 72 - 13x - 12312x - 52 + 1

Solut ion

TEACHING TIPBefore Example 7, you may wantstudents to translate the following.“Subtract 8 from 10.” “Subtract bfrom a.”

Write the phrase below as an algebraic expression. Then simplify if possible. Subtractfrom 2x - 3.4x - 2

CLASSROOM EXAMPLESubtract from

answer: -3x + 7

4x + 6.7x - 1

H e l p f u l H i n tDon’t forget to use the distributive property to multiply before adding or sub-tracting like terms.

E X A M P L E 7

Solut ion “Subtract from ” translates to Next, simplify thealgebraic expression.

Apply the distributive property.Combine like terms.

4 Next, we practice writing word phrases as algebraic expressions.

= -2x - 1 12x - 32 - 14x - 22 = 2x - 3 - 4x + 2

12x - 32 - 14x - 22.2x - 34x - 2

E X A M P L E 8Write the following phrases as algebraic expressions and simplify if possible. Let x rep-resent the unknown number.

a. Twice a number, added to 6b. The difference of a number and 4, divided by 7c. Five added to 3 times the sum of a number and 1d. The sum of twice a number, 3 times the number, and 5 times the number

Solut ion


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b. In words:

T

Translate:

c. In words:

T T T T

Translate:

Next, we simplify this expression.

Use the distributive property.Combine like terms.

d. The phrase “the sum of” means that we add.

In words:

T T T T TTranslate:

Now let’s simplify.

Combine like terms.2x + 3x + 5x = 10x

5x+3x+2x

5 timesthe number

addedto

3 timesthe number

addedto

twice anumber

= 8 + 3x 5 + 31x + 12 = 5 + 3x + 3

1x + 123 #+5

the sum ofa number

and 13 times

addedto

five

x - 47

7divided

by

the differenceof a number

and 4

CLASSROOM EXAMPLEWrite each phrase as an algebraic expression and simplify if possible.

a. Three times a number, subtractedfrom 10.

b. Five added to twice the sum of anumber and seven.

answers:

a. b. 2x + 1910 - 3x

TEACHING TIPA Group Activity for this section isavailable in the Instructor’s Resource Manual.

STUDY SKILLS REMINDERAre You Getting all the Mathematics Help that YouNeed?

Remember that in addition to your instructor, there are manyplaces to get help with your mathematics course. For example, seethe list below.

There is an accompanying video lesson for every section inthis text.The back of this book contains answers to odd-numbered ex-ercises as well as answers to every exercise in the IntegratedReviews, Chapter Reviews, Chapter Tests, and CumulativeReviews.MathPro is available with this text. It is a tutorial software pro-gram with lessons corresponding to each section in the text.There is a student solutions manual available that containsworked-out solutions to odd-numbered exercises as well as solu-tions to every exercise in the Integrated Reviews, Chapter Re-views, Chapter Tests, and Cumulative Reviews.Check with your instructor for other local resources availableto you, such as the tutoring center.

dd



MENTAL MATHIdentify the numerical coefficient of each term. See Example 1.

1. 2. 3x 3 3. x 1 4. 5. 17

6. 1.2xyz 1.2 7. 8. 9.

Indicate whether the following lists of terms are like or unlike. See Example 2.

10. like 11. unlike 12. unlike

13. like 14. like 15. unlike7.4p3q2, 6.2p3q2r8wz, 17

zwab2, -7ab2

2z, 3z2-2x2y, 6xy5y, -y

- 23

- 23

z- 53

- 5y

318

p

8

17x2y-1-y-7-7y


O b j e c t i v e s

1 Define linear equation in one variable and equivalent equations.

2 Use the addition property of equality to solve linear equations.


1 Recall from Section 1.4 that an equation is a statement that two expressionshave the same value. Also, a value of the variable that makes an equation a true state-ment is called a solution or root of the equation. The process of finding the solution ofan equation is called solving the equation for the variable. In this section we concen-trate on solving linear equations in one variable.


Linear Equation in One VariableA linear equation in one variable can be written in the form

where a, b, and c are real numbers and a Z 0.

ax + b = c

Evaluating a linear equation for a given value of the variable, as we did in Section 1.4,can tell us whether that value is a solution, but we can’t rely on evaluating an equationas our method of solving it.

Instead, to solve a linear equation in x, we write a series of simpler equations, allequivalent to the original equation, so that the final equation has the form

Equivalent equations are equations that have the same solution. This means that the“number” above is the solution to the original equation.

2 The first property of equality that helps us write simpler equivalent equationsis the addition property of equality.

x number or number x

Addition Property of EqualityIf a, b, and c are real numbers, then

are equivalent equations.a = b and a + c = b + c

This property guarantees that adding the same number to both sides of an equationdoes not change the solution of the equation. Since subtraction is defined in terms ofaddition, we may also subtract the same number from both sides without changing thesolution.

2.2 T H E A D D I T I O N P R O P E R T Y O F E Q UA L I T Y



THE ADDIT ION PROPERTY OF EQUALITY SECTION 2.2 89

We use the addition property of equality to write equivalent equations until thevariable is by itself on one side of the equation, and the equation looks like

or “number = x.”“x = number”

x 2 2

5

Solve for x.x - 7 = 10

Solut ion To solve for x, we want x alone on one side of the equation.To do this, we add 7 to bothsides of the equation.

Add 7 to both sides.Simplify.

The solution of the equation is obviously 17. Since we are writing equivalentequations, the solution of the equation is also 17.

To check, replace x with 17 in the original equation.

Replace x with 17 in the original equation.

True.

Since the statement is true, 17 is the solution or we can say that the solution set is

CONCEPT CHECKUse the addition property to fill in the blank so that the middle equation simplifies to the last equation.

x = 8 x - 5 + = 3 +

x - 5 = 3

5176. 10 = 10

17 - 7 10

x - 7 = 10

x - 7 = 10x = 17

x = 17 x - 7 + 7 = 10 + 7

x - 7 = 10

x 2 2 5 2

TEACHING TIPRemind students that they mayadd or subtract any number toboth sides of an equation, and theresult is an equivalent equation.When solving equations, we try toadd or subtract a number on bothsides so that the equivalent equa-tion is a simpler one to solve.

E X A M P L E 1

If the same weight is added to or subtracted from each side, the scale remains balanced.

Concept Check Answer:5

x 2 5

A good way to picture a true equation is as a balanced scale. Since it is balanced,each side of the scale weighs the same amount.

CLASSROOM EXAMPLESolve:

answer: 3

x - 5 = -2



Solve for y.y + 0.6 = -1.0

Solut ion

E X A M P L E 2

To get y alone on one side of the equation, subtract 0.6 from both sides of the equation.

Subtract 0.6 from both sides.

Combine like terms.

To check the proposed solution, replace y with in the original equation.-1.6-1.6,

y = -1.6

y + 0.6 - 0.6 = -1.0 - 0.6

y + 0.6 = -1.0CLASSROOM EXAMPLESolve:

answer: -1.4

y + 1.7 = 0.3

CheckReplace y with in the

original equation.True.

The solution is or we can say that the solution set is 5-1.66.-1.6

-1.0 = -1.0

-1.6 -1.6 + 0.6 -1.0

y + 0.6 = -1.0

Solve:12

= x -34

E X A M P L E 3

Solut ion To get x alone, we add to both sides.

Add to both sides.

The LCD is 4.

Add the fractions.

54

= x

24

+34

= x

12

# 22

+34

= x

34

12

+34

= x -34

+34

12

= x -34

34

CLASSROOM EXAMPLE

Solve:

answer:2924

78

= y -13

Check Original equation.

Replace x with

Subtract.

True.

The solution is 54

.

12

=12

12

24

54

. 12

54

-34

12

= x -34

H e l p f u l H i n tWe may solve an equation so that the variable is alone on either side of theequation. For example, is equivalent to x = 5

4.54 = x


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Solve for t.5t - 5 = 6 t + 2

Solut ion To solve for t, we first want all terms containing t on one side of the equation and allother terms on the other side of the equation. To do this, first subtract 5t from bothsides of the equation.

Subtract 5t from both sides.Combine like terms.

Next, subtract 2 from both sides and the variable t will be isolated.

Subtract 2 from both sides.

Check the solution, in the original equation. The solution is

Many times, it is best to simplify one or both sides of an equation before applyingthe addition property of equality.

-7.-7,

-7 = t

-5 - 2 = t + 2 - 2 -5 = t + 2

-5 = t + 2 5t - 5 - 5t = 6t + 2 - 5t

5t - 5 = 6t + 2

CLASSROOM EXAMPLESolve: answer: -2

3a + 7 = 4a + 9

Solve: 2x + 3x - 5 + 7 = 10x + 3 - 6x - 4

Solut ion

E X A M P L E 4

E X A M P L E 5

First we simplify both sides of the equation.

Combine like terms on each sideof the equation.

Next, we want all terms with a variable on one side of the equation and all numbers onthe other side.

Subtract 4x from both sides.Combine like terms.Subtract 2 from both sides

to get x alone.Combine like terms. x = -3

x + 2 - 2 = -1 - 2 x + 2 = -1

5x + 2 - 4x = 4x - 1 - 4x

5x + 2 = 4x - 1

2x + 3x - 5 + 7 = 10x + 3 - 6x - 4CLASSROOM EXAMPLESolve:

answer: -4= -2w + 3 + 7w

10w + 3 - 4w + 4


Replace x with

Multiply.True.

The solution is

If an equation contains parentheses, we use the distributive property to removethem, as before. Then we combine any like terms.

-3.

-13 = -13 -6 - 9 - 5 + 7 -30 + 3 + 18 - 4

-3. 21-32 + 31-32 - 5 + 7 101-32 + 3 - 61-32 - 4

2x + 3x - 5 + 7 = 10x + 3 - 6x - 4


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Solut ionApply the distributive

property.Multiply.Combine like terms.Add 12 to both sides.Simplify. a = 19

a - 12 + 12 = 7 + 12 a - 12 = 7

12a - 6 - 11a - 6 = 7

612a2 + 61-12 - 1111a2 - 1162 = 7

612a - 12 - 1111a + 62 = 7

CLASSROOM EXAMPLESolve: answer: 13

312x - 52 - 15x + 12 = -3

Check Check by replacing a with 19 in the original equation.

Solve: 3 - x = 7

E X A M P L E 7

Solut ion First we subtract 3 from both sides.

Subtract 3 from both sides.Simplify.

We have not yet solved for x since x is not alone. However, this equation does say thatthe opposite of x is 4. If the opposite of x is 4, then x is the opposite of 4, or If

then x = -4.-x = 4,x = -4.

-x = 4 3 - x - 3 = 7 - 3

3 - x = 7CLASSROOM EXAMPLESolve: answer: -15

5 - x = 20

TEACHING TIPAfter solving Example 7, you maywant to point out that there is morethan one way to solve this problem.

-4 = x

3 - 7 = 7 + x - 7 3 = 7 + x

3 - x + x = 7 + x

3 - x = 7


Replace x with

Add.True.

The solution is -4.

7 = 7 3 + 4 7

-4. 3 - 1-42 7ø

3 - x = 7

3 Next, we practice writing word phrases as algebraic expressions.

a. The sum of two numbers is 8. If one number is 3, find the other number.b. The sum of two numbers is 8. If one number is x, write an expression representing

the other number.c. An 8-foot board is cut into two pieces. If one piece is x feet, express the length of

the other piece in terms of x.

E X A M P L E 8

Solut ion a. If the sum of two numbers is 8 and one number is 3, we find the other number bysubtracting 3 from 8. The other number is or 5.8 - 3

3 8 3 or 5

8

b. If the sum of two numbers is 8 and one number is x, we find the other number bysubtracting x from 8. The other number is represented by 8 - x.x 8 x

8

Solve: 612a - 12 - 111a + 62 = 7

E X A M P L E 6


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In words:

Translate:

The Golden Gate Bridge is feet long.1m - 602

60-m

60 minus Length of

Verrazano-NarrowsBridge

CLASSROOM EXAMPLEThe sum of two numbers is 50. If onenumber is x, write an expression representing the other number.answer: 50 - x

c. If an 8-foot board is cut into two pieces and one piece is x feet, we find the otherlength by subtracting x from 8. The other piece is feet.18 - x2

x feet 8 x feet

8 feet

E X A M P L E 9The Verrazano-Narrows Bridge in New York City is the longest suspension bridge inNorth America. The Golden Gate Bridge in San Francisco is 60 feet shorter than theVerrazano-Narrows Bridge. If the length of the Verrazano-Narrows Bridge is m feet,express the length of the Golden Gate Bridge as an algebraic expression in m. (Source:Survey of State Highway Engineers)

Solut ion Since the Golden Gate is 60 feet shorter than the Verrazano-Narrows Bridge, we havethat its length is

STUDY SKILLS REMINDERHave You Decided to Successfully Complete thisCourse?

Ask yourself if one of your current goals is to successfully com-plete this course.

If it is not a goal of yours, ask yourself why not. One commonreason is fear of failure. Amazingly enough, fear of failure alonecan be strong enough to keep many of us from doing our best inany endeavor. Another common reason is that you simply haven’ttaken the time to make successfully completing this course one ofyour goals.

If you are taking this mathematics course, then successfullycompleting this course probably should be one of your goals. Tomake it a goal, start by writing this goal in your mathematicsnotebook. Then read or reread Section 1.1 and make acommitment to try the suggestions in this section.

If successfully completing this course is already a goal of yours,also read or reread Section 1.1 and try some of the suggestions inthis section so that you are actively working toward your goal.

Good luck and don’t forget that a positive attitude will alsomake a big difference.



MENTAL MATHSolve each equation mentally. See Examples 1 and 2.

1. 2 2. 3 3. 12

4. 18 5. 17 6. 21d - 16 = 5b - 11 = 6z + 22 = 40

n + 18 = 30x + 7 = 10x + 4 = 6



Multiplication Property of EqualityIf a, b, and c are real numbers and then

are equivalent equations.

a = b and ac = bc

c Z 0,

TEACHING TIPRemind students that a true equa-tion is like a balanced scale. Thenask them: If you double the weighton one side, what must you do tothe other side to keep it in balance?

This property guarantees that multiplying both sides of an equation by the samenonzero number does not change the solution of the equation. Since division is definedin terms of multiplication, we may also divide both sides of the equation by the samenonzero number without changing the solution.

O b j e c t i v e s

1 Use the multiplication property of equality to solve linear equations.

2 Use both the addition and multiplication properties of equality to solve linear equations.


1 As useful as the addition property of equality is, it cannot help us solve everytype of linear equation in one variable. For example, adding or subtracting a value onboth sides of the equation does not help solve

Instead, we apply another important property of equality, the multiplication propertyof equality.

52

x = 15.

2.3 T H E M U LT I P L I CAT I O N P R O P E R T Y O F E Q UA L I T Y



THE MULTIPLICATION PROPERTY OF EQUALITY SECTION 2.3 97

Solut ion To get x alone, multiply both sides of the equation by the reciprocal of which is

Multiply both sides by

Apply the associative property.

Simplify.

or

x = 6

1x = 6

a25

# 52bx =

25

# 15

25

. 25

# 52

x =25

# 15

52

x = 15

25

.52

,

CLASSROOM EXAMPLE

Solve:

answer: 21

37

x = 9

Check Replace x with 6 in the original equation.

Original equation.

Replace x with 6.

True.

The solution is 6 or we say that the solution set is

In the equation is the coefficient of x. When the coefficient of x is a

fraction, we will get x alone by multiplying by the reciprocal. When the coefficient of xis an integer or a decimal, it is usually more convenient to divide both sides by the co-efficient. (Dividing by a number is, of course, the same as multiplying by the reciprocalof the number.)

52

x = 15, 52

566 15 = 15

52

162 15

52

x = 15

E X A M P L E 2

Solve: -3x = 33

Solut ion Recall that means To get x alone, we divide both sides by the coefficient ofx, that is,

Divide both sides by

Simplify.

x = -11

1x = -11

-3. -3x

-3=

33-3

-3x = 33

-3.-3 # x.-3x


answer: -6

7x = -42


Replace x with

True.

The solution is or the solution set is 5-116.-11,

33 = 33

-11. -31-112 33

-3x = 33

Solve:52

x = 15.

E X A M P L E 1


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E X A M P L E 3

Solve:y

7= 20

Solut ion Recall that To get y alone, we multiply both sides of the equation by 7, the

reciprocal of

Multiply both sides by 7.

Simplify.

y = 140

1y = 140

7 # 17

y = 7 # 20

17

y = 20

y

7= 20

17

.

y

7=

17

y.

CLASSROOM EXAMPLE

Solve:

answer: 65

x

5= 13


Replace y with 140.

True.

The solution is 140.

20 = 20

1407

20

y

7= 20

E X A M P L E 4Solve: 3.1x = 4.96

Solut ion

Divide both sides by 3.1.

Simplify.

x = 1.6

1x = 1.6

3.1x

3.1=

4.963.1

3.1x = 4.96

CLASSROOM EXAMPLESolve: answer: -5.2

-2.6x = 13.52

Check Check by replacing x with 1.6 in the original equation. The solution is 1.6.

Solve: - 23

x = - 52

Solut ion To get x alone, we multiply both sides of the equation by the reciprocal of the co-efficient of x.

Multiply both sides by the reciprocal of

Simplify. x =154

- 23

.- 32

, - 32

# - 23

x = - 32

# - 52

- 23

x = - 52

- 32

,

E X A M P L E 5

H e l p f u l H i n tDon’t forget to multiply

both sides by - 32

.


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CLASSROOM EXAMPLE

Solve:

answer:1825

- 56

x = - 35

Check Check by replacing x with in the original equation. The solution is

2 We are now ready to combine the skills learned in the last section with theskills learned from this section to solve equations by applying more than one property.

154

.154

Solve: -z - 4 = 6

Solut ion First, get the term containing the variable alone on one side.To do so, add 4 to bothsides of the equation.

Add 4 to both sides.

Simplify.

Next, recall that means To get z alone, either multiply or divide both sides ofthe equation by In this example, we divide.

Divide both sides by the coefficient

Simplify. z = -10

-1. -z

-1=

10-1

-z = 10

-1.-1 # z.-z

-z = 10

-z - 4 + 4 = 6 + 4

-z,

CLASSROOM EXAMPLESolve: answer: 19

-x + 7 = -12

E X A M P L E 6

Check To check, replace z with in the original equation. The solution is -10.-10

E X A M P L E 7Solve: 12a - 8a = 10 + 2a - 13 - 7

Solut ion First, simplify both sides of the equation by combining like terms.

Combine like terms.

To get all terms containing a variable on one side, subtract 2a from both sides.

Subtract 2a from both sides.Simplify.

Divide both sides by 2.

Simplify. a = -5

2a

2=

-102

2a = -10 4a - 2a = 2a - 10 - 2a

4a = 2a - 10 12a - 8a = 10 + 2a - 13 - 7


answer: -2-7x + 2x = -3 + x + 20 - 5

Check by replacing a with in the original equation. The solution is

3 Next, we continue to sharpen our problem-solving skills by writing word phrases as algebraic expressions.

-5.-5Check




If x is the first of three consecutive integers, express the sum of the three integers interms of x. Simplify if possible.

E X A M P L E 8

Solut ion An example of three consecutive integers is

8 9

21

7

CLASSROOM EXAMPLEIf x is the first of two consecutive odd integers, express their sum in terms of x.Then simplify.

answer: 2x + 2 The second consecutive integer is always 1 more than the first, and the third consecu-tive integer is 2 more than the first. If x is the first of three consecutive integers, thethree consecutive integers are

x 1 x 2

21

x

Their sum isIn words:

+ +

Translate:

which simplifies to 3x + 3.

1x + 22+1x + 12+x

thirdinteger

secondinteger

firstinteger

Below are examples of consecutive even and odd integers.

Even integers:

Odd integers:

x 2 x 4

42

x10 128

x 2 x 4

42

x7 95

11234 2 3 40 5 6

2units

2units

2units

2units

H e l p f u l H i n tIf x is an odd integer, then is the next odd integer. This 2 simply meansthat odd integers are always 2 units from each other. (The same is true for evenintegers. They are always 2 units from each other.)

x + 2

c


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STUDY SKILLS REMINDERAre You Organized?

Have you ever had trouble finding a completed assignment?When it’s time to study for a test, are your notes neat and organ-ized? Have your ever had trouble reading your own mathematicshandwriting? (Be honest—I have had trouble reading my ownhandwriting before.)

When any of these things happen, it’s time to get organized.Here are a few suggestions:

Write your notes and complete your homework assignment ina notebook with pockets (spiral or ring binder). Take class notesin this notebook, and then follow the notes with your completedhomework assignment. When you receive graded papers orhandouts, place them in the notebook pocket so that you will notlose them.

Place a mark (possibly an exclamation point) beside anynote(s) that seem especially important to you. Also place a mark(possibly a question mark) beside any note(s) or homework thatyou are having trouble with. Don’t forget to see your instructor, atutor, or your fellow classmates to help you understand theconcepts or exercises you have marked.

Also, if you are having trouble reading your own handwriting,slow down and write your mathematics work clearly!

MENTAL MATHSolve each equation mentally. See Examples 2 and 3.

1. 9 2. 6 3. 2

4. 2 5. 6. -88r = -64-56x = -307t = 14

5b = 109c = 543a = 27

1. 2. 7 3. 0 4. 0 5. 12 6. 7. 8. 9. 3 10. 2 11. 2 12. 30-20-12-8-4


SOLVING L INEAR EQUATIONS SECTION 2.4 103

O b j e c t i v e s

1 Apply the general strategy for solving a linear equation.

2 Solve equations containing fractions.

3 Solve equations containing decimals.

4 Recognize identities and equations with no solution.

5 Write sentences as equations and solve.

1 We now present a general strategy for solving linear equations. One new pieceof strategy is a suggestion to “clear an equation of fractions” as a first step. Doing somakes the equation more manageable, since operating on integers is more convenientthan operating on fractions.

2.4 S O LV I N G L I N E A R E Q UAT I O N S

TEACHING TIPBefore students are shown thesteps for solving a linear equation,let them come up with their ownsteps for solving a linear equationwith your guidance. Then comparetheir steps with these. They will remember a set of steps better ifthey are involved in writing them.

Solving Linear Equations in One VariableStep 1. Multiply on both sides by the LCD to clear the equation of frac-

tions if they occur.

Step 2. Use the distributive property to remove parentheses if they occur.

Step 3. Simplify each side of the equation by combining like terms.

Step 4. Get all variable terms on one side and all numbers on the otherside by using the addition property of equality.

Step 5. Get the variable alone by using the multiplication property of equality.

Step 6. Check the solution by substituting it into the original equation.

Solve: 412x - 32 + 7 = 3x + 5

E X A M P L E 1

Solut ion There are no fractions, so we begin with Step 2.

Step 2. Apply the distributive property.

Step 3. Combine like terms.

Step 4. Get all variable terms on the same side of the equation by subtracting 3xfrom both sides, then adding 5 to both sides.

Subtract 3x from both sides.Simplify.Add 5 to both sides.Simplify.

Step 5. Use the multiplication property of equality to get x alone.


Simplify. x = 2

5x

5=

105

5x = 10 5x - 5 + 5 = 5 + 5

5x - 5 = 5 8x - 5 - 3x = 3x + 5 - 3x

8x - 5 = 3x + 5

8x - 12 + 7 = 3x + 5

412x - 32 + 7 = 3x + 5CLASSROOM EXAMPLESolve: answer: 3

513x - 12 + 2 = 12x + 6





H e l p f u l H i n tWhen checking solutions, use the original written equation.

Solve: 812 - t2 = -5t

E X A M P L E 2

Solut ion First, we apply the distributive property.

Step 2. Use the distributive property.

Step 4. To get variable terms on one side, add 8t to both sides.

Combine like terms.

Step 5. Divide both sides by 3.

Simplify.

Step 6. Check.

Original equation

Replace t with

The LCD is 3.

Subtract fractions.

True.

The solution is

2 If an equation contains fractions, we can clear the equation of fractions by mul-tiplying both sides by the LCD of all denominators. By doing this, we avoid workingwith time-consuming fractions.

163

.

- 803

= - 803

8a - 103b -

803

8a63

-163b -

803

163

. 8a2 -163b -5a16

3b

812 - t2 = -5t

163

= t

163

=3t

3

16 = 3t

16 - 8t + 8t = -5t + 8t

16 - 8t = -5t

812 - t2 = -5tCLASSROOM EXAMPLE

Solve:

answer:152

915 - x2 = -3x

Step 6. Check.

Original equation

Replace x with 2.

True.

The solution is 2 or the solution set is 526. 11 = 11

4 + 7 11

4112 + 7 11

414 - 32 + 7 6 + 5

4[2122 - 3] + 7 3122 + 5

412x - 32 + 7 = 3x + 5


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Solve:x

2- 1 =

23

x - 3

Solut ion

E X A M P L E 3

We begin by clearing fractions.To do this, we multiply both sides of the equation by theLCD of 2 and 3, which is 6.

Step 1. Multiply both sides by the LCD, 6.


Simplify.

There are no longer grouping symbols and no like terms on either side of the equa-tion, so we continue with Step 4.

To get variable terms on one side, subtractStep 4. 3x from both sides.

Simplify.Add 18 to both sides.Simplify.

Step 5. The variable is now alone, so there is no need to apply the multiplicationproperty of equality.

Step 6. Check.

Original equation

Replace x with 12.

Simplify.True.

The solution is 12. 5 = 5

6 - 1 8 - 3

122

- 1 23

# 12 - 3

x

2- 1 =

23

x - 3

12 = x -6 + 18 = x - 18 + 18

-6 = x - 18 3x - 6 - 3x = 4x - 18 - 3x

3x - 6 = 4x - 18

3x - 6 = 4x - 18

6ax

2b - 6112 = 6a2

3 xb - 6132

6ax

2- 1b = 6a2

3 x - 3b

x

2- 1 =

23

x - 3

H e l p f u l H i n tDon’t forget to multiplyeach term by the LCD.

CLASSROOM EXAMPLE

Solve:

answer: -3

52

x - 1 =32

x - 4

Solve:21a + 32

3= 6a + 2

E X A M P L E 4

Solut ion We clear the equation of fractions first.

Step 1. Clear the fraction by multiplyingboth sides by the LCD, 3.

Step 2. Next, we use the distributive property and remove parentheses.

2(a + 3) = 3(6a + 2)

3 # 21a + 323

= 316a + 22

21a + 323

= 6a + 2CLASSROOM EXAMPLE

Solve:

answer: -3

31x - 225

= 3x + 6


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answer: 9

= -0.021820.06x - 0.101x - 22

TEACHING TIPHelp your students understandthat each term must be multipliedby 100. In the case of this is accomplished by multiplying0.10 by 100. Tell your students why.

0.101x - 32,

Solve: 0.25x + 0.101x - 32 = 0.051222E X A M P L E 5

Solut ion First we clear this equation of decimals by multiplying both sides of the equation by100. Recall that multiplying a decimal number by 100 has the effect of moving the dec-imal point 2 places to the right.

0.25x + 0.101x - 32 = 0.051222

H e l p f u l H i n tBy the distributive prop-erty, 0.10 is multiplied byx and Thus to multi-ply each term here by100, we only need tomultiply 0.10 by 100.

-3.Step 1. Multiply both sides by 100.


Step 3. Combine like terms.

Step 4. Add 30 to both sides.

Combine like terms.

Step 5. Divide both sides by 35.

Step 6. To check, replace x with 4 in the original equation. The solution is 4.

4 So far, each equation that we have solved has had a single solution. However,not every equation in one variable has a single solution. Some equations have no solu-tion, while others have an infinite number of solutions. For example,

has no solution since no matter which real number we replace x with, the equation is false.

On the other hand,

has infinitely many solutions since x can be replaced by any real number and the equa-tion is always true.

real number + 6 = same real number + 6 TRUE

x + 6 = x + 6

real number + 5 = same real number + 7 FALSE

x + 5 = x + 7

x = 4

35x

35=

14035

35x = 140

35x - 30 + 30 = 110 + 30

35x - 30 = 110

25x + 10x - 30 = 110

25x + 101x - 32 = 51222 0.25x +0.101x - 32= 0.051222

Apply the distributive property.

Step 4. Subtract 6 from both sides.

Subtract 18a from both sides.

Step 5. Divide both sides by

Write the fraction in simplest form.

Step 6. To check, replace a with 0 in the original equation. The solution is 0.

3 When solving a problem about money, you may need to solve an equation con-taining decimals. If you choose, you may multiply to clear the equation of decimals.

a = 0

-16. -16a

-16=

0-16

-16a = 0 2a - 18a = 18a - 18a

2a = 18a

2a + 6 - 6 = 18a + 6 - 6

2a + 6 = 18a + 6

t


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The equation is called an identity.The next few examples illustrate spe-cial equations like these.

x + 6 = x + 6

E X A M P L E 6

Solve: -21x - 52 + 10 = -31x + 22 + x

Solut ionApply the distributive property on both sides.

Combine like terms.

Add 2x to both sides.

Combine like terms.

The final equation contains no variable terms, and there is no value for x that makesa true equation. We conclude that there is no solution to this equation. In set

notation, we can indicate that there is no solution with the empty set, or use theempty set or null set symbol, In this chapter, we will simply write no solution.¤.

5 6,20 = -6

20 = -6

-2x + 20 + 2x = -2x - 6 + 2x

-2x + 20 = -2x - 6

-2x + 10 + 10 = -3x - 6 + x -21x - 52 + 10 = -31x + 22 + x

E X A M P L E 7Solve: 31x - 42 = 3x - 12

Solut ion


The left side of the equation is now identical to the right side. Every real number maybe substituted for x and a true statement will result.We arrive at the same conclusion ifwe continue.

Add 12 to both sides.Combine like terms.Subtract 3x from both sides.

Again, one side of the equation is identical to the other side.Thus, isan identity and all real numbers are solutions. In set notation, this is

CONCEPT CHECKSuppose you have simplified several equations and obtain the following results. What can you concludeabout the solutions to the original equation?

a. b. c.

5 We can apply our equation-solving skills to solving problems written in words.Many times, writing an equation that describes or models a problem involves a directtranslation from a word sentence to an equation.

7 = -4x = 07 = 7

5all real numbers6.31x - 42 = 3x - 12

0 = 0 3x - 3x = 3x - 3x

3x = 3x 3x - 12 + 12 = 3x - 12 + 12

3x - 12 = 3x - 12

3x - 12 = 3x - 12

31x - 42 = 3x - 12

CLASSROOM EXAMPLESolve: answer: no solution

512 - x2 + 8x = 31x - 62

CLASSROOM EXAMPLE

Solve:

answer: all real numbers

= -101x + 2) - 2x

-612x + 12 - 14

FINDING AN UNKNOWN NUMBERTwice a number, added to seven, is the same as three subtracted from the number. Findthe number.

E X A M P L E 8

Concept Check Answer:

a. Every real number is a solution.b. The solution is 0.c. There is no solution.


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Solut ion Translate the sentence into an equation and solve.

In words:

Translate:

To solve, begin by subtracting x from both sides to isolate the variable term.

Subtract x from both sides.

Combine like terms.


Combine like terms.

Check the solution in the problem as it was originally stated.To do so, replace “number”in the sentence with Twice added to 7 is the same as 3 subtracted from

The unknown number is -10.

-13 = -13

21-102 + 7 = -10 - 3

“-10.”“-10”-10.

x = -10

x + 7 - 7 = -3 - 7

x + 7 = -3

2x + 7 - x = x - 3 - x

2x + 7 = x - 3

x - 3=7+2x

three subtractedfrom the number

is thesame as

sevenadded

totwice anumber

CLASSROOM EXAMPLEThree times a number, subtracted fromtwo, equals twice the number, added totwenty seven. Find the number.answer: -5

Checking Equations

We can use a calculator to check possible solutions of equations.To do this, replace thevariable by the possible solution and evaluate both sides of the equation separately.

Equation: Solution:Original equation

Replace x with 16.

Now evaluate each side with your calculator.

Evaluate left side:

Evaluate right side:

Since the left side equals the right side, the equation checks.

2 1 16 + 6 2 = or ENTER Display: 44 or 2116 + 6244

3 * 16 - 4 = or ENTER Display: 44 or 3 * 16 - 444

31162 - 4 2116 + 62 3x - 4 = 21x + 62 x = 163x - 4 = 21x + 62


H e l p f u l H i n tWhen checking solutions, go back to the original stated problem, rather than toyour equation in case errors have been made in translating to an equation.



5

Use a calculator to check the possible solutions to each equation.1. 2.

3. 4.

5. 6. 201x - 392 = 5x - 432; x = 23.2564x

4= 200x - 1116492; x = 121

-1.6x - 3.9 = -6.9x - 25.6; x = 55x - 2.6 = 21x + 0.82; x = 4.4

-3x - 7 = 3x - 1; x = -12x = 48 + 6x; x = -12



O b j e c t i v e

1 Apply the steps for problem solving.

1 In previous sections, you practiced writing word phrases and sentences as alge-braic expressions and equations to help prepare for problem solving.We now use thesetranslations to help write equations that model a problem. The problem-solving stepsgiven next may be helpful.

General Strategy for Problem Solving1. UNDERSTAND the problem. During this step, become comfortable with

the problem. Some ways of doing this are:

Read and reread the problem.Choose a variable to represent the unknown.Construct a drawing, whenever possible.Propose a solution and check. Pay careful attention to how you checkyour proposed solution. This will help when writing an equation tomodel the problem.

2. TRANSLATE the problem into an equation.3. SOLVE the equation.

4. INTERPRET the results: Check the proposed solution in the stated prob-lem and state your conclusion.

TEACHING TIPSpend some time helping studentswith step 1. We often jump to step 2before students really have achance to understand the problem.

Much of problem solving involves a direct translation from a sentence to an equation.Although we have been practicing these translations in previous sections, this sectionwill often have translations that require a placement of parentheses.

CLASSROOM EXAMPLEThree times the difference of a numberand 5 is the same as twice the number,decreased by 3. Find the number.answer: 12

E X A M P L E 1

FINDING AN UNKNOWN NUMBERTwice the sum of a number and 4 is the same as four times the number decreased by 12.Find the number.

Solut ion 1. UNDERSTAND. Read and reread the problem. If we let

“the sum of a number and 4” translates to and “four times the number”translates to “4x.”

2. TRANSLATE.

T T T T T T2 12-4x=1x + 42

12 decreased

byfour times

the numberis the

same assum of a

number and 4 twice

“x + 4”

x = the unknown number, then

2.5 A N I N T R O D U C T I O N TO P R O B L E M S O LV I N G



AN INTRODUCTION TO PROBLEM SOLVING SECTION 2.5 113

1. UNDERSTAND. Read and reread the problem. If we let

2. TRANSLATE.

T

T

T

3. SOLVE.

4. INTERPRET.

Check: If then the next odd integer Notice theirsum, as needed.

State: The area codes are 603 and 605.Note: New Hampshire’s area code is 603 and South Dakota’s area code is

605.

603 + 605 = 1208,x + 2 = 603 + 2 = 605.x = 603,

x = 603

2x

2=

12062

2x = 1206 2x + 2 - 2 = 1208 - 2

2x + 2 = 1208 x + x + 2 = 1208

1208=1x + 22+x

1208 is nextodd

integer

the sum offirstodd

integer

x + 2 = the next odd integer

x = the first odd integer, then

Some states have a single area code for the entire state. Two such states have areacodes that are consecutive odd integers. If the sum of these integers is 1208, find thetwo area codes. (Source: World Almanac, 2003)

E X A M P L E 2

CLASSROOM EXAMPLEThe sum of three consecutive even integers is 146. Find the integers.answer: 46, 48, 52

Solut ion

H e l p f u l H i n tRemember, the 2 heremeans that odd integersare 2 units apart, for ex-ample, the odd integers13 and 13 + 2 = 15.

3. SOLVE.

Apply the distributive property.Subtract 4x from both sides.



4. INTERPRET.

Check: Check this solution in the problem as it was originally stated. To do so, re-place “number” with 10. Twice the sum of “10” and 4 is 28, which is the same as 4times “10” decreased by 12.State: The number is 10.

Next, we continue to review consecutive integers.

x = 10

-2. -2x

-2=

-20-2

-2x = -20 -2x + 8 - 8 = -12 - 8

-2x + 8 = -12 2x + 8 - 4x = 4x - 12 - 4x

2x + 8 = 4x - 12

21x + 42 = 4x - 12


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E X A M P L E 3

FINDING THE LENGTH OF A BOARDA 10-foot board is to be cut into two pieces so that the longer piece is 4 times the shorter.Find the length of each piece.

Solut ion 1. UNDERSTAND the problem. To do so, read and reread the problem. You may alsowant to propose a solution. For example, if 3 feet represents the length of the shorterpiece, then is the length of the longer piece, since it is 4 times the lengthof the shorter piece.This guess gives a total board length of too long. However, the purpose of proposing a solution is not to guess correctly, but tohelp better understand the problem and how to model it.

Since the length of the longer piece is given in terms of the length of the shorterpiece, let’s let

4x = length of longer piece x = length of shorter piece, then

3 feet + 12 feet = 15 feet,4132 = 12 feet

CLASSROOM EXAMPLEAn 18-foot wire is to be cut so that thelonger piece is 5 times longer than theshorter piece. Find the length of eachpiece.answer: shorter piece

longer piece = 15 ft= 3 ft;

x feet 4x feet

10 feet

2. TRANSLATE the problem. First, we write the equation in words.

T T T T T

3. SOLVE.

Combine like terms.


4. INTERPRET.

Check: Check the solution in the stated problem. If the shorter piece of board is 2feet, the longer piece is and the sum of the two pieces is

State: The shorter piece of board is 2 feet and the longer piece of board is 8 feet.2 feet + 8 feet = 10 feet.

4 # 12 feet2 = 8 feet

x = 2

5x

5=

105

5x = 10 x + 4x = 10

10=4x+x

total lengthof board

equalslength of

longer pieceadded to

length ofshorter piece

H e l p f u l H i n tMake sure that units are included in your answer, if appropriate.

FINDING THE NUMBER OF REPUBLICAN AND DEMOCRATIC SENATORS

In a recent year, the U.S. House of Representatives had a total of 431 Democ-rats and Republicans. There were 15 more Republican representatives than

E X A M P L E 4


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1. UNDERSTAND. Read and reread the problem. Let’s propose that the studentspent 20 hours working on computers. Pay careful attention as to how his income iscalculated. For 20 hours and 10 visits, his income is morethan $575. We now have a better understanding of the problem and know that thetime working on computers is less than 20 hours.

Let’s let

working on computers. Thenof money made while working on computers 25x = amount

x = hours

201$252 + 101$202 = $700,

Solut ion 1. UNDERSTAND the problem. Read and reread the problem. Let’s suppose thatthere were 200 Democratic representatives. Since there were 15 more Republicansthan Democrats, there must have been Republicans. The total num-ber of Democrats and Republicans was then This is incorrect sincethe total should be 431, but we now have a better understanding of the problem.

In general, if we let

2. TRANSLATE the problem. First, we write the equation in words.

+

3. SOLVE.

Combine like terms.



4. INTERPRET.

Check: If there were 208 Democratic representatives, then there wereRepublican representatives. The total number of representatives is

then The results check.State: There were 208 Democratic and 223 Republican representatives in Congress.

208 + 223 = 431.208 + 15 = 223

x = 208

2x

2=

4162

2x = 416 2x + 15 - 15 = 431 - 15

2x + 15 = 431

x + 1x + 152 = 431

431=1x + 152xppppp

431equalsnumber of

Republicansadded to

number ofDemocrats

x + 15 = number of Republicans

x = number of Democrats, then

200 + 215 = 415.200 + 15 = 215

CLASSROOM EXAMPLEThrough the year 2000, the state of Cali-fornia had 22 more electoral votes forpresident than the state of Texas. If thetotal electoral votes for these two stateswas 86, find the number of electoral votes for each state.answer:

California = 54 electoral votesTexas = 32 electoral votes;

CALCULATING HOURS ON JOBA computer science major at a local university has a part-time job working on computersfor his clients. He charges $20 to come to your home or office and then $25 per hour.During one month he visited 10 homes or offices and his total income was $575. Howmany hours did he spend working on computers?

E X A M P L E 5

Solut ion

Democratic representatives. Find the number of representatives from each party.(Source: Office of the Clerk of the U.S. House of Representatives)


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CLASSROOM EXAMPLEA car rental agency charges $28 a dayand $0.15 a mile. If you rent a car for aday and your bill (before taxes) is $52,how many miles did you drive?answer: 160 miles

E X A M P L E 6FINDING ANGLE MEASURES

If the two walls of the Vietnam Veterans Memorial in Washington D.C. were connect-ed, an isosceles triangle would be formed. The measure of the third angle is 97.5° morethan the measure of either of the other two equal angles. Find the measure of the thirdangle. (Source: National Park Service)

Solut ion 1. UNDERSTAND. Read and reread the problem. We then draw a diagram (recallthat an isosceles triangle has two angles with the same measure) and let

2. TRANSLATE. Recall that the sum of the measures of the angles of a triangleequals 180.

+ 180

3. SOLVE.

Combine like terms.Subtract 97.5 from both sides.


x = 27.5

3x

3=

82.53

3x = 82.5 3x + 97.5 - 97.5 = 180 - 97.5

3x + 97.5 = 180 x + x + 1x + 97.52 = 180

=1x + 97.52x+xppppp

180equalsmeasureof thirdangle

measureof second

angle

measureof firstangle

x + 97.5 = degree measure of the third angle

x = degree measure of the second equal angle

x = degree measure of one angle

(x 97.5)

x x

2. TRANSLATE.

T T T T T

575

3. SOLVE.



Simplify.

4. INTERPRET.

Check: If the student works 15 hours and makes 10 visits, his income is

State: The student spent 15 hours working on computers.+ 101$202 = $575.

151$252

x = 15

25x

25=

37525

25x = 375 25x + 200 - 200 = 575 - 200

25x + 200 = 575

=101202+25x

575is equal tomoney made

for visitsplus

money madewhile workingon computers


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4. INTERPRET.

Check: If then the measure of the third angle is The sumof the angles is then the correct sum.State: The third angle measures 125°.( The two walls actually meet at an angle of 125 degrees 12 minutes. The measure-ment of 97.5° given in the problem is an approximation.)*

*27.5 + 27.5 + 125 = 180,

x + 97.5 = 125.x = 27.5,

CLASSROOM EXAMPLEThe measures of two angles of a triangleare equal. If the third angle measures39° more, find the measures of all threeangles.answer: 47°, 47°, 86°

TEACHING TIPA Group Activity is available forthis section in the Instructor’s Resource Manual.

STUDY SKILLS REMINDERHow Are Your Homework Assignments Going?

It is so important in mathematics to keep up with homework.Why? Many concepts build on each other. Oftentimes, your un-derstanding of a day’s lecture in mathematics depends on an un-derstanding of the previous day’s material.

Remember that completing your homework assignment involvesa lot more than attempting a few of the problems assigned.

To complete a homework assignment, remember these fourthings:

1. Attempt all of it.

2. Check it.

3. Correct it.

4. If needed, ask questions about it.



O b j e c t i v e s

1 Use formulas to solve problems.

2 Solve a formula or equation for one of its variables.

1 An equation that describes a known relationship among quantities, such as dis-tance, time, volume, weight, and money is called a formula. These quantities are repre-sented by letters and are thus variables of the formula. Here are some commonformulas and their meanings.

TEACHING TIPRemind students that the front andback covers of this text contain for-mulas that they may need and haveforgotten.

Formulas are valuable tools because they allow us to calculate measurements aslong as we know certain other measurements. For example, if we know we traveled adistance of 100 miles at a rate of 40 miles per hour, we can replace the variables d andr in the formula and find our time, t.

Formula.Replace d with 100 and r with 40.

This is a linear equation in one variable, t. To solve for t, divide both sides of the equa-tion by 40.


Simplify.

The time traveled is hours or hours.In this section we solve problems that can be modeled by known formulas. We

use the same problem-solving steps that were introduced in the previous section.Thesesteps have been slightly revised to include formulas.

2 12

52

52

= t

10040

=40t

40

100 = 40t d = rt

d = rt

degrees Fahrenheit = a95b # degrees Celsius + 32

F = a95bC + 32

Volume of a rectangular solid = length # width # heightV = lwh

distance = rate # timed = rt

Perimeter of a triangle = side a + side b + side cP = a + b + c

Simple Interest = Principal # Rate # TimeI = PRT

Area of a rectangle = length # widthA = lw

E X A M P L E 1

FINDING TIME GIVEN RATE AND DISTANCEA glacier is a giant mass of rocks and ice that flows downhill like a river. Portage Glac-ier in Alaska is about 6 miles, or 31,680 feet, long and moves 400 feet per year. Icebergs

2.6 F O R M U L A S A N D P R O B L E M S O LV I N G



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FORMULAS AND PROBLEM SOLVING SECTION 2.6 123

Solut ion 1. UNDERSTAND. Read and reread the problem. The appropriate formula neededto solve this problem is the distance formula, To become familiar with thisformula, let’s find the distance that ice traveling at a rate of 400 feet per year travelsin 100 years. To do so, we let time t be 100 years and rate r be the given 400 feet peryear, and substitute these values into the formula We then have that distance

Since we are interested in finding how long it takes iceto travel 31,680 feet, we now know that it is less than 100 years.

Since we are using the formula we let

time in years for ice to reach the lake

or speed of ice

from beginning of glacier to lake

2. TRANSLATE. To translate to an equation, we use the formula and let dis-tance and rate per year.

Let and

3. SOLVE. Solve the equation for t. To solve for t, divide both sides by 400.


Simplify.

4. INTERPRET.

Check: To check, substitute 79.2 for t and 400 for r in the distance formula andcheck to see that the distance is 31,680 feet.

State: It takes 79.2 years for the ice at the head of Portage Glacier to reach the lake.

79.2 = t

31,680

400=

400 # t

400

r = 400.d = 31,68031,680 = 400 # tΩø

d = r # t

r = 400 feetd = 31,680 feetd = rt

d = distance

r = rate

t = the

d = rt,

d = 40011002 = 40,000 feet.d = rt.

d = rt.CLASSROOM EXAMPLEA family drives from Cincinnati, Ohio toRapid City, South Dakota, a distance of1180 miles. They average a rate of 50miles per hour. How much time did theyspend driving?answer: 23.6 hr

H e l p f u l H i n tDon’t forget to includeunits, if appropriate.

are created when the front end of the glacier flows into Portage Lake. How long does ittake for ice at the head (beginning) of the glacier to reach the lake?



CALCULATING THE LENGTH OF A GARDEN

Charles Pecot can afford enough fencing to enclose a rectangular garden with aperimeter of 140 feet. If the width of his garden must be 30 feet, find the length.

Solut ion

TEACHING TIPYou may want to take a moment toremind students that perimeter ismeasured in units, area in squareunits, and volume in cubic units.

1. UNDERSTAND. Read and reread the problem. The formula needed to solve thisproblem is the formula for the perimeter of a rectangle, Before con-tinuing, let’s become familiar with this formula.

length of the rectangular garden

width of the rectangular garden

of the garden

2. TRANSLATE. To translate to an equation, we use the formula andlet perimeter and width .

Let and

3. SOLVE.

Multiply 2(30).Subtract 60 from both sides.Combine like terms.Divide both sides by 2.

4. INTERPRET.

Check: Substitute 40 for l and 30 for in the perimeter formula and check to seethat the perimeter is 140 feet.State: The length of the rectangular garden is 40 feet.

w

40 = l 80 = 2l

140 - 60 = 2l + 60 - 60 140 = 2l + 60 140 = 2l + 21302

w = 30.P = 140140 = 2l + 21302ΩT

P = 2l + 2w

w = 30 feetP = 140 feetP = 2l + 2w

P = perimeter

w = the

l = the

P = 2l + 2w.

l

w 30 feet

E X A M P L E 2

CLASSROOM EXAMPLEA wood deck is being built behind ahouse. The width of the deck is 14 feet. Ifthere is 168 square feet of decking material, find the length of the deck.answer: 12 ft

FINDING AN EQUIVALENT TEMPERATURE

The average maximum temperature for January in Algerias, Algeria, is 59° Fahrenheit.Find the equivalent temperature in degrees Celsius.

E X A M P L E 3

Solut ion 1. UNDERSTAND. Read and reread the problem. A formula that can be used tosolve this problem is the formula for converting degrees Celsius to degrees Fah-renheit, Before continuing, become familiar with this formula. UsingF = 9

5 C + 32.


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CLASSROOM EXAMPLEConvert the temperature 5°C to Fahrenheit.answer: 41°F

1. UNDERSTAND. Read and reread the problem. Recall that the formula for theperimeter of a rectangle is Draw a rectangle and guess the solution. Ifthe width of the rectangular sign is 5 feet, its length is 2 feet less than 3 times thewidth or The perimeter P of the rectangle is then

too much.We now know that the width is less than5 feet.2113 feet2 + 215 feet2 = 36 feet,

315 feet2 - 2 feet = 13 feet.

P = 2l + 2w.

13 feet

5 feet

CLASSROOM EXAMPLEThe length of a rectangle is one moremeter than 4 times its width. Find the di-mensions if the perimeter is 52 meters.answer: length: 21; width: 5

FINDING ROAD SIGN DIMENSIONS

The length of a rectangular road sign is 2 feet less than three times its width. Find thedimensions if the perimeter is 28 feet.

Solut ion

E X A M P L E 4

this formula, we let

2. TRANSLATE. To translate to an equation, we use the formula and

let degrees Fahrenheit

Let

3. SOLVE.


Combine like terms.


Simplify.

4. INTERPRET.

Check: To check, replace C with 15 and F with 59 in the formula and see that atrue statement results.State: Thus, 59° Fahrenheit is equivalent to 15° Celsius.

In the next example, we again use the formula for perimeter of a rectangle as inExample 2. In Example 2, we knew the width of the rectangle. In this example, both thelength and width are unknown.

15 = C

59.

59

# 27 =59

# 95

C

27 =95

C

59 - 32 =95

C + 32 - 32

59 =95

C + 32

F = 59.59 =95

C + 32 Substitute:

F =95

C + 32 Formula:

F = 59.

F =95

C + 32

F = temperature in degrees Fahrenheit. C = temperature in degrees Celsius, and


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3w 2

w

Let

Draw a rectangle and label it with the assigned variables.

2. TRANSLATE.

3. SOLVE.




4. INTERPRET.

Check: If the width of the sign is 4 feet, the length of the sign is This gives a perimeter of

the correct perimeter.State: The width of the sign is 4 feet and the length of the sign is 10 feet.

2 We say that the formula is solved for F because F is alone on oneside of the equation and the other side of the equation contains no F’s. Suppose that weneed to convert many Fahrenheit temperatures to equivalent degrees Celsius. In this case,it is easier to perform this task by solving the formula for C. (See Example8.) For this reason, it is important to be able to solve an equation for any one of its specifiedvariables. For example, the formula is solved for d in terms of r and t. We can alsosolve for t in terms of d and r.To solve for t, divide both sides of the equation by r.

Divide both sides by r.

Simplify.

To solve a formula or an equation for a specified variable, we use the same stepsas for solving a linear equation. These steps are listed next.

dr

= t

dr

=rtr

d = rt

d = rtd = rt

F = 95 C + 32

F = 95 C + 32

P = 214 feet2 + 2110 feet2 = 28 feet,- 2 feet = 10 feet.314 feet2

4 = w

328

=8w

8

32 = 8w 28 + 4 = 8w - 4 + 4

28 = 8w - 4 28 = 6w - 4 + 2w

28 = 213w - 22 + 2w

Substitute: 28 = 213w - 22 + 2w. Formula: P = 2l + 2w or

3w - 2 = the length of the sign. w = the width of the rectangular sign; then

Solving Equations for a Specified VariableStep 1. Multiply on both sides to clear the equation of fractions if they occur.

Step 2. Use the distributive property to remove parentheses if they occur.Step 3. Simplify each side of the equation by combining like terms.Step 4. Get all terms containing the specified variable on one side and all

other terms on the other side by using the addition property ofequality.

Step 5. Get the specified variable alone by using the multiplication proper-ty of equality.



Solve for l.V = lwh

Solut ion This formula is used to find the volume of a box. To solve for l, divide both sides by h.

Divide both sides by h.

Simplify.

Since we have l alone on one side of the equation, we have solved for l in terms of V, ,and h. Remember that it does not matter on which side of the equation we isolate thevariable.

w

V

wh= l

w V

wh=

lwh

wh

V = lwh

w

E X A M P L E 5

CLASSROOM EXAMPLESolve: for r.

answer: r =C

2p

C = 2pr

h

wl

Solve for x.y = mx + b

E X A M P L E 6

Solut ion The term containing the variable we are solving for, mx, is on the right side of the equa-tion. Get mx alone by subtracting b from both sides.

Subtract b from both sides.

Combine like terms.

Next, solve for x by dividing both sides by m.

Simplify. y - b

m= x

y - b

m=

mxm

y - b = mx

y - b = mx + b - b

y = mx + bCLASSROOM EXAMPLESolve: for x.

answer: x =y + 7

3

y = 3x - 7

Solve for .wP = 2l + 2w

Solut ion This formula relates the perimeter of a rectangle to its length and width. Find the termcontaining the variable . To get this term, 2 , alone subtract 2l from both sides.

Subtract 2l from both sides.

Combine like terms.


Simplify. P - 2l

2= w

P - 2l

2=

2w

2

P - 2l = 2w

P - 2l = 2l + 2w - 2l P = 2l + 2w

ww

E X A M P L E 7

CLASSROOM EXAMPLESolve: for a.

answer: a =P - b + c

2

P = 2a + b - c

H e l p f u l H i n tThe 2’s may not be divid-ed out here. Although 2is a factor of the denomi-nator, 2 is not a factor ofthe numerator since it isnot a factor of bothterms in the numerator.

l

w




Solve for C.F = 95 C + 32

E X A M P L E 8

The next example has an equation containing a fraction. We will first clear theequation of fractions and then solve for the specified variable.

Solut ion Clear the fraction by multiplying bothsides by the LCD.

Distribute the 5.

To get the term containing the variable C alone, subtract 160 from both sides.

Combine like terms.


Simplify. 5F - 160

9= C

5F - 160

9=

9C

9

5F - 160 = 9C

5F - 160 = 9C + 160 - 160

5F = 9C + 160

51F2 = 5195 C + 32)

F = 95 C + 32

Suppose you have just purchased ahouse. You are required by your mort-gage company to buy home-owner’s in-surance. In choosing a policy, you mustalso choose a deductible level. A de-ductible is how much you, the home-owner, pay to repair or replace a loss

before the insurance company starts paying. Typically, deductibles start at$250. However, other levels such as $500 or $1000 are also available. One wayto save money when purchasing your homeowner’s policy is to raise the de-ductible amount. However, doing so means you will have to pay more out ofpocket in case of theft or fire. Study the accompanying chart.Which amount ofdeductible would you choose? Why?

Increase Your Deductible to:

$500 save up to 12%

$1000 save up to 24%



on the cost of a homeowner’spolicy, depending on your insur-ance company.

Source: Insurance Information Institute



O b j e c t i v e s

1 Solve problems involving percents.

2 Solve problems involving distance.

3 Solve problems involving mixtures.

4 Solve problems involving interest.

This section is devoted to solving problems in the categories listed. The same problem-solving steps used in previous sections are also followed in this section. They are listedbelow for review.

2.7 F U R T H E R P R O B L E M S O LV I N G



FURTHER PROBLEM SOLVING SECTION 2.7 133

General Strategy for Problem Solving1. UNDERSTAND the problem. During this step, become comfortable with

the problem. Some ways of doing this are:Read and reread the problem.Choose a variable to represent the unknown.Construct a drawing, whenever possible.Propose a solution and check. Pay careful attention to how you check yourproposed solution.This will help writing an equation to model the problem.

2. TRANSLATE the problem into an equation.3. SOLVE the equation.


1 The first two examples involve percents.

Percent increase or percent decrease is a common way to describe how some measure-ment has increased or decreased. For example, crime increased by 8%, teachers re-ceived a 5.5% increase in salary, or a company decreased its employees by 10%. Thenext example is a review of percent increase.

CALCULATING THE COST OF ATTENDING COLLEGE

The cost of attending a public college rose from $5324 in 1990 to $8086 in 2000. Findthe percent increase. (Source: U.S. Department of Education. Note: These costs includeroom and board.)

Solut ion 1. UNDERSTAND. Read and reread the problem. Let’s guess that the percent in-crease is 20%. To see if this is the case, we find 20% of $5324 to find the increasein cost. Then we add this increase to $5324 to find the new cost. In other words,

the increase in cost. The new cost thenwould be less than the actual new cost of $8086. Wenow know that the increase is greater than 20% and we know how to check our pro-posed solution.

Let percent increase.

2. TRANSLATE. First, find the increase, and then the percent increase. The increasein cost is found by:

In words: or

Translate:

Next, find the percent increase. The percent increase or percent decrease is always apercent of the original number or in this case, the old cost.

In words:

Translate:

3. SOLVE.Divide both sides by 5324 and

round to 3 decimal places.Write as a percent. 51.9% L x

0.519 L x 2762 = 5324x

$5324#x=$2762

old costofwhat percent increaseisincrease

$2762=$5324-$8086=increase

old cost-new cost=increase

x = the

$5324 + $1064.80 = $6388.80,20% 1$53242 = 0.201$53242 = $1064.80,

E X A M P L E 1

CLASSROOM EXAMPLEIf a number decreases from 200 to 120,find the percent decrease.answer: 40%




Most of the movie screens in the United States project analog films, but the number ofcinemas using digital are increasing. Find the number of digital screens last year if aftera 175% increase, the number this year is 124. Round to the nearest whole.

E X A M P L E 2

1. UNDERSTAND. Read and reread the problem. Let’s guess a solution and see howwe would check our guess. If the number of digital screens last year was 50, wewould see if 50 plus the increase is 124; that is,

Since 137.5 is too large, we know that our guess of 50 is too large.We also have a bet-ter understanding of the problem. Let

of digital screens last year

2. TRANSLATE. To translate to an equation, we remember that

In words:

Translate:

3. SOLVE.

4. INTERPRET.

Check: Recall that x represents the number of digital screens last year. If this num-ber is approximately 45, let’s see if 45 plus the increase is close to 124. (We use theword “close” since 45 is rounded.)

which is close to 124.State: There were approximately 45 digital screens last year.

2 The next example involves distance.

45 + 175%1452 = 45 + 1.751452 = 2.751452 = 123.75,

x L 45

x =1242.75

2.75x = 124

124=1.75x+x

number ofdigital screens

this yearequalsincreaseplus

number ofdigital screen

last year

x = number

50 + 175%(50) = 50 + 1.751502 = 2.751502 = 137.50

Solut ion

4. INTERPRET.

Check: Check the proposed solution.State: The percent increase in cost is approximately 51.9%.

CLASSROOM EXAMPLEFind the original price of a suit if the saleprice is $46 after a 20% discount.answer: $57.50

FINDING TIME GIVEN RATE AND DISTANCE

Marie Antonio, a bicycling enthusiast, rode her 21-speed at an average speed of 18miles per hour on level roads and then slowed down to an average of 10 mph on thehilly roads of the trip. If she covered a distance of 98 miles, how long did the entire triptake if traveling the level roads took the same time as traveling the hilly roads?

E X A M P L E 3

Solut ion 1. UNDERSTAND the problem. To do so, read and reread the problem. The formu-la is needed. At this time, let’s guess a solution. Suppose that she spent 2hours traveling on the level roads. This means that she also spent 2 hours traveling

d = r # t


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CALCULATING PERCENT FOR A LAB EXPERIMENT

A chemist working on his doctoral degree at Massachusetts Institute of Technologyneeds 12 liters of a 50% acid solution for a lab experiment. The stockroom has only

E X A M P L E 4

CLASSROOM EXAMPLEA jet traveling at 550 mph overtakes asmall light plane traveling at 180 mphthat had a 1-hour head start. How farfrom the starting points are the planes?answer: miles267

2137

on the hilly roads, since the times spent were the same. What is her total distance?Her distance on the level road is Her distance onthe hilly roads is This gives a total distance of

not the correct distance of 98 miles. Rememberthat the purpose of guessing a solution is not to guess correctly (although this may hap-pen) but to help better understand the problem and how to model it with an equation.

We are looking for the length of the entire trip, so we begin by letting

Because the same amount of time is spent on hilly roads, then also

2. TRANSLATE. To help us translate to an equation, we now summarize the in-formation from the problem on the following chart. Fill in the rates given, thevariables used to represent the times, and use the formula to fill in thedistance column.

d = r # t

x = the time spent on hilly roads.

x = the time spent on level roads.

36 miles + 20 miles = 56 miles,rate # time = 10122 = 20 miles.

rate # time = 18122 = 36 miles.

Level 18 x 18xHilly 10 x 10x

Rate # Time Distance

Since the entire trip covered 98 miles, we have that

In words:

Translate:

3. SOLVE.

Add like terms.


4. INTERPRET the results.

Check: Recall that x represents the time spent on the level portion of the trip andalso the time spent on the hilly portion. If Marie rides for 3.5 hours at 18 mph, herdistance is If Marie rides for 3.5 hours at 10 mph, her distance is

The total distance is the re-quired distance.State: The time of the entire trip is then or 7 hours.

3 Mixture problems involve two or more different quantities being combined toform a new mixture. These applications range from Dow Chemical’s need to form achemical mixture of a required strength to Planter’s Peanut Company’s need to findthe correct mixture of peanuts and cashews, given taste and price constraints.

3.5 hours + 3.5 hours

63 miles + 35 miles = 98 miles,1013.52 = 35 miles.1813.52 = 63 miles.

3.5 = x

9828

=28x

28

98 = 28x

10x+18x=98

hilly distance+level distance=total distance


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Solut ion 1. UNDERSTAND. First, read and reread the problem a few times. Next, guess a so-lution. Suppose that we need 7 liters of the 40% solution. Then we need liters of the 70% solution. To see if this is indeed the solution, find the amount ofpure acid in 7 liters of the 40% solution, in 5 liters of the 70% solution, and in 12liters of a 50% solution, the required amount and strength.

Since liters and not 6, our guess is incorrect, but we havegained some valuable insight into how to model and check this problem.

Let

2. TRANSLATE. To help us translate to an equation, the following table summarizesthe information given. Recall that the amount of acid in each solution is found bymultiplying the acid strength of each solution by the number of liters.

12 - x = number of liters of 70% solution. x = number of liters of 40% solution; then

2.8 liters + 3.5 liters = 6.3

1210.502 or 6 liters=50%*12 liters

510.702 or 3.5 liters=70%*5 liters

710.402 or 2.8 liters=40%*7 liters

amount of pureacid

=acid strength*number of liters

12 - 7 = 5

40% and 70% solutions. How much of each solution should be mixed together to form12 liters of a 50% solution?

CLASSROOM EXAMPLEHow much 20% dye solution and 50%dye solution should be mixed to obtain 6liters of a 40% solution?answer: 2 liters of the 20% solution; 4 liters of the 50% solution

Acid

No. of Liters Strength = Amount of Acid

40% Solution x 40% 0.40x70% Solution 70%50% Solution Needed 12 50% 0.50(12)

0.70112 - x212 - x

#

(12x) liters(12x)liters x liters

12liters

x liters

40%solution

70%solution

50%solution

The amount of acid in the final solution is the sum of the amounts of acid in the twobeginning solutions.

In words:

Translate:

3. SOLVE.

Apply the distributive property.Combine like terms.Subtract 8.4 from both sides.Divide both sides by

4. INTERPRET.

Check: To check, recall how we checked our guess.State: If 8 liters of the 40% solution are mixed with or 4 liters of the 70%solution, the result is 12 liters of a 50% solution.

12 - 8

-0.3. x = 8 -0.3x = -2.4

-0.3x + 8.4 = 6 0.4x + 8.4 - 0.7x = 6

0.40x + 0.70112 - x2 = 0.501122

0.501122=0.70112 - x2+0.40x

acid in 50%mixture

=acid in 70%

solution+

acid in 40%solution



1. UNDERSTAND. Read and reread the problem. Next, guess a solution. Supposethat Rajiv invested $8000 in the 7% fund and the rest, $12,000, in the fund paying 9%.To check, find his interest after one year. Recall the formula, so the inter-est from the The interest from the

The sum of the interests is Our guess is incorrect, since the sum of the interests is not $1550, but we now have abetter understanding of the problem.

Let

2. TRANSLATE. We apply the simple interest formula and organize ourinformation in the following chart. Since there are two different rates of interest andtwo different amounts invested, we apply the formula twice.

I = PRT

20,000 - x = amount of money in the account paying 9%.

The rest of the money is $20,000 less x or

x = amount of money in the account paying 7%.

$560 + $1080 = $1640.= $12,00010.092112 = $1080.9% fund7% fund = $800010.072112 = $560.

I = PRT,

4 The next example is an investment problem.

FINDING THE INVESTMENT AMOUNT

Rajiv Puri invested part of his $20,000 inheritance in a mutual funds account that pays 7%simple interest yearly and the rest in a certificate of deposit that pays 9% simple interestyearly. At the end of one year, Rajiv’s investments earned $1550. Find the amount heinvested at each rate.

E X A M P L E 5

Solut ion

Principal Rate Time Interest

7% Fund x 0.07 1 x(0.07)(1) or 0.07x9% Fund 0.09 1

or Total 20,000 1550

0.09120,000 - x2120,000 - x210.09211220,000 - x

##

The total interest earned, $1550, is the sum of the interest earned at 7% and the in-terest earned at 9%.

In words:

Translate:

3. SOLVE.

Apply the distributive property.Combine like terms.Subtract 1800 from both sides.Divide both sides by

4. INTERPRET.

Check: If then or 7500. These solutionsare reasonable, since their sum is $20,000 as required. The annual interest on $12,500at 7% is $875; the annual interest on $7500 at 9% is $675, and State: The amount invested at 7% is $12,500.The amount invested at 9% is $7500.

$875 + $675 = $1550.

20,000 - x = 20,000 - 12,500x = 12,500,

0.02. x = 12,500 -0.02x = -250

1800 - 0.02x = 1550 0.07x + 1800 - 0.09x = 1550

0.07x + 0.09120,000 - x2 = 1550

1550=0.09120,000 - x2+0.07x

totalinterest

=interestat 9%

+interestat 7%

CLASSROOM EXAMPLEThomas Mason invested part of his$10,000 inheritance in a savings planthat pays 4% simple interest yearly andthe rest in a certificate of deposit thatpays 6% simple interest yearly. At theend of one year, the investments earned$520. Find the amount invested at eachrate.answer: $4000 @ 4%; $6000 @ 6%


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Suppose you are a personal incometax preparer. Your clients, Jose and Fe-licia Fernandez, are filing jointly Form1040 as their individual income tax re-turn. You know that medical expensesmay be written off as an itemized de-duction if the expenses exceed 7.5% of

their adjusted gross income. Furthermore, only the portion of medical ex-penses that exceed 7.5% of their adjusted gross income can be deducted. Isthe Fernandez family eligible to deduct their medical expenses? Explain.

Internal Revenue Service

Adjusted Gross Income . . . $33,650

Form 1040

Fernandez FamilyDeductible Medical Expenses

Medical billsDental billsPrescription drugsMedical Insurance premiums

$1025$ 325$ 360$1200

10. 55 mph 20. $8500 @ 10%; $17,000 @ 12%


SOLVING L INEAR INEQUALIT IES SECTION 2.8 141

O b j e c t i v e s

1 Define linear inequality in one variable.

2 Graph solution sets on a number line and use interval notation.

3 Solve linear inequalities.

4 Solve compound inequalities.

5 Solve inequality applications.

1 In Chapter 1, we reviewed these inequality symbols and their meanings:

means “is less than” means “is less than or equal to”means “is greater than” means “is greater than or equal to”Ú7

…6

This definition and all other definitions, properties, and steps in this section also holdtrue for the inequality symbols, and

2 A solution of an inequality is a value of the variable that makes the inequalitya true statement. The solution set is the set of all solutions. For the inequality replacing x with any number less than 3, that is, to the left of 3 on a number line, makesthe resulting inequality true.This means that any number less than 3 is a solution of theinequality

Since there are infinitely many such numbers, we cannot list all the solutions ofthe inequality. We can use set notation and write

5x | x 6 36. Recall that this is readq q ()*

the suchset of that x is less than 3.all x

q

x 6 3.

x 6 3,

… .7 , Ú ,

Equations Inequalities

x

4- 6 7 1

x

4- 6 = 1

12 … 7 - 3y12 = 7 - 3y5n - 6 7 145n - 6 = 14x … 3x = 3

2.8 S O LV I N G L I N E A R I N E Q UA L I T I E S

0 1 2 3 4 5 6

A linear inequality is similar to a linear equation except that the equality symbolis replaced with an inequality symbol.

Linear Inequality in One VariableA linear inequality in one variable is an inequality that can be written in theform

where a, b, and c are real numbers and a is not 0.

ax + b 6 c




We can also picture the solutions on a number line. If we use open/closed-circlenotation, the graph of looks like the following.5x ƒ x 6 36

71 2 3 4 5 6

In this text, a convenient notation, called interval notation, will be used to write solu-tion sets of inequalities. To help us understand this notation, a different graphing nota-tion will be used. Instead of an open circle, we use a parenthesis; instead of a closedcircle, we use a bracket. With this new notation, the graph of now looks like5x ƒ x 6 36

71 2 3 4 5 6

CLASSROOM EXAMPLEGraph . Write the solution in interval notation.answer: 1-3, q2

x 7 -3

Graph Then write the solutions in interval notation.x Ú -1.

0 1 2 31234

We place a bracket at since the inequality symbol is and is greater than orequal to Then we shade to the right of -1.-1.

-1Ú-1

E X A M P L E 1

Solut ion

1234

TEACHING TIPRemind students that and

or are not numbers. Sincewe approach and butnever arrive, we always placeparentheses about them.

- q ,qq+ q

- q

and can be represented in interval notation as The symbol read as “nega-tive infinity,” does not indicate a number, but does indicate that the shaded arrow to theleft never ends. In other words, the interval includes all numbers less than 3.

Picturing the solutions of an inequality on a number line is called graphing the so-lutions or graphing the inequality, and the picture is called the graph of the inequality.

To graph or simply shade the numbers to the left of 3 andplace a bracket at 3 on the number line. The bracket indicates that 3 is a solution: 3 isless than or equal to 3. In interval notation, we write (- q , 3].

x … 3,5x ƒ x … 36

1- q , 32- q ,1- q , 32.

x 7

7653 4(5, )

x 5

67 589(, 7)

0 1 2 3 4 5 6

In interval notation, this is

3 When solutions of a linear inequality are not immediately obvious, they arefound through a process similar to the one used to solve a linear equation. Our goal isto get the variable alone, and we use properties of inequality similar to properties ofequality.

[-1, q2.

H e l p f u l H i n tWhen writing an inequality in interval notation, it may be easier to first graphthe inequality, then write it in interval notation. To help, think of the numberline as approaching to the left and or to the right. Then simplywrite the interval notation by following your shading from left to right.

q+ q- q


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CLASSROOM EXAMPLESolve Graph the solutionset and write it in interval notation.answer: [-5, q2

x - 6 Ú -11.

Solve for x. Graph the solution set and write it in interval notation.x + 4 … -6

To solve for x, subtract 4 from both sides of the inequality.

Original inequality.


The solution set is 1- q , -10].

x … -10 x + 4 - 4 … -6 - 4

x + 4 … -6

E X A M P L E 2

Solut ion

345678

This property also holds true for subtracting values, since subtraction is definedin terms of addition. In other words, adding or subtracting the same quantity from bothsides of an inequality does not change the solution of the inequality.

1112 910 78 6

TEACHING TIPAsk students to give a few specificsolutions to Example 2. Stress thatall numbers less than or equal to

are solutions.-10An important difference between linear equations and linear inequalities is

shown when we multiply or divide both sides of an inequality by a nonzero real num-ber. For example, start with the true statement and multiply both sides by 2. Aswe see below, the resulting inequality is also true.

True.Multiply both sides by 2.

True.

But if we start with the same true statement and multiply both sides by the resulting inequality is not a true statement.

True.


False.

Notice, however, that if we reverse the direction of the inequality symbol, the resultinginequality is true.

False.

True.

This demonstrates the multiplication property of inequality.

-12 7 -16

-12 6 -16

-12 6 -16

2. -2162 6 -2182 6 6 8

-2,6 6 8

12 6 16

2162 6 2182 6 6 8

6 6 8

Addition Property of InequalityIf a, b, and c are real numbers, then

are equivalent inequalities.

a 6 b and a + c 6 b + c

H e l p f u l H i n tNotice that any number less than or equal to is a solution to Forexample, solutions include

-10, -200, -11 12

, -7p, -2130, -50.3

x … -10.-10


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Multiplication Property of Inequality1. If a, b, and c are real numbers, and c is positive, then


2. If a, b, and c are real numbers, and c is negative, then


a 6 b and ac 7 bc

a 6 b and ac 6 bc

CLASSROOM EXAMPLESolve answer: [-4, q2

-3x … 12.

Solve Graph the solution set and write it in interval notation.-2x … -4.

Solut ion Remember to reverse the direction of the inequality symbol when dividing by a nega-tive number.

Divide both sides by and reverse the direction of the inequality sign.

Simplify.

The solution set is graphed as shown.[2, q2 x Ú 2

2 -2x

-2Ú

-4-2

-2x … -4

0 1 2 3 4 5 61

Solve Graph the solution set and write it in interval notation.2x 6 -4.

Solut ionDivide both sides by 2.Do not reverse the direction of the inequality sign.

Simplify. x 6 -2

2x

26

-42

2x 6 -4

E X A M P L E 3

H e l p f u l H i n tDon’t forget to reversethe direction of the inequality sign.

012345

E X A M P L E 4

H e l p f u l H i n tDo not reverse the inequality sign.

Because division is defined in terms of multiplication, this property also holdstrue when dividing both sides of an inequality by a nonzero number. If we multiply ordivide both sides of an inequality by a negative number, the direction of the inequalitysign must be reversed for the inequalities to remain equivalent.

H e l p f u l H i n tWhenever both sides of an inequality are multiplied or divided by a negativenumber, the direction of the inequality symbol must be reversed to form anequivalent inequality.

0 1 21234

The solution set is graphed as shown.1- q , -22

CLASSROOM EXAMPLESolve answer: 1- q , -32

5x 6 -15.

234



CONCEPT CHECKFill in the blank with or

a. Since then

b. Since then

c. If then 2a ____ 2b.

d. If then

The following steps may be helpful when solving inequalities. Notice that thesesteps are similar to the ones given in Section 2.4 for solving equations.

a

-3 _____

b

-3.a Ú b,

a 6 b,

5-7

_____ -2-7

.5 Ú -2,

31-82_____31-42.-8 6 -4,

Ú .6 , 7 , … ,


a. b. c. d. …6…6

Solving Linear Inequalities in One VariableStep 1. Clear the inequality of fractions by multiplying both sides of the in-

equality by the lowest common denominator (LCD) of all fractionsin the inequality.

Step 2. Remove grouping symbols such as parentheses by using the distrib-utive property.

Step 3. Simplify each side of the inequality by combining like terms.

Step 4. Write the inequality with variable terms on one side and numberson the other side by using the addition property of inequality.

Step 5. Get the variable alone by using the multiplication property of inequality.

H e l p f u l H i n tDon’t forget that if both sides of an inequality are multiplied or divided by anegative number, the direction of the inequality sign must be reversed.

E X A M P L E 5Solve Graph the solution set and write it in interval notation.-4x + 7 Ú -9.

Solut ionSubtract 7 from both sides.Simplify.

Divide both sides by and reversethe direction of the inequality sign.

Simplify.

The solution set is graphed as shown.1- q , 4]

x … 4

4 -4x

-4…

-16-4

-4x Ú -16 -4x + 7 - 7 Ú -9 - 7

-4x + 7 Ú -9

CLASSROOM EXAMPLESolve answer: [8, q2

-3x + 11 … -13.

6 7 8 9 10

3 4 5 6 7 8



Solve Graph the solution set and write it in interval notation.2x + 7 … x - 11.

Solut ionSubtract x from both sides.Combine like terms.Subtract 7 from both sides.Combine like terms.

The graph of the solution set is shown.1- q , -18]

x … -18 x + 7 - 7 … -11 - 7

x + 7 … -11 2x + 7 - x … x - 11 - x

2x + 7 … x - 11

E X A M P L E 6

CLASSROOM EXAMPLESolve answer: [0, q2

5x + 3 Ú 4x + 3.

19181716151420

0 1 212

Solve Graph the solution set and write it in interval notation.-5x + 7 6 21x - 32.

Solut ion


Subtract 2x from both sides.

Combine like terms.


Combine like terms.

Divide both sides by and reverse the direction of the inequality sign.

Simplify.

The graph of the solution set is shown.1137 , q2

x 7137

7 -7x

-77

-13-7

-7x 6 -13

-7x + 7 - 7 6 -6 - 7

-7x + 7 6 -6

-5x + 7 - 2x 6 2x - 6 - 2x

-5x + 7 6 2x - 6

-5x + 7 6 21x - 32

E X A M P L E 7

0 1 2 3 412

œ

CLASSROOM EXAMPLESolve

answer: 1- q , 122-6x - 3 7 -41x + 12.

12 10 2q

Solve Graph the solution set and write it in intervalnotation.

21x - 32 - 5 … 31x + 22 - 18.

E X A M P L E 8

Solut ionApply the distributive property.Combine like terms.Subtract 3x from both sides.Add 11 to both sides. -x … -1

-x - 11 … -12 2x - 11 … 3x - 12

2x - 6 - 5 … 3x + 6 - 18

21x - 32 - 5 … 31x + 22 - 18


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CLASSROOM EXAMPLE

Solve

answer: 1- q , 6]

31x + 52 - 1 Ú 51x - 12 + 7.

3 2 1 0 1 2 3

Divide both sides by and reversethe direction of the inequality sign.

Simplify.

The graph of the solution set is shown.[1, q2 x Ú 1

1 -x

-1Ú

-1-1

Graph Write the solutions in interval notation.2 6 x … 4.

Solut ion Graph all numbers greater than 2 and less than or equal to 4. Place a parenthesis at 2,a bracket at 4, and shade between.

0 1 2 3 4 5 61

E X A M P L E 9

CLASSROOM EXAMPLEGraph Write the solutions in interval notation.answer: [-2, 02

-2 … x 6 0.

10123

When we solve a simple inequality, we isolate the variable on one side of the in-equality. When we solve a compound inequality, we isolate the variable in the middlepart of the inequality. Also, when solving a compound inequality, we must perform thesame operation to all three parts of the inequality: left, middle, and right.

E X A M P L E 1 0Solve Graph the solution set and write it in interval notation.-1 … 2x - 3 6 5.

CLASSROOM EXAMPLESolve answer: 1-5, 1]

-24 6 5x + 1 … 6.

4 3 25 1 0 1 26

0 1 2 3 412

Solut ionAdd 3 to all three parts.Combine like terms.

Divide all three parts by 2.

Simplify. 1 … x 6 4

22

…2x

26

82

2 … 2x 6 8 -1 + 3 … 2x - 3 + 3 6 5 + 3

-1 … 2x - 3 6 5

The graph of the solution set [1, 4) is shown.

7 83 4 5 6

4 Inequalities containing one inequality symbol are called simple inequalities,while inequalities containing two inequality symbols are called compound inequalities.Acompound inequality is really two simple inequalities in one. The compound inequality

This can be read “x is greater than 3 and less than 5.”A solution of a compound inequality is a value that is a solution of both of the

simple inequalities that make up the compound inequality. For example,

To graph place parentheses at both 3 and 5 and shade between.3 6 x 6 5,

4 12

is a solution of 3 6 x 6 5 since 3 6 4 12

and 4 12

6 5.

3 6 x 6 5 means 3 6 x and x 6 5

0 1 2 3 4 5 6





Solve Graph the solution set and write it in interval notation.3 …3x

2+ 4 … 5.

Solut ion

Multiply all three parts by 2 toclear the fraction.

Distribute.

Subtract 8 from all three parts.

Divide all three parts by 3.

Simplify.

The graph of the solution set is shown.[-23, 23]

-23

… x …23

-23

…3x

3…

23

-2 … 3x … 2

6 … 3x + 8 … 10

2132 … 2a3x

2+ 4b … 2152

3 …3x

2+ 4 … 5

E X A M P L E 1 1

CLASSROOM EXAMPLE

Solve

answer: [-152 , 522

-4 …2x

5- 1 6 0.

ei

0 1 2 312

ss

5 Problems containing words such as “at least,” “at most,” “between,” “no morethan,” and “no less than” usually indicate that an inequality should be solved instead ofan equation. In solving applications involving linear inequalities, use the same proce-dure you use to solve applications involving linear equations.

E X A M P L E 1 2STAYING WITHIN BUDGET

Marie Chase and Jonathan Edwards are having their wedding reception at the GalleryReception Hall. They may spend at most $2000 for the reception. If the reception hallcharges a $100 cleanup fee plus $36 per person, find the greatest number of people thatthey can invite and still stay within their budget.

Solut ion 1. UNDERSTAND. Read and reread the problem. Next, guess a solution. If 40 peopleattend the reception, the cost is Let number of people who attend the reception.

2. TRANSLATE.

In words:

Translate:

3. SOLVE.


Divide both sides by 36. x … 52 79

36x … 1900 100 + 36x … 2000

2000…36x+100

$2000must be less than

or equal tocost per person+cleanup fee

x = the$100 + $361402 = $100 + $1440 = $1540.


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CLASSROOM EXAMPLEA couple will spend at most $3000 for acatered retirement party. If the cateringcompany changes a one-time fee of$200 plus $32 per person, find thegreatest number of people that can at-tend.answer: 87

MENTAL MATHSolve each of the following inequalities.

1. 2. 3. 4. x … 79x … 63x Ú 82x Ú 16x 6 54x 6 20x 7 25x 7 10

4. INTERPRET.

Check: Since x represents the number of people, we round down to the nearestwhole, or 52. Notice that if 52 people attend, the cost is

If 53 people attend, the cost iswhich is more than the given $2000.

State: Marie Chase and Jonathan Edwards can invite at most 52 people to the reception.

$100 + $361532 = $2008,$100 + $361522 = $1972.


3.1 T H E R E C TA N G U L A R C O O R D I N AT E S Y S T E M

164 CHAPTER 3 GRAPHING

O b j e c t i v e s

1 Define the rectangular coordinate system and plot ordered pairs of numbers.

2 Graph paired data to create a scatter diagram.

3 Determine whether an ordered pair is a solution of an equation in two variables.

4 Find the missing coordinate of an ordered pair solution, given one coordinate of the pair.

1 In Section 1.9, we learned how to read graphs. Example 4 in Section 1.9 pre-sented the graph below showing the relationship between time since smoking a ciga-rette and pulse rate. Notice in this graph that there are two numbers associated witheach point of the graph. For example, we discussed earlier that 15 minutes after “light-ing up,” the pulse rate is 80 beats per minute. If we agree to write the time first and thepulse rate second, we can say there is a point on the graph corresponding to theordered pair of numbers (15, 80). A few more ordered pairs are listed alongside theircorresponding points.

In general, we use this same ordered pair idea to describe the location of a pointin a plane (such as a piece of paper).We start with a horizontal and a vertical axis. Eachaxis is a number line, and for the sake of consistency we construct our axes to intersectat the 0 coordinate of both. This point of intersection is called the origin. Notice thatthese two number lines or axes divide the plane into four regions called quadrants.Thequadrants are usually numbered with Roman numerals as shown. The axes are notconsidered to be in any quadrant.

TEACHING TIPIf your class is arranged in rows,begin this lesson by discussingways to describe the location ofdesks in the classroom. Point outthat you need to define a referencepoint, that each location needs arow value and a column value andthat you need to agree on whichvalue to state first. Then connectthese ideas to the origin and or-dered pairs. Test students’ under-standing by asking them to find thelocation of their desk. Then havegroups of students stand up whomeet certain criteria.

100

90

80

70

60

50

40

30

20

10

0

Pul

se R

ate

(hea

rtbe

ats

per

min

ute)

Time (minutes)405 0

(0, 60)

(5, 95)

(15, 80)

5 10 15 20 25 30 35

(40, 70)

112345

12345

2 3 4 512345

Quadrant II Quadrant I

origin

Quadrant III Quadrant IV

x-axis

y-axis



THE RECTANGULAR COORDINATE SYSTEM SECTION 3.1 165

It is helpful to label axes, so we label the horizontal axis the x-axis and the vertical axisthe y-axis. We call the system described above the rectangular coordinate system.

Just as with the pulse rate graph, we can then describe the locations of points byordered pairs of numbers. We list the horizontal x-axis measurement first and the ver-tical y-axis measurement second.

To plot or graph the point corresponding to the ordered pair

(a, b)

we start at the origin. We then move a units left or right (right if a is positive, left if a isnegative). From there, we move b units up or down (up if b is positive, down if b is neg-ative). For example, to plot the point corresponding to the ordered pair (3, 2), we startat the origin, move 3 units right, and from there move 2 units up. (See the figure below.)The x-value, 3, is called the x-coordinate and the y-value, 2, is called the y-coordinate.From now on, we will call the point with coordinates (3, 2) simply the point (3, 2). Thepoint is graphed below also.1-2, 52

112345

12345

2 3 4 512345

(3, 2)(2, 3)

x

y

TEACHING TIPRemind students that ordered pairvalues are given in alphabeticalorder. (x, y).

TEACHING TIPThroughout this section, remindstudents of the contents of thisHelpful Hint.

Concept Check Answer:The graph of point lies inquadrant II and the graph of point

lies in quadrant IV. They arenot in the same location.11, -52

1-5, 12 H e l p f u l H i n tDon’t forget that each ordered pair corresponds to exactly one point in theplane and that each point in the plane corresponds to exactly one ordered pair.

112345

12345

2 3 4 512345

(3, 2)

(2, 5)

2 units up

2 units left

3 units right5 units up

x-axis

y-axis

Does the order in which the coordinates are listed matter? Yes! Notice that thepoint corresponding to the ordered pair (2, 3) is in a different location than the pointcorresponding to (3, 2). These two ordered pairs of numbers describe two differentpoints of the plane.

CONCEPT CHECKIs the graph of the point in the same location as the graph of the point Explain.11, -52?1-5, 12



Solut ion Point (5, 3) lies in quadrant I.Point lies in quadrant II.Point lies in quadrant III.Point lies in quadrant IV.

Points (0, 0), (0, 2), and lie on axes, so they are not in any quadrant.a0, -5 12b1-5, 02,

11, -221-2, -421-5, 32

CLASSROOM EXAMPLEOn a single coordinate system, plot eachordered pair.

a. (4, 2) b.c. d.e. (0, 3) f. (3, 0)

g. h.

answer:

a -2 12

, 0b10, -42

1-5, 1212, -221-1, -32

112345

12345

2 3 4 512345

(0, 2)

(0, 0)

(1, 2)

(5, 3) (5, 3)

(2, 4)

(5, 0)x

y

(0, 5q)

The table gives the annual net sales for Wal-Mart Stores for the years shown. (Source:Wal-Mart Stores, Inc.)

E X A M P L E 1On a single coordinate system, plot each ordered pair. State in which quadrant, if any,each point lies.

a. (5, 3) b. c. d.

e. (0, 0) f. (0, 2) g. h. a0, -5 12b1-5, 02

11, -221-2, -421-5, 32

From Example 1, notice that the y-coordinate of any point on the x-axis is 0. Forexample, the point lies on the x-axis. Also, the x-coordinate of any point on they-axis is 0. For example, the point (0, 2) lies on the y-axis.

CONCEPT CHECKFor each description of a point in the rectangular coordinate system, write an ordered pair that represents it.

a. Point A is located three units to the left of the y-axis and five units above the x-axis.b. Point B is located six units below the origin.

2 Data that can be represented as an ordered pair is called paired data. Manytypes of data collected from the real world are paired data. For instance, the annualmeasurement of a child’s height can be written as an ordered pair of the form (year,height in inches) and is paired data. The graph of paired data as points in the rectangu-lar coordinate system is called a scatter diagram. Scatter diagrams can be used to lookfor patterns and trends in paired data.

1-5, 02

E X A M P L E 2Concept Check Answer:a. b. 10, -621-3, 52

y

x

(0, 4)

(3, 0)

(1, 3)

(5, 1)

(2, 2)

(0, 3)(4, 2)

(2q, 0)


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Wal-Mart Net SalesYear (in billions of dollars)

1997 105

1998 118

1999 138

2000 165

2001 191

2002 218

2003 245

CLASSROOM EXAMPLEThe table gives the number of tornadoesthat have occurred in the United Statesfor the years shown. (Source: Storm Pre-diction Center, National Weather Service)

Year Tornadoes

1995 12341996 11731997 11481998 14241999 13432000 997

Create a scatter diagram of the paireddata.answer:

a. Write this paired data as a set of ordered pairs of the form (year, sales in billionsof dollars).

b. Create a scatter diagram of the paired data.c. What trend in the paired data does the scatter diagram show?

Solut ion a. The ordered pairs are (1997, 105), (1998, 118), (1999, 138), (2000, 165), (2001, 191),(2002, 218), and (2003, 245).

b. We begin by plotting the ordered pairs. Because the x-coordinate in each orderedpair is a year, we label the x-axis “Year” and mark the horizontal axis with theyears given. Then we label the y-axis or vertical axis “Net Sales (in billions of dol-lars).” In this case it is convenient to mark the vertical axis in multiples of 20. Sinceno net sale is less than 100, we use the notation to skip to 100, then proceed bymultiples of 20.

Net

Sal

es(i

n bi

llion

s of

dol

lars

)

1997 1998 1999Year2000 2001

220

200

180

160

140

120

100

240

260

2002 2003

MARTWAL ®

c. The scatter diagram shows that Wal-Mart net sales steadily increased over the years1997–2003.

3 Let’s see how we can use ordered pairs to record solutions of equations con-taining two variables. An equation in one variable such as has one solution,which is 4: the number 4 is the value of the variable x that makes the equation true.

An equation in two variables, such as has solutions consisting of twovalues, one for x and one for y. For example, and is a solution of

because, if x is replaced with 3 and y with 2, we get a true statement.2x + y = 8y = 2x = 3

2x + y = 8,

x + 1 = 5

1200

1400

1100

1000

1300

1500

900

Year

Num

ber

of T

orna

does

0

U.S. Tornadoes

1995 1996 1998 20001997 1999



CLASSROOM EXAMPLEDetermine whether each ordered pair is asolution of the equation a. (0, 12) b.answer: a. no b. yes

11, -923x - y = 12.

Complete the following ordered pair solutions for the equation

a. (0, ) b. ( , 6) c. 1-1, 23x + y = 12.

True.

The solution and can be written as (3, 2), an ordered pair of numbers. Thefirst number, 3, is the x-value and the second number, 2, is the y-value.

In general, an ordered pair is a solution of an equation in two variables if replac-ing the variables by the values of the ordered pair results in a true statement.

y = 2x = 3

8 = 8

2132 + 2 = 8

2x + y = 8

Determine whether each ordered pair is a solution of the equation

a. (6, 0) b. (0, 3) c. a1, - 52b

x - 2y = 6.

Solut ion a. Let and in the equation

Replace x with 6 and y with 0.

Simplify.

True.

(6, 0) is a solution, since is a true statement.

b. Let and

Replace x with 0 and y with 3.

False.

(0, 3) is not a solution, since is a false statement.

c. Let and in the equation.

Replace x with 1 and y with

True.

is a solution, since is a true statement.

4 If one value of an ordered pair solution of an equation is known, the othervalue can be determined. To find the unknown value, replace one variable in the equa-tion by its known value. Doing so results in an equation with just one variable that canbe solved for the variable using the methods of Chapter 2.

6 = 6A1, - 52 B

6 = 6

1 + 5 = 6

52. 1 - 2a -

52b = 6

x - 2y = 6

y = - 52x = 1

-6 = 6

-6 = 6

0 - 6 = 6

0 - 2132 = 6

x - 2y = 6

y = 3.x = 0

6 = 6

6 = 6

6 - 0 = 6

6 - 2102 = 6

x - 2y = 6

x - 2y = 6.y = 0x = 6

E X A M P L E 3

E X A M P L E 4


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Solut ion a. In the ordered pair (0, ), the x-value is 0. Let in the equation and solve for y.

Replace x with 0.

The completed ordered pair is (0, 12).

b. In the ordered pair ( , 6), the y-value is 6. Let in the equation and solve for x.

Replace y with 6.



The ordered pair is (2, 6).

c. In the ordered pair the x-value is Let in the equation and solvefor y.

Replace x with


The ordered pair is

Solutions of equations in two variables can also be recorded in a table of values,as shown in the next example.

1-1, 152. y = 15

-3 + y = 12

1. 31-12 + y = 12

3x + y = 12

x = -1-1.1-1, 2,

x = 2

3x = 6

3x + 6 = 12

3x + y = 12

y = 6

y = 12

0 + y = 12

3102 + y = 12

3x + y = 12

x = 0

CLASSROOM EXAMPLEComplete each ordered pair so that it is asolution to the equation

a. (0, ) b. ( , 3) c.answer:a. (0, 4) b. (2, 3) c. 1-4, 62

1-4, 2x + 2y = 8.

E X A M P L E 5

Complete the table for the equation y = 3x.

x y

a.b. 0c. -9

-1

CLASSROOM EXAMPLEComplete the table for the equation

x y

a.b. 0c. 10answer:

x ya. 6b. 0 0c. 10-5

-3

-3

y = -2x.

Solut ion a. Replace x with in the equation and solve for y.

Let

The ordered pair is

b. Replace y with 0 in the equation and solve for x.

Let


The ordered pair is (0, 0).

0 = x

y 0. 0 = 3x

y = 3x

1-1, -32. y = -3

x 1. y = 31-12 y = 3x

-1




Complete the table for the equation y = 3.

x y

0

-5

-2

Solut ion The equation is the same as Replace x with andwe have or Notice that no matter what valuewe replace x by, y always equals 3. The completed table is shown on the right.

By now, you have noticed that equations in two variables often have more thanone solution. We discuss this more in the next section.

A table showing ordered pair solutions may be written vertically or horizontallyas shown in the next example.

y = 3.01-22 + y = 3-20x + y = 3.y = 3

CLASSROOM EXAMPLEComplete the table for the equation

x y

04

answer:

x y

55 05 4

-2

-2

x = 5.

x y

3

0 3

3-5

-2

FINDING THE VALUE OF A COMPUTER

A computer was recently purchased for a small business for $2000. The business man-ager predicts that the computer will be used for 5 years and the value in dollars y of thecomputer in x years is Complete the table.y = -300x + 2000.

Solut ion

x 0 1 2 3 4 5

y

To find the value of y when x is 0, replace x with 0 in the equation. We use this sameprocedure to find y when x is 1 and when x is 2.

When When When

We have the ordered pairs (0, 2000), (1, 1700), and (2, 1400). This means that in 0 yearsthe value of the computer is $2000, in 1 year the value of the computer is $1700, and in 2years the value is $1400. To complete the table of values, we continue the procedure for

and x = 5.x = 4,x = 3,

y = 1400y = 1700y = 2000

y = -600 + 2000y = -300 + 2000y = 0 + 2000

y = -300 # 2 + 2000y = -300 # 1 + 2000y = -300 # 0 + 2000

y = -300x + 2000y = -300x + 2000y = -300x + 2000

x 2,x 1,x 0,

x y

0 0

-9-3

-3-1

c. Replace y with in the equation and solve for x.

Let


The ordered pair is The completed table is shown to the left.1-3, -92. -3 = x

y 9. -9 = 3x

y = 3x

-9

E X A M P L E 6

E X A M P L E 7


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When When When

The completed table is

y = 500y = 800y = 1100

y = -1500 + 2000y = -1200 + 2000y = -900 + 2000

y = -300 # 5 + 2000y = -300 # 4 + 2000y = -300 # 3 + 2000

y = -300x + 2000y = -300x + 2000y = -300x + 2000

x 5,x 4,x 3,CLASSROOM EXAMPLEA company purchased a fax machine for$400 that will be used for 7 years. Thevalue in dollars y of the machine in xyears is Completethe table.

x 1 2 3 4 5 6 7

y

answer:

x 1 2 3 4 5 6 7

y 350 300 250 200 150 100 50

y = -50x + 400.

x 0 1 2 3 4 5

y 2000 1700 1400 1100 800 500

The ordered pair solutions recorded in the completed table for the example above aregraphed below. Notice that the graph gives a visual picture of the decrease in value ofthe computer.

2000

1800

1600

1400

1200

1000

800

600

400

200

0

Val

ue o

f Com

pute

r(d

olla

rs)

3 540 1 2Time (years)

x y

0 20001 17002 14003 11004 8005 500

Suppose you own asmall mail-order busi-ness. One of your em-ployees brings you thisgraph showing the ex-pected retail revenuefrom on-line shopping

sales on the Internet. Should you consider expanding yourbusiness to include taking orders over the Internet? Ex-plain your reasoning. What other factors would you want toconsider?

Rev

enue

from

Int

erne

t Sal

es(i

n bi

llion

s of

dol

lars

)

Expected Retail Revenuefrom On-line Shopping

Year1999

2.4

4.8

8.0

12.2

17.4

2000 2001 2002 2003

20

18

1614

12

10

02

4

6

8

Source: Forrester Research Inc.

Computer Value



MENTAL MATH

STUDY SKILLS REMINDERAre You Satisfied with Your Performance on a Particular Quiz or Exam?

If not, analyze your quiz or exam like you would a good mysterynovel. Look for common themes in your errors.

Were most of your errors a result of• Carelessness? If your errors were careless, did you turn in your

work before the allotted time expired? If so, resolve to use theentire time allotted next time. Any extra time can be spentchecking your work.

• Running out of time? If so, make a point to better manageyour time on your next exam. A few suggestions are to workany questions that you are unsure of last and to check yourwork after all questions have been answered.

• Not understanding a concept? If so, review that concept andcorrect your work. Remember next time to make sure that allconcepts on a quiz or exam are understood before the exam.

Give two ordered pair solutions for each of the following linear equations.

1. 2. 3. 4. y = -2x = 3x + y = 6x + y = 10

1. answers may vary: Ex. (5, 5), (7, 3) 2. answers may vary: Ex. (0, 6), (6, 0) 3. answers mayvary: Ex. (3, 5), (3, 0) 4. answers may vary: Ex. 10, -22, 11, -22



O b j e c t i v e s

1 Identify linear equations.

2 Graph a linear equation by finding and plotting ordered pair solutions.

1 In the previous section, we found that equations in two variables may havemore than one solution. For example, both (6, 0) and are solutions of the equa-tion In fact, this equation has an infinite number of solutions. Other solu-tions include and If we graph these solutions, notice that apattern appears.

1-2, -42.10, -32, 14, -12,x - 2y = 6.

12, -22

3.2 G R A P H I N G L I N E A R E Q UAT I O N S

112345

12345

2 3 6 74 5123

(2, 4)

(6, 0)

(4, 1)(2, 2)

(0, 3)

x

y

TEACHING TIP Group ActivityBegin this section by graphing 4linear equations on 4 sheets ofgraph paper using sticky dots.Draw axes on each sheet of paperand label with one of the followingequations:

and

Divide the class into 9 groups. As-sign each group an integer from to 4 to use as their x-value. Haveeach group find the y-value foreach equation which correspondsto their x-value and plot each pointusing a sticky dot. When the graphsare completed, ask what similari-ties and differences they notice inthe graphs.

-4

y = -4.x + y = 2, 2x - y = 6, x = 5,

These solutions all appear to lie on the same line, which has been filled in below. It canbe shown that every ordered pair solution of the equation corresponds to a point onthis line, and every point on this line corresponds to an ordered pair solution. Thus, wesay that this line is the graph of the equation x - 2y = 6.

112345

12345

2 3 6 74 5123 x

y

x 2y 6



GRAPHING L INEAR EQUATIONS SECTION 3.2 177

The equation is called a linear equation in two variables and the graph ofevery linear equation in two variables is a line.

x - 2y = 6

The form is called standard form.Ax + By = C

Linear Equation in Two VariablesA linear equation in two variables is an equation that can be written in the form

where A, B, and C are real numbers and A and B are not both 0. The graphof a linear equation in two variables is a straight line.

Ax + By = C

Examples of Linear Equations in Two Variables

Before we graph linear equations in two variables, let’s practice identifying theseequations.

y = 7y =13

x + 2-2x = 7y2x + y = 8

H e l p f u l H i n tNotice in the form the understood exponent on both x and y is 1.Ax + By = C,

E X A M P L E 1Identify the linear equations in two variables.

a. b. c. d. x = 5x + y2 = 9y = -2xx - 1.5y = -1.6

Solut ion a. This is a linear equation in two variables because it is written in the formwith and

b. This is a linear equation in two variables because it can be written in the form

Add 2x to both sides.

c. This is not a linear equation in two variables because y is squared.

d. This is a linear equation in two variables because it can be written in the form

Add

2 From geometry, we know that a straight line is determined by just two points.Graphing a linear equation in two variables, then, requires that we find just two of itsinfinitely many solutions. Once we do so, we plot the solution points and draw the lineconnecting the points. Usually, we find a third solution as well, as a check.

0 # y . x + 0y = 5

x = 5

Ax + By = C.

2x + y = 0 y = -2x

Ax + By = C.

C = -1.6.A = 1, B = -1.5,Ax + By = C

TEACHING TIPFor contrast, you may want to givesome examples of equations in twovariables that are not linear. For in-stance,

x2 = 6y + 4y = 3x

y + 9 = 1x

3x -1y

= 5

CLASSROOM EXAMPLEIdentify the linear equation in two variables.

a. b.

c.

answer: a. no b. yes c. yes

12

x - 3.4y = 1

y = 7y2 = x + 1


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Graph the linear equation 2x + y = 5.

Solut ion

CLASSROOM EXAMPLE

Graph:

answer:

x + 3y = 6

Find three ordered pair solutions of To do this, choose a value for onevariable, x or y, and solve for the other variable. For example, let Then

becomes

Replace x with 1.

Multiply.


Since when the ordered pair (1, 3) is a solution of Next, let

Replace x with 0.

The ordered pair (0, 5) is a second solution.

y = 5

0 + y = 5

2102 + y = 5

2x + y = 5

x = 0.2x + y = 5.x = 1,y = 3

y = 3

2 + y = 5

2112 + y = 5

2x + y = 5

2x + y = 5x = 1.

2x + y = 5.

The two solutions found so far allow us to draw the straight line that is the graphof all solutions of However, we find a third ordered pair as a check. Let

Replace y with



The third solution is These three ordered pair solutions are listed in table formas shown. The graph of is the line through the three points.2x + y = 5

13, -12. x = 3

2x = 6

2x - 1 = 5

1. 2x + 1-12 = 5

2x + y = 5

y = -1.2x + y = 5.

E X A M P L E 2

(6, 0)(0, 2)

(3, 1)x

y

8

8

11234

1234

65

2 3 4 512345

(0, 5)

(1, 3)

(3, 1)x

y

2x y 5

x y

1 30 53 -1

TEACHING TIPPoint out the arrowheads on thegraph in Example 2. See if the stu-dents understand their meaning.Remind them that when they aregraphing an equation, they areillustrating all the solutions of theequation.


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H e l p f u l H i n tAll three points should fall on the same straight line. If not, check your orderedpair solutions for a mistake.

E X A M P L E 3Graph the linear equation -5x + 3y = 15.

CLASSROOM EXAMPLEGraph: answer:

-2x + 4y = 8

Solut ion Find three ordered pair solutions of

Let Let Let

The ordered pairs are (0, 5), and The graph of is theline through the three points.

-5x + 3y = 151-2, 532.1-3, 02, y = 5

3 x = -3 y = 5

3y = 5 -5x = 15 3y = 15

10 + 3y = 15 -5x + 0 = 15 0 + 3y = 15

-51-22 + 3y = 15 -5x + 3 # 0 = 15 -5 # 0 + 3y = 15

-5x + 3y = 15 -5x + 3y = 15 -5x + 3y = 15

x 2.y 0.x 0.

-5x + 3y = 15.

11234

1234

65

2 3 4 512345

(0, 5)

(3, 0)x

y

5x 3y 15

(2, f)

x y

0 50

53 = 1

23-2

-3

y

x

(2, 3)

(0, 2)

(2, 1)

E X A M P L E 4Graph the linear equation y = 3x.

Solut ion To graph this linear equation, we find three ordered pair solutions. Since this equationis solved for y, choose three x values.

Next, graph the ordered pair solutions listed in the table above and draw a line throughthe plotted points as shown on the next page. The line is the graph of Everypoint on the graph represents an ordered pair solution of the equation and every or-dered pair solution is a point on this line.

y = 3x.

If x = -1, y = 3 # -1 = -3.

If x = 0, y = 3 # 0 = 0.

If x = 2, y = 3 # 2 = 6.x y

2 60 0

-3-1


y = 2x

(0, 0)(2, 4)

(3, 6)

x

y

8

8


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11234

123

67

45

2 3 4 512345

(2, 6)

(1, 3)

(0, 0)x

y

y 3x

Graph the linear equation y = - 13 x.

Solut ion Find three ordered pair solutions, graph the solutions, and draw a line through the plot-ted solutions. To avoid fractions, choose x values that are multiples of 3 to substitute inthe equation. When a multiple of 3 is multiplied by the result is an integer. See thecalculations shown to the right of the table below.

If x = -3, then y = -13

# -3 = 1.

If x = 0, then y = -13

# 0 = 0.

If x = 6, then y = -13

# 6 = -2.

- 13,CLASSROOM EXAMPLE

Graph: answer:

y = - 12 x

x y

6

0 0

1-3

-2

112345

12345

2 3 6 74 51234(6, 2)

(0, 0)(3, 1)

x

y

y ax

E X A M P L E 5

Let’s compare the graphs in Examples 4 and 5. The graph of tilts upward(as we follow the line from left to right) and the graph of tilts downward (aswe follow the line from left to right). Also notice that both lines go through the originor that (0, 0) is an ordered pair solution of both equations. In general, the graphs of

and are of the form where m is a constant. The graph of anequation in this form always goes through the origin (0, 0) because when x is 0,becomes y = m # 0 = 0.

y = mxy = mxy = -

13 xy = 3x

y = - 13 x

y = 3x

(0, 0)

(6, 3)

(4, 2)

x

y

8

8


x y

0 61 9

-3-3


Graph the linear equation and compare this graph with the graph ofin Example 4.y = 3x

y = 3x + 6

E X A M P L E 6

Solut ion Find ordered pair solutions, graph the solutions, and draw a line through the plotted so-lutions. We choose x values and substitute in the equation

If x = 1, then y = 3112 + 6 = 9.

If x = 0, then y = 3102 + 6 = 6.

If x = -3, then y = 31-32 + 6 = -3.

y = 3x + 6.

11234

6789

10

12345

2 3 4 512345

(3, 3)

6 units up

(0, 6) (1, 9)

x

y

y 3xy 3x 6

CLASSROOM EXAMPLEGraph the linear equation and compare this graph with the graphof answer:

Same as the graph of exceptthat the graph of is moved3 units upward.

y = 2x + 3y = 2x

y = 2x.

y = 2x + 3

The most startling similarity is that both graphs appear to have the same upward tilt aswe move from left to right. Also, the graph of crosses the y-axis at the origin,while the graph of crosses the y-axis at 6. In fact, the graph of is the same as the graph of moved vertically upward 6 units.

Notice that the graph of crosses the y-axis at 6.This happens becausewhen becomes The graph contains the point (0, 6),which is on the y-axis.

In general, if a linear equation in two variables is solved for y, we say that it iswritten in the form The graph of this equation contains the point (0, b)because when is y = m # 0 + b = b.x = 0, y = mx + b

y = mx + b.

y = 3 # 0 + 6 = 6.y = 3x + 6x = 0,y = 3x + 6

y = 3xy = 3x + 6y = 3x + 6

y = 3x

(0, 3)(2, 7)

(2, 1) x

y

8

8

E X A M P L E 7ESTIMATING THE NUMBER OF MEDICAL ASSISTANTS

One of the occupations expected to have the most growth in the next few years is med-ical assistant.The number of people y (in thousands) employed as medical assistants inthe United States can be estimated by the linear equation where x isthe number of years after the year 1995. (Source: based on data from the Bureau ofLabor Statistics)

Graph the equation and use the graph to predict the number of medical assis-tants in the year 2010.

y = 31.8x + 180,

The graph of crosses the y-axis at (0, b).y = mx + b

We will review this again in Section 3.5.Linear equations are often used to model real data as seen in the next example.


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To use the graph to predict the number of medical assistants in the year 2010, weneed to find the y-coordinate that corresponds to (15 years after 1995 is theyear 2010.) To do so, find 15 on the x-axis. Move vertically upward to the graphed lineand then horizontally to the left. We approximate the number on the y-axis to be 655.Thus in the year 2010, we predict that there will be 655 thousand medical assistants.(The actual value, using 15 for x, is 657.)

x = 15.

Num

ber

of M

edic

alA

ssis

tant

s (i

n th

ousa

nds)

Years after 19954 6 8 9 107 11 13 14 151221 3

350300

250200

150

0

600550

700650

500450

400

5

Solut ion

CLASSROOM EXAMPLEUse the graph in Example 7 to predict thenumber of medical assistants in 2004.answer: 465 thousand

x y

0 1802 243.67 402.6

To graph choose x-values and substitute in the equation.

If x = 7, then y = 31.8172 + 180 = 402.6.

If x = 2, then y = 31.8122 + 180 = 243.6.

If x = 0, then y = 31.8102 + 180 = 180.

y = 31.8x + 180,

In this section, we begin an optional study of graphing calculators and graphing soft-ware packages for computers. These graphers use the same point plotting techniquethat was introduced in this section. The advantage of this graphing technology is, ofcourse, that graphing calculators and computers can find and plot ordered pair solu-tions much faster than we can. Note, however, that the features described in theseboxes may not be available on all graphing calculators.

The rectangular screen where a portion of the rectangular coordinate systemis displayed is called a window. We call it a standard window for graphing when boththe x- and y-axes show coordinates between and 10. This information is oftendisplayed in the window menu on a graphing calculator as

The scale on the x-axis is one unit per tick mark.

The scale on the y-axis is one unit per tick mark.

To use a graphing calculator to graph the equation press the

key and enter the keystrokes The top row should now read 2 x + 3 . Y= y = 2x + 3,

Yscl = 1 Ymax = 10 Ymin = -10

Xscl = 1 Xmax = 10 Xmin = -10

-10

Graphing Calculator Explorations



Next press the key, and the display should look like this:

Use a standard window and graph the following linear equations. (Unless otherwisestated, use a standard window when graphing.)

1. 2.

3. 4.

5. 6. y =29

x -223

y = - 310

x +325

y = -1.3x + 5.2y = 2.5x - 7.9y = -x + 5y = -3x + 7

GRAPH Y1 = 2x + 3.

10

10

10

10


INTERCEPTS SECTION 3.3 185

11

6789

2345

1

2 3 6 74 5123

(2, 0)

(0, 8)

x

y

y 4x 8

x-intercept

y-intercept

O b j e c t i v e s

1 Identify intercepts of a graph.

2 Graph a linear equation by finding and plotting intercepts.

3 Identify and graph vertical and horizontal lines.

1 In this section, we graph linear equations in two variables by identifying inter-cepts. For example, the graph of is shown below. Notice that this graphcrosses the y-axis at the point This point is called the y-intercept. Likewise, thegraph crosses the x-axis at (2, 0), and this point is called the x-intercept.

10, -82.y = 4x - 8

TEACHING TIPRemind students that all points onthe x-axis have a y-value of 0, andall points on the y-axis have an x-value of 0. Also, the point (0, 0)lies on both axes.

3.3 I N T E R C E P T S

59.–60. See graphing answer section.




H e l p f u l H i n tIf a graph crosses the x-axis at and the y-axis at (0, 7), then

x-intercept y-intercept

Notice that for the y-intercept, the x-value is 0 and for the x-intercept, the y-value is 0.

Note: Sometimes in mathematics, you may see just the number 7 stated as they-intercept, and stated as the x-intercept.-3

qq1-3, 02(1)1*

10, 72()*

1-3, 02

Identify the x- and y-intercepts.

a. b.

E X A M P L E 1

c. d.

e.

2

4

2

5

1

4

1

3

3

5

2 424 15 3 513 x

y

2

4

2

5

1

4

1

3

3

5

2 424 15 3 513 x

y

2

4

2

5

1

4

1

3

3

5

2 424 15 3 513 x

y

2

4

2

5

1

4

1

3

3

5

2 424 15 3 513 x

y

2

4

2

5

1

4

1

3

3

5

2 424 15 3 513 x

y

H e l p f u l H i n tNotice that any time (0, 0) is a point of a graph,then it is an x-interceptand a y-intercept.

Solut ion

TEACHING TIPSome questions you may want toask your students: Can the same point ever be the

x-intercept point and the y-intercept point of a graph?

Excluding the line whatis the maximum number of x-intercepts a line will have?

Will a line always have at leastone y-intercept?

y = 0,

a. The graph crosses the x-axis at so the x-intercept is The graph crossesthe y-axis at 2, so the y-intercept is (0, 2).

b. The graph crosses the x-axis at and so the x-intercepts are andThe graph crosses the y-axis at 1, so the y-intercept is (0, 1).1-1, 02. 1-4, 02-1,-4

1-3, 02.-3,


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CLASSROOM EXAMPLEIdentify the x- and y-intercepts.answer:

x-intercept: (2, 0); y-intercept: 10, -42

y

x

Finding x- and y-interceptsTo find the x-intercept, let and solve for x.

To find the y-intercept, let and solve for y.x = 0

y = 0

E X A M P L E 2Graph by finding and plotting intercepts.x - 3y = 6

Solut ion Let to find the x-intercept and let to find the y-intercept.

The x-intercept is (6, 0) and the y-intercept is We find a third ordered pairsolution to check our work. If we let then Plot the points (6, 0),and The graph of is the line drawn through these points, as shown.x - 3y = 613, -12. 10, -22,x = 3.y = -1,

10, -22. y = -2 x = 6

-3y = 6 x - 0 = 6 0 - 3y = 6 x - 3102 = 6 x - 3y = 6 x - 3y = 6

Let x = 0 Let y = 0

x = 0y = 0

c. The x-intercept and the y-intercept are both (0, 0).

d. The x-intercept is (2, 0). There is no y-intercept.

e. The x-intercepts are and (3, 0).The y-intercepts are and (0, 2).

2 Given the equation of a line, intercepts are usually easy to find since one coor-dinate is 0.

One way to find the y-intercept of a line, given its equation, is to let sincea point on the y-axis has an x-coordinate of 0.To find the x-intercept of a line, let since a point on the x-axis has a y-coordinate of 0.

y = 0,x = 0,

10, -12,1-1, 02

x y

6 003 -1

-2

112345

12345

2 3 6 74 5123

(0, 2)(3, 1)

(6, 0)

x

y

x 3y 6

CLASSROOM EXAMPLEGraph by finding and plotting its intercepts.answer:

2x - y = 4

y

x(2, 0)

(0, 4)

Graph by plotting intercepts.x = -2y

E X A M P L E 3

Solut ion Let to find the x-intercept and to find the y-intercept.

0 = y x = 00 = -2y x = -2102 x = -2y x = -2y

Let x = 0 Let y = 0

x = 0y = 0


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Both the x-intercept and y-intercept are (0, 0). In other words, when then which gives the ordered pair (0, 0).Also, when then which gives the same or-dered pair (0, 0).This happens when the graph passes through the origin. Since two pointsare needed to determine a line, we must find at least one more ordered pair that satisfies

Let to find a second ordered pair solution and let as a checkpoint.

The ordered pairs are (0, 0), and Plot these points to graph x = -2y.1-2, 12.12, -12,x = -2 x = 2x = -2112 x = -21-12

Let y = 1 Let y = -1

y = 1y = -1x = -2y.

x = 0,y = 0,y = 0,x = 0,


CLASSROOM EXAMPLEGraph by finding and plotting itsintercepts.answer:

y = 3x

112345

12345

2 3 4 512345

(2, 1)

(2, 1) (0, 0)x

y

x 2y

x y

0 02

1-2-1

Graph 4x = 3y - 9.

Solut ion Find the x- and y-intercepts, and then choose to find a third checkpoint.

Solve for x. Solve for y. Solve for y.

or

or

The ordered pairs are (0, 3), and The equation isgraphed as follows.

3 The equation where c is a real number constant, is a linear equation intwo variables because it can be written in the form The graph of this equa-tion is a vertical line as shown in the next example.

x + 0y = c.x = c,

4x = 3y - 9A2, 5 23 B .A -2

14, 0 B ,

y = 5 23

173

= y

17 = 3y3 = y-2 14

x = - 94

8 = 3y - 99 = 3y4x = -94122 = 3y - 94 # 0 = 3y - 94x = 3102 - 9

Let x = 2Let x = 0Let y = 0

x = 2

CLASSROOM EXAMPLEGraph answer:

3x = -5y + 10.

y

x(0, 0)

E X A M P L E 4

y

x(0, 2) (3a, 0)

67

1123

12345

2 3 4 512345

(0, 3)

x

y

4x 3y 9 (2, 5s)

(2~, 0 )

x y

00 32 5

23

-2 14


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E X A M P L E 5Graph x = 2.

Solut ion The equation can be written as For any y-value chosen, notice thatx is 2. No other value for x satisfies Any ordered pair whose x-coordinateis 2 is a solution of We will use the ordered pair solutions (2, 3), (2, 0), and

to graph x = 2.12, -32 x + 0y = 2.x + 0y = 2.

x + 0y = 2.x = 2

112345

12345

2 3 4 512345

(2, 3)

(2, 0)

(2, 3)

x

y

x 2x y

2 32 02 -3


x = -3.y

x(3, 0)

Vertical LinesThe graph of where c is a real number, is a vertical line with x-intercept (c, 0).

x = c,

The graph is a vertical line with x-intercept (2, 0). Note that this graph has no y-interceptbecause x is never 0.

x

y

x c

(c, 0)

The equation can be written as For any x-value chosen, y is If we choose 4, 1, and as x-values, the ordered pair solutions are and Use these ordered pairs to graph The graph is a horizontal linewith y-intercept and no x-intercept.10, -32 y = -3.1-2, -32. 14, -32, 11, -32,-2

-3.0x + y = -3.y = -3

Graph y = -3.

Solut ion

E X A M P L E 6

112345

12345

2 3 4 512345

(2, 3) (1, 3)

(4, 3)

x

y

y 3

x y

41

-3-2-3-3


y = 4.y

x

(0, 4)


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Horizontal LinesThe graph of where c is a real number, is a horizontal line with y-intercept (0, c).

y = c,

y

y c

(0, c)x

You may have noticed that to use the key on a grapher to graph an equation,the equation must be solved for y. For example, to graph we solve thisequation for y.



Simplify.

To graph or press the key and enter

Graph each linear equation.

1. 2.

3. 4.

5. 6. 5.9x - 0.8y = -10.4-2.2x + 6.8y = 15.5-4x + 6y = 123x + 7y = 21-2.61y = xx = 3.78y

Y1 = - 23

x +73

Y= y = - 23

x +73

,2x + 3y = 7

y = - 23

x +73

3y

3= -

2x

3+

73

3y = -2x + 7

2x + 3y = 7

2x + 3y = 7, Y=


10

10

10

10

2x 3y 7 or y xs g



MENTAL MATHAnswer the following true or false.

1. The graph of is a horizontal line. false

2. All lines have an x-intercept and a y-intercept. false

3. The graph of contains the point (0, 0). true

4. The graph of has an x-intercept of (5, 0) and a y-intercept of (0, 5). true

5. The graph of contains the point (5, 1). false

6. The graph of is a horizontal line. truey = 5

y = 5x

x + y = 5

y = 4x

x = 2


SLOPE AND RATE OF CHANGE SECTION 3.4 193

O b j e c t i v e s

1 Find the slope of a line given two points of the line.

2 Find the slopes of horizontal and vertical lines.

3 Compare the slopes of parallel and perpendicular lines.

4 Solve applications of slope.

1 Thus far, much of this chapter has been devoted to graphing lines. You haveprobably noticed by now that a key feature of a line is its slant or steepness. In mathe-matics, the slant or steepness of a line is formally known as its slope. We measure theslope of a line by the ratio of vertical change to the corresponding horizontal change aswe move along the line.

On the line below, for example, suppose that we begin at the point (1, 2) and moveto the point (4, 6). The vertical change is the change in y-coordinates: or 4 units.The corresponding horizontal change is the change in x-coordinates:The ratio of these changes is

slope =change in y 1vertical change2

change in x 1horizontal change2 =43

4 - 1 = 3 units.6 - 2

3.4 S L O P E A N D R AT E O F C H A N G E

TEACHING TIPBegin this lesson by having stu-dents compare and contrast thegraphs of three lines with the samey-intercept and different slopes.Ask how we could describe theirdifferences. Possible lines to use

y = -4x + 3

y = 3

y = 2x + 3

6 7

67

1123

12345

2 3 4 5123

(4, 6)

Horizontal change is4 1 3 units

Vertical change is6 2 4 units

(1, 2)

x

y




If we let be then and Also, let be sothat and Then, by the definition of slope,

The slope of the line is - 83

.

=-83

= - 83

=-3 - 5

2 - 1-12

m =y2 - y1

x2 - x1

y2 = -3.x2 = 212, -321x2, y22y1 = 5.x1 = -11-1, 52,1x1, y12

Find the slope of the line through and Graph the line.12, -32.1-1, 52Solut ion

CLASSROOM EXAMPLEFind the slope of the line through

and Graph the line.answer: m = -

23

14, -12.1-2, 32

Slope of a LineThe slope m of the line containing the points and is given by

m =riserun

=change in y

change in x=

y2 - y1

x2 - x1, as long as x2 Z x1

1x2, y221x1, y12

E X A M P L E 1

y

x

(2, 3)

(4, 1)

The slope of this line, then, is for every 4 units of change in y-coordinates, there is acorresponding change of 3 units in x-coordinates.

43:

H e l p f u l H i n tIt makes no difference what two points of a line are chosen to find its slope.Theslope of a line is the same everywhere on the line.

To find the slope of a line, then, choose two points of the line. Label the two x-co-

ordinates of two points, and (read “x sub one” and “x sub two”), and label the cor-responding y-coordinates and

The vertical change or rise between these points is the difference in the y-coordi-nates: The horizontal change or run between the points is the difference of thex-coordinates: The slope of the line is the ratio of to and we

traditionally use the letter m to denote slope .m =y2 - y1

x2 - x1

x2 - x1,y2 - y1x2 - x1.y2 - y1.

y2.y1

x2x1

x2

y1

y2

x1

x2 x1 horizontalchange or run

y2 y1 verticalchange or rise

x

y

(x1, y1)

(x2, y2)

6

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(2, 3)

(1, 5)

x

y

run: 3

rise: 8


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TEACHING TIPFor Example 1, have students placetheir finger on and make avertical and horizontal move to

Then have them writetheir move as a ratio of the verticalmove to the horizontal move usingthe following convention: upwardmoves are positive, downwardmoves are negative, rightwardmoves are positive, leftward movesare negative. Now have them puttheir finger on and move to

using a vertical and hori-zontal move. Have them write thismove as a ratio. How did the ratioschange? Are the ratios equal?

1-1, 5212, -32

12, -32.1-1, 52


The order in which the x- and y- values are used must be the same.

m =5 - 6

3 - 1-22 =-15

= - 15

m = 32

Find the slope of the line through and (2, 4). Graph the line.1-1, -22Solut ion y-value

corresponding x-value

The slope is the same if we begin with the other y-value.

y-value

corresponding x-value

The slope is 2.

CONCEPT CHECKWhat is wrong with the following slope calculation for the points (3, 5) and

Notice that the slope of the line in Example 1 is negative, whereas the slope of theline in Example 2 is positive. Let your eye follow the line with negative slope from left

m =5 - 6

-2 - 3=

-1-5

=15

1-2, 62?

˚

m =4 - 1-222 - 1-12 =

63

= 2

Ω

˚m =

-2 - 4-1 - 2

=-6-3

= 2

Ω

H e l p f u l H i n tWhen finding slope, it makes no difference which point is identified as and which is identified as Just remember that whatever y-value is firstin the numerator, its corresponding x-value is first in the denominator.Anotherway to calculate the slope in Example 1 is:

Same slope as found in Example 1.;m =y2 - y1

x2 - x1=

5 - 1-32-1 - 2

=8

-3 or -

83

1x2, y22.1x1, y12

E X A M P L E 2

CONCEPT CHECKThe points (4, 4), and (10, 13) all lie on the same line. Work with a partner and ver-ify that the slope is the same no matter which points are used to find slope.

10, -22,1-2, -52,

CLASSROOM EXAMPLEFind the slope of the line through

and (3, 5). Graph the line.

answer: m = 45

1-2, 12

y

x

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(2, 4)

(1, 2)

x

y




Find the slope of the line x = 5.

Solut ion Recall that the graph of is a vertical line with x-intercept (5, 0).

To find the slope, find two ordered pair solutions of Solutions of must have an x-value of 5. Let’s use points (5, 0) and (5, 4), which are on the line.

m =y2 - y1

x2 - x1=

4 - 05 - 5

=40

x = 5x = 5.

x = 5

CLASSROOM EXAMPLEFind the slope of answer: m = 0

y = 1.

112345

12345

2 3 4 512345 x

y

y 1

Any two points of a horizontal line will have the same y-values. This means thatthe y-values will always have a difference of 0 for all horizontal lines. Thus, all horizon-tal lines have a slope 0.

E X A M P L E 4

CLASSROOM EXAMPLEFind the slope of answer: slope is undefined

x = -2.

to right and notice that the line “goes down.” Following the line with positive slopefrom left to right, notice that the line “goes up.” This is true in general.

y

Negative slope

Goes down

x

y

Positive slope

Goes

up

x

TEACHING TIPMake sure that students under-stand they must follow a line fromleft to right to determine whetherthe slope is positive or negative.

H e l p f u l H i n tTo decide whether a line“goes up” or “goesdown”, always follow theline from left to right.

2 If a line tilts upward from left to right, its slope is positive. If a line tilts down-ward from left to right, its slope is negative. Let’s now find the slopes of two speciallines, horizontal and vertical lines.

E X A M P L E 3Find the slope of the line y = -1.

Solut ion Recall that is a horizontal line with y-intercept To find the slope, findtwo ordered pair solutions of Solutions of must have a y-value of Let’s use points and which are on the line.

The slope of the line is 0. The graph of is given below.y = -1y = -1

m =y2 - y1

x2 - x1=

-1 - 1-12-3 - 2

=0

-5= 0

1-3, -12,12, -12 -1.y = -1y = -1.10, -12.y = -1


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112345

12345

2 3 4 512345 x

y

x 5

H e l p f u l H i n tSlope of 0 and undefined slope are not the same. Vertical lines have undefinedslope or no slope, while horizontal lines have a slope of 0.

Since is undefined, we say the slope of the vertical line is undefined.

Any two points of a vertical line will have the same x-values. This means that thex-values will always have a difference of 0 for all vertical lines. Thus all vertical lineshave undefined slope.

x = 540

Here is a general review of slope.

Summary of Slope

Slope m of the line through and is given by the equation

m =y2 - y1

x2 - x1.

1x2, y221x1, y12

Upwardline

y

Positive slope: m 0

x

Downwardline

y

Negative slope: m 0

x

Zero slope: m 0

c

y

Horizontal liney c

x

Undefined slope or no slope

x

y

Vertical linex c

c

3 Two lines in the same plane are parallel if they do not intersect. Slopes of linescan help us determine whether lines are parallel. Parallel lines have the same steep-ness, so it follows that they have the same slope.



Is the line passing through the points and parallel to the line passingthrough the points (5, 4) and (9, 7)?

1-2, 321-6, 02

Solut ion To see if these lines are parallel, we find and compare slopes. The line passing throughthe points and has slope

m =3 - 0

-2 - 1-62 =34

1-2, 321-6, 02

H e l p f u l H i n tHere are examples of numbers that are negative (opposite) reciprocals.

Number Negative Reciprocal Their product is

or -5 # 15

= - 55

= -115

- 51

-5

23

# - 32

= - 66

= -1- 32

23

1.

H e l p f u l H i n tHere are a few important facts about vertical and horizontal lines.

• Two distinct vertical lines are parallel.

• Two distinct horizontal lines are parallel.

• A horizontal line and a vertical line are always perpendicular.

TEACHING TIPSome questions you might want toask your students:

Can parallel lines have the samey-intercept?

Can perpendicular lines havethe same y-intercept?

Do perpendicular lines alwayshave the same y-intercept?

E X A M P L E 5

Parallel LinesNonvertical parallel lines have the same slope.

x

y

How do the slopes of perpendicular lines compare? Two lines that intersect atright angles are said to be perpendicular. The product of the slopes of two perpendicu-lar lines is -1.

Perpendicular LinesIf the product of the slopes of two lines is the linesare perpendicular.

(Two nonvertical lines are perpendicular if the slopesof one is the negative reciprocal of the slope of theother.)

-1,

y

slope a

a1slope


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CLASSROOM EXAMPLEIs the line passing through and(7, 1) parallel to the line passing through(0, 8) and (10, 9)?answer: no

1-4, 02

Find the slope of a line perpendicular to the line passing through the points and (2, 2).

1-1, 72E X A M P L E 6

CLASSROOM EXAMPLEFind the slope of a line perpendicular tothe line through and answer: -

25

1-1, 42.1-5, -62

Solut ion First, let’s find the slope of the line through and (2, 2). This line has slope

The slope of every line perpendicular to the given line has a slope equal to the negativereciprocal of

negative reciprocal

The slope of a line perpendicular to the given line has slope

4 Slope can also be interpreted as a rate of change. In other words, slope tells ushow fast y is changing with respect to x. To see this, lets look at a few of the many real-world applications of slope. For example, the pitch of a roof, used by builders and archi-tects, is its slope. The pitch of the roof on the left is This means that the roofrises vertically 7 feet for every horizontal 10 feet. The rate of change for the roof is 7vertical feet (y) per 10 horizontal feet (x).

The grade of a road is its slope written as a percent. A 7% grade, as shown below,means that the road rises (or falls) 7 feet for every horizontal 100 feet. (Recall that

) Here, the slope of gives us the rate of change. The road rises (in our dia-gram) 7 vertical feet (y) for every 100 horizontal feet (x).

71007% = 7

100.

710 1rise

run2.

35

.

qq-

53 or - a -

35b =

35

m =2 - 7

2 - 1-12 =-53

1-1, 72

The line passing through the points (5, 4) and (9, 7) has slope

Since the slopes are the same, these lines are parallel.

m =7 - 49 - 5

=34

10 feet

7 feet

Î pitch

7 feet

100 feet

7% grade7100

E X A M P L E 7

FINDING THE GRADE OF A ROAD

At one part of the road to the summit of Pikes Peak, the road rises at a rate of 15 ver-tical feet for a horizontal distance of 250 feet. Find the grade of the road.

Solut ion Recall that the grade of a road is its slope written as a percent.

grade =riserun

=15250

= 0.06 = 6%

250 feet15 feet

The grade is 6%.

CLASSROOM EXAMPLEFind the grade of the road:

answer: 15%

20 feet

3 feet


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This means that the rate of change of a phone call is 11 cents per 1 minute or the costof the phone call increases 11 cents per minute.

m =81 - 485 - 2

=333

=111

centsminute

Suppose you own a house that you are try-ing to sell. You want a real estate agency tohandle the sale of your house. You begin byscanning your local newspaper to find likelycandidates, and spot the two ads shown.

Which real estate agency would you wantto check into first? Explain. What other fac-tors would you want to consider?

E X A M P L E 8FINDING THE SLOPE OF A LINE

The following graph shows the cost y (in cents) of an in-state long-distance telephonecall in Massachusetts where x is the length of the call in minutes. Find the slope of theline and attach the proper units for the rate of change.

Solut ion Use (2, 48) and (5, 81) to calculate slope.

Cos

t of C

all (

in c

ents

)

Length of Call (in minutes)4 6 x

y

21 3

(2, 48)

(5, 81)

60

70

50

40

90

100

80

30

20

10

5

CLASSROOM EXAMPLEThe points (3, 59) and (6, 92) also lie onthis graph for Example 8. Find the slopethrough these points.answer: m = 11


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It is possible to use a grapher to sketch the graph of more than one equation on thesame set of axes.This feature can be used to confirm our findings from Section 3.2when we learned that the graph of an equation written in the form

has a y-intercept of b. For example, graph the equations and

on the same set of axes. To do so, press the key and enter the

equations on the first three lines.

The screen should look like:

Notice that all three graphs appear to have the same positive slope. The graph of

is the graph of moved 7 units upward with a y-intercept of 7.

Also, the graph of is the graph of moved 4 units downward with

a y-intercept of

Graph the equations on the same set of axes. Describe the similarities and differencesin their graphs.

1.

2.

3.

4. y = - 34

x, y = - 34

x - 5, y = - 34

x + 6

y =14

x; y =14

x + 5, y =14

x - 8

y = -4.9x, y = -4.9x + 1, y = -4.9x + 8

y = 3.8x, y = 3.8x - 3, y = 3.8x + 9

-4.

y = 25 xy = 2

5 x - 4

y = 25 xy = 2

5 x + 7

Y3 = a25bx - 4

Y2 = a25bx + 7

Y1 = a25bx

Y= y = 25 x - 4

y = 25 x, y = 2

5 x + 7,

y = mx + b

.y2 Wx 7 y1 Wx

y3 Wx 4

10

10

10 10


MENTAL MATHDecide whether a line with the given slope is upward, downward, horizontal, or vertical.

1. upward 2. downward 3. horizontal 4. m is undefined. verticalm = 0m = -3m =76



THE SLOPE- INTERCEPT FORM SECTION 3.5 207

3.5 T H E S L O P E - I N T E R C E P T F O R M

O b j e c t i v e s

1 Use the slope-intercept form to find the slope and the y-intercept of a line.

2 Use the slope-intercept form to determine whether two lines are parallel, perpendicular,or neither.

3 Use the slope-intercept form to write an equation of a line.

4 Use the slope-intercept form to graph a linear equation.

1 In Section 3.4, we learned that the slant of a line is called its slope. Recall thatthe slope of a line m is the ratio of vertical change (change in y) to the correspondinghorizontal change (change in x) as we move along the line.

H e l p f u l H i n tDon’t forget that the slope of a line is the same no matter which ordered pairsof the line are used to calculate slope.

Find the slope of the line whose equation is y =34

x + 6.

Solut ion

CLASSROOM EXAMPLEFind the slope of the line whose equationis

answer: m = 59

y =59

x + 3.

To find the slope of this line, find any two ordered pair solutions of the equation. Let’sfind and use intercept points as our two ordered pair solutions. Recall from Section 3.2that the graph of an equation of the form has y-intercept (0, b).

This means that the graph of has a y-intercept of (0, 6) as shown below.

To find the y-intercept, let To find the x-intercept, let

Then becomes Then becomes




The y-intercept is (0, 6), as expected. The x-intercept is

Use the points (0, 6) and to find the slope. Then

The slope of the line is 34

.

m =change in y

change in x=

0 - 6-8 - 0

=-6-8

=34

1-8, 021-8, 02.

-8 = x

-24 = 3x

0 = 3x + 24 y = 6

0 =34

x + 6 or y =34

# 0 + 6 or

y =34

x + 6y =34

x + 6

y = 0.x = 0.

y =34

x + 6

y = mx + b

E X A M P L E 1




Find the slope and the y-intercept of the line whose equation is 5x + y = 2.

Solut ion Write the equation in slope-intercept form by solving the equation for y.


The coefficient of x, is the slope and the constant term, 2, is the y-value of the y-in-tercept, (0, 2).

-5,

y = -5x + 2

5x + y = 2CLASSROOM EXAMPLEFind the slope and the y-intercept of

answer: y-intercept (0, 7)m = 4,-4x + y = 7,

Find the slope and the y-intercept of the line whose equation is 3x - 4y = 4.

E X A M P L E 2

E X A M P L E 3

Solut ion Write the equation in slope-intercept form by solving for y.



Simplify.

The coefficient of x, is the slope, and the y-intercept is

2 The slope-intercept form can be used to determine whether two lines are parallelor perpendicular. Recall that nonvertical parallel lines have the same slope and differenty-intercepts.Also, nonvertical perpendicular lines have slopes whose product is -1.

10, -12.34

,

y =34

x - 1

4. -4y

-4=

-3x

-4+

4-4

-4y = -3x + 4

3x - 4y = 4CLASSROOM EXAMPLEFind the slope and the y-intercept of

answer: y-intercept 10, -42m = 72,

7x - 2y = 8.


TEACHING TIPStress this concept to students inwords: If a linear equation in twovariables is solved for y, the coeffi-cient of x is the slope, and the con-stant term is the y-intercept.

Slope-Intercept FormWhen a linear equation in two variables is written in slope-intercept form,

m is the slope of the line and (0, b) is the y-intercept of the line.

y = mx + b

Analyzing the results of Example 1, you may notice a striking pattern:

The slope of is the same as the coefficient of x.

Also, as mentioned earlier, the y-intercept is (0, 6). Notice that the y-value 6 is thesame as the constant term.

When a linear equation is written in the form not only is (0, b) the

y-intercep ºt of the line, but m is its slope.The form is appropriately called

the slope-intercept form.

y = mx + b q q

slope y-intercept10, b2

y = mx + b,

34

,y =34

x + 6


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E X A M P L E 4

Determine whether the graphs of and are parallel lines,

perpendicular lines, or neither.

2x + 10y = 30y = - 15

x + 1

Solut ion The graph of is a line with slope and with y-intercept (0, 1). To find

the slope and the y-intercept of the graph of write this equation inslope-intercept form. To do this, solve the equation for y.



The graph of this equation is a line with slope and y-intercept (0, 3). Because both

lines have the same slope but different y-intercepts, these lines are parallel.

CONCEPT CHECKWrite the equations of any three parallel lines.

3 The slope-intercept form can also be used to write the equation of a line givenits slope and y-intercept.

- 15

y = -15

x + 3

10y = -2x + 30

2x + 10y = 30

2x + 10y = 30,

- 15

y = - 15

x + 1

CLASSROOM EXAMPLEDetermine whether each pair of lines isparallel, perpendicular, or neither.a.

b.

answer: a. neither b. perpendicular

y = 25 x - 3

5

5x + 2y = 1

x + y = 52x + y = 5

Concept Check Answer:For example,y = 2x - 3, y = 2x - 1, y = 2x

112345

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2 3 4 512345

Slope

x

y

Slope

y-intercept

y-intercept

y Qx 3

y Qx 1

We are given the slope and the y-intercept. Let and and write the equa-

tion in slope-intercept form, y = mx + b.

b = -3,m =14

Find an equation of the line with y-intercept and slope of 14

.10, -32E X A M P L E 5

Solut ion


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Let and

Simplify.

4 We now use the slope-intercept form of the equation of a line to graph a linearequation.

y =14

x - 3

b 3.m 14

y =14

x + 1-32 y = mx + b


Use the slope-intercept form to graph the equation 4x + y = 1.

Solut ion First, write the given equation in slope-intercept form.

The graph of this equation will have slope and y-intercept (0, 1). To graph this line,first plot the intercept point (0, 1).To find another point of the graph, use the slope

which can be written as Start at the point (0, 1) and move 4 units down (since the

numerator of the slope is ); then move 1 unit to the right (since the denominator ofthe slope is 1). We arrive at the point The line through (0, 1) and is thegraph of 4x + y = 1.

11, -3211, -32.-4

-41

.

-4,-4

y = -4x + 1

4x + y = 1CLASSROOM EXAMPLEGraph answer:

3x + y = 2.

(0, 2)

x

y

E X A M P L E 7

Since the equation is written in slope-intercept form the slope

of its graph is and the y-intercept is To graph, begin by plotting the intercept

From this point, find another point of the graph by using the slope and recalling

that slope is . Start at the intercept point and move 3 units up since the numerator

of the slope is 3; then move 5 units to the right since the denominator of the slope is 5.

Stop at the point (5, 1). The line through and (5, 1) is the graph of y =35

x - 2.10, -22

riserun

35

10, -22.10, -22.3

5

y = mx + b,y =35

x - 2

Graph: y =35

x - 2.


y = 35 x - 4.

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2 3 4 512345

(5, 1)

(0, 2)

x

y

y-intercept

3 units up

5 units rightm 5

3

TEACHING TIP

If you’d like, show students thatany slope equivalent to gives thesame line. For example, use Count 6 units up and 10 units tothe right, and the same line is obtained.

m = 610.

35

E X A M P L E 6

(0, 4)

x

y

Solut ion

CLASSROOM EXAMPLEFind an equation of the line with

y-intercept and slope of

answer: y = 35 x - 4

35

.10, -42


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(1, 3)

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12345

2 3 4 512345

(0, 1)

x

y

1 unit right

4 units down

m 14

H e l p f u l H i n t

In Example 7, if we interpret the slope of as we arrive at for a

second point. Notice that this point is also on the line.

1-1, 524-1

,-4

A grapher is a very useful tool for discovering patterns. To discover the change inthe graph of a linear equation caused by a change in slope, try the following. Use astandard window and graph a linear equation in the form Recall thatthe graph of such an equation will have slope m and y-intercept b.

First graph To do so, press the key and enter Notice that this graph has slope 1 and that the y-intercept is 3. Next, on the same setof axes, graph and by pressing and entering

and

Notice the difference in the graph of each equation as the slope changes from 1to 2 to 3. How would the graph of appear? To see the change in thegraph caused by a change in negative slope, try graphing and on the same set of axes.

Use a grapher to graph the following equations. For each exercise, graph the firstequation and use its graph to predict the appearance of the other equations. Thengraph the other equations on the same set of axes and check your prediction.

1. 2.

3. 4.

5. 6. y = 3x - 1; y = -3x - 1y = -7x + 5; y = 7x + 5

y = x + 1; y =54

x + 1, y =52

x - 1y =12

x + 2; y =34

x + 2, y = x + 2

y = -x; y = -5x, y = -10xy = x; y = 6x, y = -6x

y = -3x + 3y = -x + 3, y = -2x + 3,

y = 5x + 3

Y3 = 3x + 3.Y2 = 2x + 3 Y= y = 3x + 3y = 2x + 3

Y1 = x + 3. Y= y = x + 3.

y = mx + b.


10

10

10

10

y3 3x 3y2 2x 3

y1 x 3



MENTAL MATHIdentify the slope and the y-intercept of the graph of each equation.

1. 2. 3.

4. 5. 6. m = - 14

; a0, 35by = -

14

x +35

m =57

; 10, -42y =57

x - 4m = -1, a0, - 29by = -x -

29

m = 1; a0, 13by = x +

13

m = -7; 10, 32y = -7x + 3m = 2; 10, -12y = 2x - 1

STUDY SKILLS REMINDERTips for Studying for an Exam

To prepare for an exam, try the following study techniques.

Start the study process days before your exam.Make sure that you are current and up-to-date on your assign-ments.If there is a topic that you are unsure of, use one of the manyresources that are available to you. For example,

See your instructor.Visit a learning resource center on campus where math tutors are available.Read the textbook material and examples on the topic.View a videotape on the topic.

Reread your notes and carefully review the Chapter High-lights at the end of the chapter.Work the review exercises at the end of the chapter and checkyour answers. Make sure that you correct any missed exercises.If you have trouble on a topic, use a resource listed above.

Find a quiet place to take the Chapter Test found at the end ofthe chapter. Do not use any resources when taking this sampletest. This way you will have a clear indication of how preparedyou are for your exam. Check your answers and make surethat you correct any missed exercises.

Get lots of rest the night before the exam. It’s hard to showhow well you know the material if your brain is foggy fromlack of sleep.

Good luck and keep a positive attitude.


O b j e c t i v e s

1 Use the point-slope form to find an equation of a line given its slope and a point of the line.

2 Use the point-slope form to find an equation of a line given two points of the line.

3 Find equations of vertical and horizontal lines.

4 Use the point-slope form to solve problems.

1 From the last section, we know that if the y-intercept of a line and its slope aregiven, then the line can be graphed and an equation of this line can be found. Our goal inthis section is to answer the following: If any two points of a line are given, can an equa-tion of the line be found? To answer this question, first let’s see if we can find an equationof a line given one point (not necessarily the y-intercept) and the slope of the line.

Suppose that the slope of a line is and the line contains the point (2, 4). Forany other point with ordered pair (x, y) to be on the line, its coordinates must satisfythe slope equation

Now multiply both sides of this equation by

This equation is a linear equation whose graph is a line that contains the point (2, 4)and has slope This form of a linear equation is called the point-slope form.

In general, when the slope of a line and any point on the line are known, theequation of the line can be found.To do this, use the slope formula to write the slope ofa line that passes through points (x, y) and We have

Multiply both sides of this equation by to obtain

slopeq

y - y1 = m1x - x12x - x1

y - y1

x - x1= m

1x1, y12.

-3.

y - 4 = -31x - 22x - 2.

y - 4

x - 2= -3

-3

THE POINT-SLOPE FORM SECTION 3.6 215

6 7

678

112

12345

2 3 4 5123

(x, y)

(2, 4)

x

y

x 2

y 4

Point-Slope Form of the Equation of a LineThe point-slope form of the equation of a line is wherem is the slope of the line and is a point on the line.1x1, y12

y - y1 = m1x - x12,

Find an equation of the line passing through with slope Write the equationin standard form: Ax + By = C.

-2.1-1, 52

Solut ion Since the slope and a point on the line are given, use point-slope formto write the equation. Let and 1-1, 52 = 1x1, y12.m = -2y - y1 = m1x - x12

E X A M P L E 1

3.6 T H E P O I N T- S L O P E F O R M



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y

y c

(0, c) xx

y

x c

(c, 0)

Horizontal LineVertical Line

3 Recall from Section 3.3 that:

CLASSROOM EXAMPLEFind an equation of the line that passesthrough with slope answer: 3x + y = 2

-3.12, -42

E X A M P L E 2

Find an equation of the line through (2, 5) and Write the equation in standardform.

1-3, 42.

Solut ion First, use the two given points to find the slope of the line.

Next, use the slope and either one of the given points to write the equation in point-

slope form. We use (2, 5). Let and

Use point-slope form.

Let and

Multiply both sides by 5 to clear fractions.

Use the distributive property and simplify.Subtract x from both sides.

Add 25 to both sides. -x + 5y = 23

-x + 5y - 25 = -2 5y - 25 = x - 2

51y - 52 = 5 # 15

1x - 22m

1

5.x1 2, y1 5, y - 5 =

15

1x - 22 y - y1 = m1x - x12

m =15

.x1 = 2, y1 = 5,

15

m =4 - 5

-3 - 2=

-1-5

=15

CLASSROOM EXAMPLEFind an equation of the line through (1, 3) and answer: 5x + 4y = 17

15, -22.

H e l p f u l H i n tMultiply both sides of the equation by and it becomes

Both and are in standard form, andthey are equations of the same line.

x - 5y = -23-x + 5y = 23x - 5y = -23.-1,-x + 5y = 23

Let and Simplify.Use the distributive property.Add 5 to both sides.Add 2x to both sides.

In standard form, the equation is

2 We may also find the equation of a line given any two points of the line, as seenin the next example.

2x + y = 3.

2x + y = 3 y = -2x + 3

y - 5 = -2x - 2 y - 5 = -21x + 12 1x1, y12 11, 52.m 2 y - 5 = -2[x - 1-12]

y - y1 = m1x - x12


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Find an equation of the vertical line through 1-1, 52.Solut ion The equation of a vertical line can be written in the form so an equation for a

vertical line passing through is x = -1.1-1, 52 x = c,

112345

12345

2 3 4 512345 x

y

x 1

Find an equation of the line parallel to the line and passing through 1-2, -32.y = 5

112345

12345

2 3 4 512345

(2, 3)

x

y

y 3

y 5

Solut ion Since the graph of is a horizontal line, any line parallel to it is also horizontal.The equation of a horizontal line can be written in the form An equation for thehorizontal line passing through is y = -3.1-2, -32 y = c.

y = 5

E X A M P L E 3

CLASSROOM EXAMPLEFind an equation of the vertical linethrough answer: x = -7

1-7, -22.

E X A M P L E 4

CLASSROOM EXAMPLEFind an equation of the line parallel tothe line and passing through(10, 4).answer: y = 4

y = -3

PREDICTING THE SALES OF FRISBEES

The Whammo Company has learned that by pricing a newly released Frisbee at $6,sales will reach 2000 Frisbees per day. Raising the price to $8 will cause the sales to fallto 1500 Frisbees per day.

a. Assume that the relationship between sales price and number of Frisbees sold is lin-ear and write an equation describing this relationship. Write the equation in slope-intercept form.

b. Predict the daily sales of Frisbees if the price is $7.50.

4 Problems occurring in many fields can be modeled by linear equations in twovariables. The next example is from the field of marketing and shows how consumerdemand of a product depends on the price of the product.

E X A M P L E 5


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TEACHING TIPHelp students develop a strategyfor various scenarios. Ask them thefollowing: How would you find theequation if you knew the slope and y-intercept? two points? a point and the slope? a point and the y-intercept? the x-intercept and the

y-intercept?How much information must begiven in order to determine a line?

Á

Solut ion a. First, use the given information and write two ordered pairs. Ordered pairs will be inthe form (sales price, number sold) so that our ordered pairs are (6, 2000) and (8, 1500).Use the point-slope form to write an equation.To do so, we find the slope of the linethat contains these points.

Next, use the slope and either one of the points to write the equation in point-slopeform. We use (6, 2000).

Use point-slope form.

Let and

Use the distributive property.

Write in slope-intercept form.

b. To predict the sales if the price is $7.50, we find y when

Let

If the price is $7.50, sales will reach 1625 Frisbees per day.

The preceding example may also be solved by using ordered pairs of the form(number sold, sales price).

y = 1625

y = -1875 + 3500

x = 7.50. y = -25017.502 + 3500 y = -250x + 3500

x = 7.50.

y = -250x + 3500

y - 2000 = -250x + 1500

m 250.x1 6, y1 2000, y - 2000 = -2501x - 62 y - y1 = m1x - x12

m =2000 - 1500

6 - 8=

500-2

= -250

CLASSROOM EXAMPLEA pool company learned that by pricinga new pool toy at $10, local sales willreach 200 a week. Lowering the price to$9 will cause sales to rise to 250 aweek.a. Assume that the relationship is linear

and write the equation in slope-inter-cept form. Use ordered pairs of theform (sales price, number sold).

b. Predict the weekly sales of the toy ifthe price is $7.50.

answer: a.b. 325

y = -50x + 700

Forms of Linear Equations

Standard form of a linear equation.

A and B are not both 0.

Slope-intercept form of a linear equation.

The slope is m and the y-intercept is (0, b).

Point-slope form of a linear equation.

The slope is m and is a point on the line.

Horizontal line

The slope is 0 and the y-intercept is (0, c).

Vertical line

The slope is undefined and the x-intercept is (c, 0).

Parallel and Perpendicular LinesNonvertical parallel lines have the same slope.

The product of the slopes of two nonvertical perpendicular lines is -1.

x = c

y = c

1x1, y12y - y1 = m1x - x12

y = mx + b

Ax + By = C



MENTAL MATHThe graph of each equation below is a line. Use the equation to identify the slope and a point of the line.

1. answers may vary, Ex. (4, 8) 2.

3. answers may vary, Ex. 4.

5. answers may vary, Ex. 6. y =37

1x + 421-1, 02m = 25;y =

25

1x + 12y + 6 = -71x - 22110, -32m = -2;y + 3 = -21x - 102y - 1 = 51x - 22m = 3;y - 8 = 31x - 42

Suppose you are a phys-ical therapist. At weeklysessions, you are admin-istering treadmill treat-ment to a patient. In thefirst phase of the treat-ment plan, the patient is

to walk on a treadmill at a speed of 2 mph until fatigue sets in.Once the patient is able to walk 10 minutes at this speed, hewill be ready for the next phase of treatment. However, youmay change the treatment plan if it looks like it will take longerthan 10 weeks to build up to 10 minutes of walking at 2 mph.

You record and plot the patient’s progress on his chart for5 weeks. Do you think you may need to change the first phaseof treatment? Explain your reasoning. If not, when do you thinkthe patient will be ready for the next phase of treatment?

600

550

500

450

400

350

300

250

200

150

100

0

Tim

e (i

n se

cond

s)

Weeks54321 1098760

Patient Progress: Treadmill at 2 mph

2. answers may vary, Ex. (2, 1) 4. answers may vary, Ex. 6. answers may vary, Ex. 1-4, 02m = 37;12, -62m = -7;m = 5;




Find the domain and the range of the relation 510, 22, 13, 32, 1-1, 02, 13, -226.Solut ion The domain is the set of all x-values or and the range is the set of all y-values,

or 5-2, 0, 2, 36.5-1, 0, 36,

CLASSROOM EXAMPLEFind the domain and range of the rela-tion answer:domain: range: 50, 1, 5, 665-3, 4, 76;51-3, 52, 1-3, 12, 14, 62, 17, 026.

3.7 F U N C T I O N S

O b j e c t i v e s

1 Identify relations, domains, and ranges.

2 Identify functions.

3 Use the vertical line test.

4 Use function notation.

1 In previous sections, we have discussed the relationships between two quanti-ties. For example, the relationship between the length of the side of a square x and itsarea y is described by the equation These variables x and y are related in thefollowing way: for any given value of x, we can find the corresponding value of y bysquaring the x-value. Ordered pairs can be used to write down solutions of thisequation. For example, (2, 4) is a solution of and this notation tells us thatthe x-value 2 is related to the y-value 4 for this equation. In other words, when thelength of the side of a square is 2 units, its area is 4 square units.

A set of ordered pairs is called a relation.The set of all x-coordinates is called thedomain of a relation, and the set of all y-coordinates is called the range of a relation.Equations such as are also called relations since equations in two variables de-fine a set of ordered pair solutions.

y = x2

y = x2,

y = x2.

E X A M P L E 1



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TEACHING TIPRemind students that all functionsare relations, but not all relationsare functions.

FUNCTIONS SECTION 3.7 223

Which of the following relations are also functions?

a. b. 510, -22, 11, 52, 10, 32, 17, 72651-1, 12, 12, 32, 17, 32, 18, 626Solut ion

CLASSROOM EXAMPLEWhich of the following relations are alsofunction?a.b.answer:a. function b. not a function

511, 42, 16, 62, 11, -32, 17, 526512, 52, 1-3, 72, 14, 52, 10, -126

a. Although the ordered pairs (2, 3) and (7, 3) have the same y-value, each x-value isassigned to only one y-value so this set of ordered pairs is a function.

b. The x-value 0 is assigned to two y-values, and 3, so this set of ordered pairs is nota function.

-2

Relations and functions can be described by a graph of their ordered pairs.

FunctionA function is a set of ordered pairs that assigns to each x-value exactly one y-value.

E X A M P L E 2

2 Some relations are also functions.

Which graph is the graph of a function?

a. b.

Solut ion a. This is the graph of the relation Each x-coordinate has exactly one y-coordinate, so this is the graph of a function.

b. This is the graph of the relation The x-coordinate 1is paired with two y-coordinates, 2 and 3, so this is not the graph of a function.

3 The graph in Example 3(b) was not the graph of a function because the x-coordinate 1 was paired with two y-coordinates, 2 and 3. Notice that when an x-coordi-nate is paired with more than one y-coordinate, a vertical line can be drawn that will in-tersect the graph at more than one point. We can use this fact to determine whether arelation is also a function. We call this the vertical line test.

51-2, -32, 11, 22, 11, 32, 12, -126.51-4, -22, 1-2, -121-1, -12, 11, 226.

CLASSROOM EXAMPLEWhich graph is the graph of a function?a.

b.

answer:

a. function b. not a function

112345

12345

2 3 4 512345 x

y

112345

12345

2 3 4 512345 x

y

E X A M P L E 3

y

x

y

x


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Solut ion

CLASSROOM EXAMPLEDetermine whether each graph is a function.a.

b.

answer: a. yes b. no

y

Not a function

a. This graph is the graph of a function since no vertical line will intersect this graphmore than once.

b. This graph is also the graph of a function; no vertical line will intersect it more thanonce.

c. This graph is not the graph of a function. Vertical lines can be drawn that intersectthe graph in two points. An example of one is shown.

TEACHING TIPShow students many visual exam-ples of this to help them remember.

Recall that the graph of a linear equation is a line, and a line that is not verticalwill pass the vertical line test. Thus, all linear equations are functions except those ofthe form which are vertical lines.x = c,

x

y y y y

E X A M P L E 4

y

x

y

x

d. This graph is not the graph of a function.A vertical line can be drawn that intersectsthis line at every point.

This vertical line test works for all types of graphs on the rectangular coordinate system.

112345

12345

2 3 4 512345

(1, 3)

(1, 2)

x

y x-coordinate 1 pairedwith two y-coordinates,

2 and 3.

Not the graphof a function.

Vertical Line TestIf a vertical line can be drawn so that it intersects a graph more than once,the graph is not the graph of a function.

Use the vertical line test to determine whether each graph is the graph of a function.

a. b. c. d.


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Which of the following linear equations are functions?

a. b. c. d. x = -1y = 5y = 2x + 1y = x

Solut ion a, b, and c are functions because their graphs are nonvertical lines. d is not a functionbecause its graph is a vertical line.

Examples of functions can often be found in magazines, newspapers, books, andother printed material in the form of tables or graphs such as that in Example 6.

CLASSROOM EXAMPLEWhich linear equations are also functions?a. b.c.answer: a. yes b. no c. yes

y = 2x = -5y = -x + 1

The graph shows the sunrise time for Indianapolis, Indiana, for the year. Use this graphto answer the questions.

E X A M P L E 5

E X A M P L E 6

9 A.M.

8 A.M.

7 A.M.

6 A.M.

5 A.M.

4 A.M.

3 A.M.

2 A.M.

1 A.M.

Tim

e

Apr

Sunrise

MonthJune Nov DecJulyMay OctSeptAugJan Feb Mar

a. Approximate the time of sunrise on February 1.b. Approximately when does the sun rise at 5 A.M.?c. Is this the graph of a function?

Solut ion a. To approximate the time of sunrise on February 1, we find the mark on the horizon-tal axis that corresponds to February 1. From this mark, we move vertically upwarduntil the graph is reached. From that point on the graph, we move horizontally tothe left until the vertical axis is reached. The vertical axis there reads 7 A.M.

CLASSROOM EXAMPLEUse the graph in Example 6 to answerthe questions.a. Approximate the time of sunrise on

March 1.b. Approximate the date(s) when the

sun rises at 6 A.M.answer:a. 6:30 A.M.b. middle of March and middle of

October

9 A.M.

8 A.M.

7 A.M.

6 A.M.

5 A.M.

4 A.M.

3 A.M.

2 A.M.

1 A.M.

Tim

e

Apr

Sunrise

MonthJune Nov DecJulyMay OctSeptAugJan Feb Mar


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We often use letters such as f, g, and h to name functions. For example, the symbolmeans function of x and is read “f of x.” This notation is called function notation.

The equation can be written as using function notation,and these equations mean the same thing. In other words,

The notation means to replace x with 1 and find the resulting y or functionvalue. Since

then

This means that, when y or and we have the ordered pair (1, 3). Nowlet’s find f(2), f(0), and

Ordered Pair: (2, 5) (0, 1) 1-1, -12

= -1= 1= 5

= -2 + 1= 0 + 1= 4 + 1

f1-12 = 21-12 + 1f102 = 2102 + 1f122 = 2122 + 1

f1x2 = 2x + 1f1x2 = 2x + 1f1x2 = 2x + 1

f1-12. f1x2 = 3,x = 1,

f112 = 2112 + 1 = 3

f1x2 = 2x + 1

f(1)y = f1x2.f1x2 = 2x + 1y = 2x + 1

f(x)

TEACHING TIPTake time to help students masterthe fact that, for example, if

then the correspondingordered pair is (7, 10).f172 = 10,

112345

12345

2 3 4 512345 x

y

y 2x 1

b. To approximate when the sun rises at 5 A.M., we find 5 A.M. on the time axis and movehorizontally to the right. Notice that we will reach the graph twice, corresponding totwo dates for which the sun rises at 5 A.M. We follow both points on the graph verti-cally downward until the horizontal axis is reached.The sun rises at 5 A.M. at approx-imately the end of the month of April and the middle of the month of August.

c. The graph is the graph of a function since it passes the vertical line test. In otherwords, for every day of the year in Indianapolis, there is exactly one sunrise time.

4 The graph of the linear equation passes the vertical line test, so wesay that is a function. In other words, gives us a rule for writingordered pairs where every x-coordinate is paired with one y-coordinate.

y = 2x + 1y = 2x + 1y = 2x + 1



Given find the following. Then write down the corresponding orderedpairs generated.a. g(2) b. c. g(0)g1-22

g1x2 = x2 - 3,

Solut ion a. b. c.

Ordered Pair: (2, 1)

We now practice finding the domain and the range of a function. The domain ofour functions will be the set of all possible real numbers that x can be replaced by. Therange is the set of corresponding y-values.

10, -321-2, 12 g102 = -3 gives g1-22 = 1 gives g122 = 1 gives

= -3 = 1 = 1

= 0 - 3 = 4 - 3 = 4 - 3

g102 = 02 - 3 g1-22 = 1-222 - 3 g122 = 22 - 3

g1x2 = x2 - 3 g1x2 = x2 - 3 g1x2 = x2 - 3

H e l p f u l H i n tNote that f(x) is a special symbol in mathematics used to denote a function.The symbol f(x) is read “f of x.” It does not mean ( f times x).f # x

E X A M P L E 7

E X A M P L E 8Find the domain of each function.

a. b. f1x2 = 2x + 1g1x2 =1x

Solut ion a. Recall that we cannot divide by 0 so that the domain of g(x) is the set of all realnumbers except 0. In interval notation, we can write

b. In this function, x can be any real number. The domain of f(x) is the set of all realnumbers, or in interval notation.

CONCEPT CHECKSuppose that the value of f is 7 when the function is evaluated at 2. Write this situation in function notation.

1- q , q21- q , 02 ´ 10, q2.

E X A M P L E 9Find the domain and the range of each function graphed. Use interval notation.

a. b.

112345

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2 3 4 512345

(4, 5)

(3, 1)

x

y

6 7112345

12345

2 3 4 5123

(3, 2)

x

y

Concept Check Answer:f(2) 7

CLASSROOM EXAMPLE

Given , find the followingand list the generated ordered pair.

a. b. c.

answer: a.b.c. f(0) = 1; (0, 1)

f(-3) = 10; (-3, 10)f(1) = 2; (1, 2)

f(0)f(-3)f(1)

f(x) = x2 + 1

CLASSROOM EXAMPLE

Find the domain.

a.

b.

answer: a. all real numbers or b. all real numbers except 2 or

(- q , q) ´ (2, q)

(- q , q)

g(x) =1

x - 2

f(x) = 5x - 2

CLASSROOM EXAMPLE

Find the domain and range. Use intervalnotation.

a. b.

answer:

a. domain: [2, 3]; range: [1, 2]

b. domain ; range [-3, q ](- q , q)

y

x

(2, 1)(3, 2)

x

y

(2, 3)


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Suppose youare the proper-ty and groundsmanager for asmall companyin Portland,Oregon. A re-

cent heat wave has caused employees to ask for air-conditioning in the building. As you begin to researchair-conditioning installation, you discover this graphof average high temperatures for Portland. Becausethe building is surrounded by trees, the current venti-lation system (without air-conditioning) is able to keepthe temperature 10°F cooler than the outside temper-ature on a warm day. A comfortable temperaturerange for working is 68°F to 74°F. Would you recom-mend installing air-conditioning in the building? Ex-plain your reasoning. What other factors would youwant to consider?


90

80

70

60

50

40

30

20

10

0

Tem

pera

ture

(deg

rees

Fah

renh

eit)

3

Average Monthly High Temperature: Portland, Oregon

Month (1 January)5 10 121164 9870 1 2

Source: The Weather Channel Enterprises, Inc.

1. domain: range:

2. domain: range: -6, -2, 4-2, 1, 3 ;

5-7, 0, 4, 1065-7, 0, 2, 106;

a. b.

112345

1234

2 3 512345 4 x

5

y

Range:[1, 5]

Domain[3, 4]

Solut ion

6 7112345

12345

2 3 4 5123

(3, 2)

x

y

Range:[2, )

Domain:All real numbers

or (, )


244 CHAPTER 4 SOLVING SYSTEMS OF L INEAR EQUATIONS AND INEQUALIT IES

4.1 S O LV I N G S Y S T E M S O F L I N E A R E Q UAT I O N S B Y G R A P H I N G

O b j e c t i v e s

1 Determine if an ordered pair is a solution of a system of equations in two variables.

2 Solve a system of linear equations by graphing.

3 Without graphing, determine the number of solutions of a system.

1 A system of linear equations consists of two or more linear equations. In thissection, we focus on solving systems of linear equations containing two equations intwo variables. Examples of such linear systems are

A solution of a system of two equations in two variables is an ordered pair of numbersthat is a solution of both equations in the system.

e3x - 3y = 0 x = 2y e x - y = 0

2x + y = 10 ey = 7x - 1

y = 4

E X A M P L E 1Which of the following ordered pairs is a solution of the given system?

First equationSecond equation

a. (12, 6) b. 10, -22

e2x - 3y = 6 x = 2y

TEACHING TIPClassroom ActivityIf the desks in your classroom arearranged in columns and rows, con-sider beginning this lesson with astudent demonstration. Tell stu-dents you will use the floor of theclassroom as a coordinate planewith the back right corner as theorigin and the rows and columns asunit measures of y and x, respec-tively. Then have students repre-senting stand up, that is, thestudents whose row value equalstheir column value. While they re-main standing, have students rep-resenting stand up, that is,the students in row 3. Then ask ifanyone is a member of both equa-tions. What happens at that point?

y = 3

y = x

Solut ion If an ordered pair is a solution of both equations, it is a solution of the system.

a. Replace x with 12 and y with 6 in both equations.

First equation Second equation

Let and Let and

Simplify. True

True

Since (12, 6) is a solution of both equations, it is a solution of the system.

b. Start by replacing x with 0 and y with in both equations.


Let and Let and

Simplify. False

True

While is a solution of the first equation, it is not a solution of the second equa-tion, so it is not a solution of the system.

2 Since a solution of a system of two equations in two variables is a solutioncommon to both equations, it is also a point common to the graphs of both equations.Let’s practice finding solutions of both equations in a system—that is, solutions of asystem—by graphing and identifying points of intersection.

10, -22 6 = 6

0 = -4 0 + 6 6

y 2 .x 00 21-22y 2 .x 0 2102 - 31-22 6

x = 2y 2x - 3y = 6

-2

6 = 6

12 = 12 24 - 18 6

y 6 .x 12 12 2162y 6 .x 12 21122 - 3162 6

x = 2y 2x - 3y = 6

CLASSROOM EXAMPLEDetermine whether is a solutionof the system

answer: no

e2x - y = 8x + 3y = 4

13, -22




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SOLVING SYSTEMS OF L INEAR EQUATIONS BY GRAPHING SECTION 4.1 245

E X A M P L E 2Solve the system of equations by graphing.

e -x + 3y = 10x + y = 2

Solut ion

CLASSROOM EXAMPLESolve the system of equations by graphing

answer:consistent, independent12, -42,

e -3x + y = -10x - y = 6

112345

12345

2 3 4 512345

(1, 3)

x

y

x 3y 10

x y 2

H e l p f u l H i n tThe point of intersectiongives the solution of thesystem.

x y

0

22 4

-4

103

-x + 3y = 10

x y

0 22 01 1

x + y = 2

The two lines appear to intersect at the point To check, we replace x with and y with 3 in both equations.


Let and Let and

Simplify. True

True

checks, so it is the solution of the system.1-1, 32 10 = 10

2 = 2 1 + 9 10

y 3 .x 1 -1 + 3 2y 3 .x 1 -1-12 + 3132 10

x + y = 2 -x + 3y = 10

-11-1, 32 .

A system of equations that has at least one solution as in Example 2 is said to bea consistent system. A system that has no solution is said to be an inconsistent system.

On a single set of axes, graph each linear equation.

TEACHING TIPBefore discussing Example 2, dis-cuss the various scenarios using thefollowing graph.

Name a point not on line l or linem (S).Name a point on line l but not online m (T).Name a point on line m but not online l (Q).Name a point on both line l andline m (R).Then state that a point is a solutionof a system of equations only whenit is on the graphs of all the equa-tions in the system. Finally, ask,“How can we determine if a pointwill be on the graph without graph-ing the equation?” (Substitute thepoint into the equation and see ifwe get a true statement.)

Solve the following system of equations by graphing.

e2x + y = 72y = -4x

H e l p f u l H i n tNeatly drawn graphs can help when you are estimating the solution of a sys-tem of linear equations by graphing.

In the example above, notice that the two lines intersected in a point.This means that the system has 1 solution.

y

x(2, 4)x y 6

3x y 10

8

4

x

y

T

lm

Q

S

R

E X A M P L E 3



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answer:no solution, inconsistent, independent

e3x - y = 66x = 2y

Solut ion Graph the two lines in the system.

112345

12345

2 3 4 512345

(1, 5)

(1, 2)

(0, 0)

x

y

2y 4x 2x y 7

(3q, 0)

The lines appear to be parallel. To confirm this, write both equations in slope-interceptform by solving each equation for y.


Subtract 2x from Divide both sides by 2.both sides.

Recall that when an equation is written in slope-intercept form, the coefficient of x isthe slope. Since both equations have the same slope, but different y-intercepts, thelines are parallel and have no points in common. Thus, there is no solution of the sys-tem and the system is inconsistent.

In Examples 2 and 3, the graphs of the two linear equations of each system aredifferent. When this happens, we call these equations independent equations. If thegraphs of the two equations in a system are identical, we call the equations dependentequations.

-2,

y 2x

2y

2=

-4x

2 y = 2x 7

2y = -4x 2x + y = 7

6x 2y

3x y 6

x

y

8

8

TEACHING TIPClassroom ActivityBefore discussing Example 3, do astudent demonstration using thefloor of the classroom as a coordi-nate plane with the back right cor-ner as the origin. Have studentsrepresenting stand up andstudents representing standup. Then ask: “What is the solutionof this system of equations?” Fol-low up by asking, “Is it possible forthese two equations to intersect ifwe had a bigger room with moredesks? Why not?”

x = 4x = 2

E X A M P L E 4Solve the system of equations by graphing.

e x - y = 3 -x + y = -3

Solut ion Graph each line.

112345

12345

2 3 4 512345

(4, 1)(3, 0)x

y

x y 3x y 3


answer:infinite number of solutions, consistent,dependent

e3x + 4y = 129x + 12y = 36

y

x

3x 4y 129x 12y 36


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x

y

One point of intersection: one solution

Consistent system (at least one solution)Independent equations (graphs of equations differ)

3 You may have suspected by now that graphing alone is not an accurate way tosolve a system of linear equations. For example, a solution of is unlikely to beread correctly from a graph. The next two sections present two accurate methods ofsolving these systems. In the meantime, we can decide how many solutions a system hasby writing each equation in the slope-intercept form.

A12 , 29 B

E X A M P L E 5Without graphing, determine the number of solutions of the system.

L 12

x - y = 2

x = 2y + 5

Solut ion First write each equation in slope-intercept form.


Add y to both sides. Subtract 5 from both sides.

Divide both sides by 2. x

2-

52

=2y

2

x - 5 = 2y 12

x = y + 2

x = 2y + 5 12

x - y = 2

x

y

Parallel lines: no solution

Inconsistent system (no solution)Independent equations (graphs of equations differ)

x

y

Same line: infinite number of solutions

Consistent system (at least one solution)Dependent equations (graphs of equations identical)

These graphs appear to be identical. To confirm this, write each equation in slope-intercept form.


Subtract x from Add x to both sides.both sides.


The equations are identical and so must be their graphs.The lines have an infinite num-ber of points in common. Thus, there is an infinite number of solutions of the systemand this is a consistent system. The equations are dependent equations.

As we have seen, three different situations can occur when graphing the two linesassociated with the equations in a linear system:

y x 31 .

-y

-1=

-x

-1+

3-1

y x 3 -y = -x + 3

-x + y = -3 x - y = 3

CLASSROOM EXAMPLEDetermine the number of solutions of thesystem.

answer: no solution

e3x - y = 6x = 1

3 y


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Subtract 2 from Simplify.both sides.

The slope of each line is but they have different y-intercepts. This tells us that thelines representing these equations are parallel. Since the lines are parallel, the systemhas no solution and is inconsistent.

12

,

12

x -52

= y 12

x - 2 = y

E X A M P L E 6Determine the number of solutions of the system.

e3x - y = 4x + 2y = 8

Solut ion Once again, the slope-intercept form helps determine how many solutions this systemhas.


Add y to both sides. Subtract 2y fromboth sides.

Subtract 4 from Subtract 8 from

both sides. both sides.

Divide both sidesby

Simplify.

The slope of the second line is whereas the slope of the first line is 3. Since theslopes are not equal, the two lines are neither parallel nor identical and must intersect.Therefore, this system has one solution and is consistent.

- 12

,

- 12

x + 4 = y

2 .

x

-2-

8-2

=-2y

-2

x - 8 = -2y 3x - 4 = y

x = -2y + 8 3x = y + 4

x + 2y = 8 3x - y = 4

A graphing calculator may be used to approximate solutions of systems of equations.For example, to approximate the solution of the system

first graph each equation on the same set of axes. Then use the intersect feature ofyour calculator to approximate the point of intersection.

The approximate point of intersection is

Solve each system of equations. Approximate the solutions to two decimal places.

1. 2.

3. 4. e -3.6x - 8.6y = 10-4.5x + 9.6y = -7.7

e4.3x - 2.9y = 5.68.1x + 7.6y = -14.1

ey = 4.25x + 3.89y = -1.88x + 3.21

ey = -2.68x + 1.21y = 5.22x - 1.68

1-0.82, 1.232 .

ey = -3.14x - 1.35y = 4.88x + 5.25,


CLASSROOM EXAMPLEDetermine the number of solutions of thesystem.

answer: one solution

ex + 3y = -14x - y = 10



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MENTAL MATHEach rectangular coordinate system shows the graph of the equations in a system of equations. Use each graph to deter-mine the number of solutions for each associated system. If the system has only one solution, give its coordinates. (Thecoordinates will be integers.)

1. 2. 3.

no solution infinite number of solutions

4. 5. 6.

1 solution, (3, 4) no solution infinite number of solutions

7. 8.

1 solution, (3, 2) 1 solution, 10, -32

1 solution, 1-1, 32

2

4

2

5

1

4

1

3

3

5

2 424 15 3 513 x

y

6 7

67

2

21

4

1

3

3

5

2 42 1 3 513 x

y

4

8

4

10

2

8

2

6

6

10

4 848 210 6 1026 x

y

4

8

4

10

2

8

2

6

6

10

4 848 210 6 1026 x

y

2

4

2

5

1

4

1

3

3

5

2 424 15 3 513 x

y

2

4

2

5

1

4

1

3

3

5

2 424 15 3 513 x

y

2

4

2

5

1

4

1

3

3

5

2 424 15 3 513 x

y

2

4

2

5

1

4

1

3

3

5

2 424 15 3 513 x

y


SOLVING SYSTEMS OF L INEAR EQUATIONS BY SUBSTITUTION SECTION 4.2 251

Solut ion The second equation in this system is This tells us that x and have thesame value. This means that we may substitute for x in the first equation.

First equation

Substitute for x since x y 2 .y 2 2 1y + 222

+ y = 10

2x + y = 10

y + 2y + 2x = y + 2.

4.2 S O LV I N G S Y S T E M S O F L I N E A R E Q UAT I O N S B Y S U B S T I T U T I O N

O b j e c t i v e

1 Use the substitution method to solve a system of linear equations.

1 As we stated in the preceding section, graphing alone is not an accurate way tosolve a system of linear equations. In this section, we discuss a second, more accuratemethod for solving systems of equations.This method is called the substitution methodand is introduced in the next example.

E X A M P L E 1Solve the system:

First equationSecond equation

e2x + y = 10x = y + 2

CLASSROOM EXAMPLESolve the system:

answer: (5, 1)

e2x + 3y = 13x = y + 4

ø




H e l p f u l H i n tDon’t forget the distrib-utive property.

Check We check to see that (4, 2) satisfies both equations of the original system.

First Equation Second Equation

Let and

True True

The solution of the system is (4, 2).

A graph of the two equations shows the two lines intersecting at the point (4, 2).

4 = 4 10 = 10

y 2 .x 4 4 2 + 2 2142 + 2 10

x = y + 2 2x + y = 10

Notice that this equation now has one variable, y. Let’s now solve this equation for y.


Combine like terms.



Now we know that the y-value of the ordered pair solution of the system is 2. Tofind the corresponding x-value, we replace y with 2 in the equation andsolve for x.

Let

The solution of the system is the ordered pair (4, 2). Since an ordered pair solutionmust satisfy both linear equations in the system, we could have chosen the equation

to find the corresponding x-value. The resulting x-value is the same.2x + y = 10

x = 4

y 2 . x = 2 + 2

x = y + 2

x = y + 2

y = 2

3y = 6

3y + 4 = 10

2y + 4 + y = 10

21y + 22 + y = 10

1123

1234

765

2 3 4 765123

(4, 2)

x

y

2x y 10

x y 2

To solve a system of equations by substitution, we first need an equation solvedfor one of its variables.



Solut ion


answer: A13 , 4 Be3x + y = 5

3x - 2y = -7

H e l p f u l H i n tDon’t forget to insertparentheses when substi-tuting for x.7 - 2y

We choose one of the equations and solve for x or y.We will solve the first equation forx by subtracting 2y from both sides.

First equation

Subtract 2y from both sides.

Since we now substitute for x in the second equation and solve for y.

Second equation

Let


Simplify.



To find x, we let in the equation

Let

The solution is Check the solution in both equations of the original system.

The following steps may be used to solve a system of equations by the substitu-tion method.

a6, 12b .

x = 6

x = 7 - 1

y 1

2. x = 7 - 2a1

2b

x = 7 - 2y

x = 7 - 2y .y =12

2 . y =12

-2y = -1

14 - 2y = 13

14 - 4y + 2y = 13

x 72y . 217 - 2y2 + 2y = 13

2x + 2y = 13

7 - 2yx = 7 - 2y ,

x = 7 - 2y

x + 2y = 7

Solving a System of Two Linear Equations by the Substitution MethodStep 1. Solve one of the equations for one of its variables.

Step 2. Substitute the expression for the variable found in step 1 into theother equation.

Step 3. Solve the equation from step 2 to find the value of one variable.

Step 4. Substitute the value found in step 3 in any equation containing bothvariables to find the value of the other variable.

Step 5. Check the proposed solution in the original system.

Concept Check Answer:No, the solution will be an ordered pair.

CONCEPT CHECKAs you solve the system you find that Is this the solution of the system?y = -5.e2x + y = -5

x - y = 5

E X A M P L E 2Solve the system:

ex + 2y = 72x + 2y = 13




E X A M P L E 4

Solve the system: L 12

x - y = 3

x = 6 + 2y

Solut ion


answer: infinite number of solutions

c -x + 3y = 6

y =13

x + 2

The second equation is already solved for x in terms of y.Thus we substitute forx in the first equation and solve for y.

First equation

Let

Arriving at a true statement such as indicates that the two linear equationsin the original system are equivalent. This means that their graphs are identical and

3 = 3

3 = 3

3 + y - y = 3

x 6 2y . 12

16 + 2y2$'%'&

- y = 3

12

x - y = 3

6 + 2y

H e l p f u l H i n tWhen solving a system of equations by the substitution method, begin by solvingan equation for one of its variables. If possible, solve for a variable that has a co-efficient of 1 or This way, we avoid working with time-consuming fractions.-1.

Solut ion To avoid introducing fractions, we will solve the second equation for y.

Second equation

Next, substitute for y in the first equation.

First equation

To find the corresponding y-value, substitute for x in the equation Then or The solution of the system is Check this solu-tion in both equations of the system.

1-2, 02 .y = 0.y = 31-22 + 6y = 3x + 6.-2

x = -2

-2x

-2=

4-2

-2x = 4

-2x - 18 = -14

7x - 9x - 18 = -14

7x - 3 13x + 62$'%'&

= -14

7x - 3yø

= -14

3x + 6

y = 3x + 6

-3x + y = 6CLASSROOM EXAMPLESolve the system:

answer: 10, -32e5x - 2y = 6

-3x + y = -3

Solve the system: e7x - 3y = -14-3x + y = 6

E X A M P L E 3

ø


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there are an infinite number of solutions of the system.Any solution of one equation isalso a solution of the other.


TEACHING TIPAsk students, “What types of linearsystems have an infinite number ofsolutions?” Have students name anequation that is equivalent to

Have students usesubstitution to solve the resultingsystem. Because a true statementresults, any solution of one equa-tion is a solution of the other.

y = 3x + 5.

6 7112345

12345

2 3 4 5123 x

y

qx y 3x 6 2y

Use substitution to solve the system.

e 6x + 12y = 5-4x - 8y = 0

E X A M P L E 5

Solut ion Choose the second equation and solve for y.

Second equationAdd 4x to both sides.


Simplify.

Now replace y with in the first equation.

First equation

Let

Simplify.Combine like terms.

The false statement indicates that this system has no solution and is inconsistent.The graph of the linear equations in the system is a pair of parallel lines.

0 = 5

0 = 5 6x + 1-6x2 = 5

y 12

x . 6x + 12 a - 12

xb = 5

6x + 12y = 5

- 12

x

y = - 12

x

8 . -8y

-8=

4x

-8

-8y = 4x -4x - 8y = 0


answer: no solution

e2x - 3y = 6-4x + 6y = 5

TEACHING TIPAsk students, “What types of linearsystems have no solutions?” Havethem name an equation whosegraph is a line parallel to

Now use substitutionto solve the resulting system. Be-cause a false statement results, nosolution of one equation will satisfy the other.

y = 3x + 5.

112345

12345

2 3 4 512345 x

y

6x 12y 5

4x 8y 0

CONCEPT CHECKDescribe how the graphs of the equations in a system appear if the system has

a. no solution b. one solution c. an infinite number of solutions


a. parallel lines,b. intersect at one point,c. identical graphs


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MENTAL MATHGive the solution of each system. If the system has no solution or an infinite number of solutions, say so. If the system as 1 solu-tion, find it.

1. 2. 3.

When solving, you When solving, you When solving, youobtain (1, 4) obtain no solution obtain infinite number of solutions

4. 5. 6.

When solving, you When solving, you When solving, youobtain (5, 0) obtain (0, 0) obtain infinite number of solutions0 = 0x = 0y = 0

ey = -2x + 54x + 2y = 10

e x + y = 07x - 7y = 0

e5x + 2y = 25x = y + 5

0 = 00 = 34x = 1

e 4x - y = 17-8x + 2y = -34

e 4x - y = 17-8x + 2y = 0

e y = 4x

-3x + y = 1

STUDY SKILLS REMINDERAre You Satisfied with Your Performance on a Partic-ular Quiz or Exam?

If not, don’t forget to analyze your quiz or exam and look forcommon errors.

Were most of your errors a result of

Carelessness? If your errors were careless, did you turn in yourwork before the allotted time expired? If so, resolve to use theentire time allotted next time. Any extra time can be spentchecking your work.

Running out of time? If so, make a point to better manageyour time on your next exam. A few suggestions are to workany questions that you are unsure of last and to check yourwork after all questions have been answered.

Not understanding a concept? If so, review that concept andcorrect your work. Remember next time to make sure that allconcepts on a quiz or exam are understood before the exam.



E X A M P L E 1

Solve the system: ex + y = 7x - y = 5

Solut ion Since the left side of each equation is equal to the right side, we add equal quantities byadding the left sides of the equations together and the right sides of the equations to-gether. If we choose wisely, this adding gives us an equation in one variable, x, which wecan solve for x.

The x-value of the solution is 6. To find the corresponding y-value, let in eitherequation of the system. We will use the first equation.

First equationLet

Solve for y.

Simplify.

The solution is (6, 1). Check this in both equations.

First Equation Second Equation

Let and True True

Thus, the solution of the system is (6, 1) and the graphs of the two equations intersectat the point (6, 1) as shown on the top of the next page.

5 = 5 7 = 7y 1 .x 6 6 - 1 5 6 + 1 7

x - y = 5 x + y = 7

y = 1

y = 7 - 6

x 6 . 6 + y = 7 x + y = 7

x = 6

First equationSecond equationAdd the equations.


x + y = 7 x - y = 5 2x = 12

x = 6

CLASSROOM EXAMPLE

Solve the system:

answer: (9, 4)

ex + y = 13x - y = 5

O b j e c t i v e

1 Use the addition method to solve a system of linear equations.

1 We have seen that substitution is an accurate way to solve a linear system. An-other method for solving a system of equations accurately is the addition or eliminationmethod. The addition method is based on the addition property of equality: addingequal quantities to both sides of an equation does not change the solution of the equa-tion. In symbols,

if A = B and C = D , then A + C = B + D .

4.3 S O LV I N G S Y S T E M S O F L I N E A R E Q UAT I O N S B Y A D D I T I O N



SOLVING SYSTEMS OF L INEAR EQUATIONS BY ADDIT ION SECTION 4.3 259

6 7 8112345

12345

2 3 4 512

(6, 1)

x

y

x y 7

x y 5

E X A M P L E 2

Solve the system: e -2x + y = 2-x + 3y = -4

Solut ion If we simply add the two equations, the result is still an equation in two variables. How-ever, our goal is to eliminate one of the variables. Notice what happens if we multiplyboth sides of the first equation by which we are allowed to do by the multiplicationproperty of equality. The system

Now add the resulting equations and the y-variable is eliminated.

Add.Divide both sides by 5.

To find the corresponding y-value, let in any of the preceding equationscontaining both variables. We use the first equation of the original system.

First equation

Let

The solution is Check this ordered pair in both equations of the originalsystem.

In Example 2, the decision to multiply the first equation by was no accident.To eliminate a variable when adding two equations, the coefficient of the variable inone equation must be the opposite of its coefficient in the other equation.

-3

1-2, -22 . y = -2

4 + y = 2x 2 . -21-22 + y = 2

-2x + y = 2

x = -2

6x - 3y = -6 -x + 3y = -4 5x = -10

x = -2

e -31-2x + y2 = -3122-x + 3y = -4

simplifies to e 6x - 3y = -6-x + 3y = -4

-3,CLASSROOM EXAMPLE

Solve the system:

answer: 1-1, 42e 2x - y = -6

-x + 4y = 17

H e l p f u l H i n tBe sure to multiply both sides of an equation by a chosen number when solvingby the addition method. A common mistake is to multiply only the side con-taining the variables.




Solve the system: e 2x - y = 78x - 4y = 1

Solut ion

H e l p f u l H i n tDon’t forget to multiplyboth sides by -4.

Multiply both sides of the first equation by and the resulting coefficient of x is the opposite of 8, the coefficient of x in the second equation. The system becomes

Now add the resulting equations.

False

When we add the equations, both variables are eliminated and we have afalse statement. This means that the system has no solution. The graphs of these equa-tions are parallel lines.

0 = -27,

-8x + 4y = -28 8x - 4y = 1

0 = -27

e -412x - y2 = -41728x - 4y = 1

simplifies to e -8x + 4y = -288x - 4y = 1

-8,-4

CLASSROOM EXAMPLE

Solve the system:

answer: no solution

e x - 3y = -2-3x + 9y = 5

E X A M P L E 4

Solve the system: e3x - 2y = 2-9x + 6y = -6

Solut ion

CLASSROOM EXAMPLE

Solve the system:

answer: infinite number of solutions

e2x + 5y = 1-4x - 10y = -2

TEACHING TIPContinue to remind students tomake sure that they multiply bothsides of an equation by the samenonzero number. If they don’t, thenew equation is not equivalent tothe old equation.

First we multiply both sides of the first equation by 3, then we add the resulting equations.

Add the equations.True

Both variables are eliminated and we have a true statement. Whenever youeliminate a variable and get the equation the system has an infinite number ofsolutions. The graphs of these equations are identical.

CONCEPT CHECKSuppose you are solving the system

You decide to use the addition method by multiplying both sides of the second equation by 2. In whichof the following was the multiplication performed correctly? Explain.

a. b. 4x - 8y = 64x - 8y = 3

e3x + 8y = -52x - 4y = 3

0 = 0,0 = 0,

e 313x - 2y2 = 3122-9x + 6y = -6

simplifies to e 9x - 6y = 6-9x + 6y = -6 0 = 0

Concept CheckAnswer: b

E X A M P L E 3

E X A M P L E 5

Solve the system: e3x + 4y = 135x - 9y = 6

Solut ion We can eliminate the variable y by multiplying the first equation by 9 and the secondequation by 4.


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SOLVING SYSTEMS OF L INEAR EQUATIONS BY ADDIT ION SECTION 4.3 261

Add the equations.

To find the corresponding y-value, we let in any equation in this example con-taining two variables. Doing so in any of these equations will give The solutionto this system is (3, 1). Check to see that (3, 1) satisfies each equation in the originalsystem.

If we had decided to eliminate x instead of y in Example 5, the first equationcould have been multiplied by 5 and the second by Try solving the original systemthis way to check that the solution is (3, 1).

The following steps summarize how to solve a system of linear equations by theaddition method.

-3.

y = 1.x = 3

e913x + 4y2 = 91132415x - 9y2 = 4162 simplifies to

e27x + 36y = 11720x - 36y = 24

47x = 141

x = 3

CLASSROOM EXAMPLE

Solve the system:

answer: (1, 2)

e4x + 5y = 143x - 2y = -1


a. multiply the second equation by 4b. possible answer:

multiply the firstequation by and the secondequation by 7

-2

Solving a System of Two Linear Equations by the Addition MethodStep 1. Rewrite each equation in standard form

Step 2. If necessary, multiply one or both equations by a nonzero num-ber so that the coefficients of a chosen variable in the system areopposites.

Step 3. Add the equations.

Step 4. Find the value of one variable by solving the resulting equationfrom Step 3.

Step 5. Find the value of the second variable by substituting the valuefound in Step 4 into either of the original equations.

Step 6. Check the proposed solution in the original system.

Ax + By = C .

CONCEPT CHECKSuppose you are solving the system

by the addition method.

a. What step(s) should you take if you wish to eliminate x when adding the equations?b. What step(s) should you take if you wish to eliminate y when adding the equations?

e -4x + 7y = 6x + 2y = 5

E X A M P L E 6

Solve the system: d -x -y

2=

52

- x

2+

y

4= 0

Solut ion We begin by clearing each equation of fractions.To do so, we multiply both sides of thefirst equation by the LCD 2 and both sides of the second equation by the LCD 4. Then

TEACHING TIPBefore attempting Example 6, re-mind students to use what theyknow to make this system as easyas possible to solve. Then ask howExample 6 could be written as asimpler system.


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CLASSROOM EXAMPLE

Solve the system:

answer: A - 172 , -

32 B

d - x

3+ y =

43

x

2-

52

y = - 12

the system

Now we add the resulting equations in the simplified system.

Add.

To find y, we could replace x with in one of the equations with two variables.

Instead, let’s go back to the simplified system and multiply by appropriate factors toeliminate the variable x and solve for y. To do this, we multiply the first equation in thesimplified system by Then the system

Add.

Check the ordered pair in both equations of the original system. The solu-

tion is a - 54

, - 52b .

a - 54

, - 52b

e -11-2x - y2 = -1152-2x + y = 0

simplifies to

e 2x + y = -5-2x + y = 0 2y = -5

y = - 52

-1.

- 54

-2x - y = 5-2x + y = 0-4x = 5 x = -

54

d 2a -x -y

2b = 2a5

2b

4a - x

2+

y

4b = 4102

simplifies to e -2x - y = 5-2x + y = 0

Suppose you have been offered two similar positions as a sales associate. Inone position, you would be paid a monthly salary of $1500 plus a 2% commis-sion on all sales you make during the month. In the other position, you wouldbe paid a monthly salary of $500 plus a 6% commission on all sales you makeduring the month. Which position would you choose? Explain your reasoning.Would knowing that the sales positions were at a car dealership affect yourchoice? What if the positions were at a shoe store?


SYSTEMS OF L INEAR EQUATIONS AND PROBLEM SOLVING SECTION 4.4 265

4.4 S Y S T E M S O F L I N E A R E Q UAT I O N S A N D P R O B L E M S O LV I N G

O b j e c t i v e

1 Use a system of equations to solve problems.

1 Many of the word problems solved earlier using one-variable equations canalso be solved using two equations in two variables. We use the same problem-solvingsteps that have been used throughout this text.The only difference is that two variablesare assigned to represent the two unknown quantities and that the problem is translat-ed into two equations.

Problem-Solving Steps1. UNDERSTAND the problem. During this step, become comfortable with

the problem. Some ways of doing this are to

Read and reread the problem.Choose two variables to represent the two unknowns.Construct a drawing, if possible.Propose a solution and check. Pay careful attention to how you checkyour proposed solution. This will help when writing equations to modelthe problem.

2. TRANSLATE the problem into two equations.3. SOLVE the system of equations.




1. UNDERSTAND. Read and reread the problem. Suppose that one number is 20.If their sum is 37, the other number is 17 because Is their difference21? No; Our proposed solution is incorrect, but we now have a betterunderstanding of the problem.

Since we are looking for two numbers, we let

number

number

2. TRANSLATE. Since we have assigned two variables to this problem, we translateour problem into two equations.In words:

T T TTranslate:

In words:

T T T

Translate:

3. SOLVE. Now we solve the system

Notice that the coefficients of the variable y are opposites. Let’s then solve by theaddition method and begin by adding the equations.

Now we let in the first equation to find y.

First equation

4. INTERPRET. The solution of the system is (29, 8).Check: Notice that the sum of 29 and 8 is the required sum. Theirdifference is the required difference.State: The numbers are 29 and 8.

29 - 8 = 21,29 + 8 = 37,

y = 37 - 29 = 8 29 + y = 37 x + y = 37

x = 29

Add the equations.


x + y = 37 x - y = 212x = 58

x =582

= 29

ex + y = 37x - y = 21

21=x - y

21istwo numbers

whose difference

37=x + y

37istwo numberswhose sum

y = second

x = first

20 - 17 = 3.20 + 17 = 37.


FINDING UNKNOWN NUMBERS

Find two numbers whose sum is 37 and whose difference is 21.

Solut ion

E X A M P L E 1

CLASSROOM EXAMPLEFind two numbers whose sum is 50 andwhose difference is 22.answer: 36 and 14

TEACHING TIPConsider beginning Example 1 byhaving students list some numberswhose sum is 37. Ask them, “Howmany such pairs exist?” Then havethem list some numbers whose dif-ference is 21. Again ask them,“How many such pairs exist?”Point out that finding numbers thatsatisfy both conditions could take awhile using the list approach. Solv-ing the problem with a system ofequations can be more efficient.



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1. UNDERSTAND. Read and reread the problem and guess a solution. Let’s sup-pose that the price of an adult’s ticket is $50 and the price of a child’s ticket is $40. Tocheck our proposed solution, let’s see if admission for 4 adults and 2 children is $374.Admission for 4 adults is 4($50) or $200 and admission for 2 children is 2($40) or $80.This gives a total admission of not the required $374. Againthough, we have accomplished the purpose of this process. We have a better under-standing of the problem. To continue, we let

price of an adult’s ticket and

price of a child’s ticket

2. TRANSLATE. We translate the problem into two equations using both variables.

In words:

T T T T T

Translate:

In words:

T T T T T

Translate:

3. SOLVE. We solve the system.

Since both equations are written in standard form, we solve by the addition method.First we multiply the second equation by so that when we add the equations weeliminate the variable A. Then the system

-2

e4A + 2C = 3742A + 3C = 285

285=3C+2A

$285isadmission for

3 childrenand

admissionfor 2 adults

374=2C+4A

$374isadmission for

2 childrenand

admissionfor 4 adults

C = the

A = the

$200 + $80 = $280,


E X A M P L E 2

SOLVING A PROBLEM ABOUT PRICES

The Cirque du Soleil show Alegria is performing locally. Matinee admission for 4adults and 2 children is $374, while admission for 2 adults and 3 children is $285.

a. What is the price of an adult’s ticket?b. What is the price of a child’s ticket?c. Suppose that a special rate of $1000 is offered for groups of 20 persons. Should a

group of 4 adults and 16 children use the group rate? Why or why not?

CLASSROOM EXAMPLEAdmission prices at a local weekend fairwere $5 for children and $7 for adults.The total money collected was $3379,and 587 people attended the fair. Howmany children and how many adults attended the fair?answer:365 children and 222 adults

Solut ion


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FINDING RATES

As part of an exercise program, Albert and Louis started walking each morning. Theylive 15 miles away from each other and decided to meet one day by walking towardone another. After 2 hours they meet. If Louis walks one mile per hour faster than Al-bert, find both walking speeds.

To find A, we replace C with 49 in the first equation.

First equationLet

4. INTERPRET.Check: Notice that 4 adults and 2 children will pay

the required amount. Also, the price for2 adults and 3 children is the requiredamount.

State: Answer the three original questions.

a. Since the price of an adult’s ticket is $69.

b. Since the price of a child’s ticket is $49.

c. The regular admission price for 4 adults and 16 children is

This is $60 more than the special group rate of $1000, so they should request the grouprate.

= $1060

41$692 + 161$492 = $276 + $784

C = 49,

A = 69,

21$692 + 31$492 = $138 + $147 = $285,41$692 + 21$492 = $276 + $98 = $374,

A =2764

=69 or $69,the adult’s ticket price.

4A = 276

4A + 98 = 374

C 49. 4A + 21492 = 374 4A + 2C = 374

C =-196-4

= 49 or $49, the children’s ticket price.

-4C = -196

e 4A + 2C = 374-212A + 3C2 = -212852 simplifies to

e 4A + 2C = 374-4A - 6C = -570 -4C = -196

E X A M P L E 3

15 miles

2y 2x

Solut ion 1. UNDERSTAND. Read and reread the problem. Let’s propose a solution anduse the formula to check. Suppose that Louis’s rate is 4 miles per hour.Since Louis’s rate is 1 mile per hour faster, Albert’s rate is 3 miles per hour. Tocheck, see if they can walk a total of 15 miles in 2 hours. Louis’s distance is

and Albert’s distance is rate Their totaltime = 3122 = 6 miles .= 4122 = 8 milesrate # time

d = r # t

Add the equations.

CLASSROOM EXAMPLETwo cars are 550 miles apart and travel-ing toward each other. They meet in 5hours. If one car’s speed is 5 miles perhour faster than the other car’s speed,find the speed of each car.answer:One car’s speed is mph and theother car’s speed is mph.57.5

52.5

TEACHING TIPBefore beginning Example 3, re-view the relationship between dis-tance, rate, and time by askingstudents the following: Suppose a car traveled at 30

mph for 4 hours, what distancedid it travel? (120 miles)

Suppose a car traveled 400miles in 8 hours, what was itsaverage speed? (50 mph)

Suppose a car traveled 240miles at 60 miles per hour, howlong did it take? (4 hours)

Describe the relationship betweendistance, rate, and time as an equation.


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distance is not the required 15 miles. Now that wehave a better understanding of the problem, let’s model it with a system of equations.

First, we let

rate in miles per hour

rate in miles per hour

Now we use the facts stated in the problem and the formula to fill in the fol-lowing chart.

d = rt

y = Louis’s

x = Albert’s

8 miles + 6 miles = 14 miles ,

Albert x 2 2x

Louis y 2 2y

r # t = d

2. TRANSLATE. We translate the problem into two equations using both variables.

In words:

T T T

Translate:

In words:

T T TTranslate:

3. SOLVE. The system of equations we are solving is

Let’s use substitution to solve the system since the second equation is solved for y.

First equation

Replace y with

4. INTERPRET. Albert’s proposed rate is 3.25 miles per hour and Louis’s pro-posed rate is 4.25 miles per hour.Check: Use the formula and find that in 2 hours, Albert’s distance is(3.25)(2) miles or 6.5 miles. In 2 hours, Louis’s distance is (4.25)(2) miles or 8.5 miles.The total distance walked is or 15 miles, the given distance.State: Albert walks at a rate of 3.25 miles per hour and Louis walks at a rate of4.25 miles per hour.

6.5 miles + 8.5 miles

d = rt

y = x + 1 = 3.25 + 1 = 4.25

x =134

= 3.25

4x = 13 2x + 2x + 2 = 15

x 1 . 2x + 21x + 12$%&

= 15

2x + 2y = 15

e2x + 2y = 15y = x + 1

x + 1=y

1 mile per hourfaster than Albert’s

isLouis’s

rate

15=2y+2x

15=Louis’sdistance

+Albert’sdistance

TEACHING TIPBefore beginning Example 4, youmay want to have students make atable showing the amounts of saltand water in 1, 5, 10, 25, 50, and 100liters of a 5% saline solution.



E X A M P L E 4FINDING AMOUNTS OF SOLUTIONS

Eric Daly, a chemistry teaching assistant, needs 10 liters of a 20% saline solution (saltwater) for his 2 P.M. laboratory class. Unfortunately, the only mixtures on hand are a5% saline solution and a 25% saline solution. How much of each solution should hemix to produce the 20% solution?

Solut ion

CLASSROOM EXAMPLEA pharmacist, needs 15 liters of a 40%alcohol solution. She currently has avail-able a 20% solution and an 80% solu-tion. How many liters of each must sheuse to make the needed 15 liters of 40%alcohol solution?answer:10 liters of the 20% alcohol solution and5 liters of the 80% alcohol solution

x y or10 liters

5% salinesolution

25% salinesolution

20% salinesolution

x liters

y litersy litersx liters

Now we use a table to organize the given data.

Concentration Rate Liters of Solution Liters of Pure Salt

First Solution 5% x 0.05x

Second Solution 25% y 0.25y

Mixture Needed 20% 10 (0.20)(10)

2. TRANSLATE. We translate into two equations using both variables.

In words:

T T TTranslate: 10=y+x

10=liters of 25%

solution+

liters of 5%solution

1. UNDERSTAND. Read and reread the problem. Suppose that we need 4 liters ofthe 5% solution. Then we need of the 25% solution. To see if thisgives us 10 liters of a 20% saline solution, let’s find the amount of pure salt in eachsolution.

T T T5% solution: 0.05 4 liters 0.2 liters

25% solution: 0.25 6 liters 1.5 liters20% solution: 0.20 10 liters 2 liters

Since not 2 liters, our proposed solution is incor-rect. But we have gained some insight into how to model and check this problem.We let

of liters of 5% solution

of liters of 25% solutiony = number

x = number

0.2 liters + 1.5 liters = 1.7 liters ,

amount ofpure salt

=amount ofsolution

*concentration

rate

10 - 4 = 6 liters


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Concept Check Answer:b

In words:

T T TTranslate:

3. SOLVE. Here we solve the system

To solve by the addition method, we first multiply the first equation by and thesecond equation by 100. Then the system

Add.

To find y, we let in the first equation of the original system.

Let

4. INTERPRET. Thus, we propose that Eric needs to mix 2.5 liters of 5% saline so-lution with 7.5 liters of 25% saline solution.Check: Notice that the required number of liters. Also, the sum ofthe liters of salt in the two solutions equals the liters of salt in the required mixture:

State: Eric needs 2.5 liters of the 5% saline solution and 7.5 liters of the 25% solution.

CONCEPT CHECKSuppose you mix an amount of a 30% acid solution with an amount of a 50% acid solution. Which ofthe following acid strengths would be possible for the resulting acid mixture?

a. 22% b. 44% c. 63%

0.125 + 1.875 = 2

0.0512.52 + 0.2517.52 = 0.2011022.5 + 7.5 = 10,

y = 7.5x 2.5. 2.5 + y = 10

x + y = 10

x = 2.5

e -251x + y2 = -25110210010.05x + 0.25y2 = 100122

simplifiesto

e -25x - 25y = -250 5x + 25y = 200

-20x = -50

x = 2.5

-25

ex + y = 100.05x + 0.25y = 2

10.2021102=0.25y+0.05x

salt inmixture

=salt in 25%

solution+

salt in 5%solution


GRAPHING L INEAR INEQUALIT IES SECTION 4.5 275

O b j e c t i v e

1 Graph a linear inequality in two variables.

1 In the next section, we continue our work with systems by solving systems oflinear inequalities. Before that section, we first need to learn to graph a single linear in-equality in two variables.

Recall that a linear equation in two variables is an equation that can be written inthe form where A, B, and C are real numbers and A and B are not both0.The definition of a linear inequality is the same except that the equal sign is replacedwith an inequality sign.

A linear inequality in two variables is an inequality that can be written in one ofthe forms:

Ax + By 6 C Ax + By … C

Ax + By 7 C Ax + By Ú C

Ax + By = C

4.5 G R A P H I N G L I N E A R I N E Q UA L I T I E S




where A, B, and C are real numbers and A and B are not both 0. Just as for linear equa-tions in x and y, an ordered pair is a solution of an inequality in x and y if replacing thevariables by coordinates of the ordered pair results in a true statement.

To graph a linear inequality in two variables, we will begin by graphing a relatedequation. For example, to graph we begin by graphing

The linear equation is graphed next. Recall that all points on the linecorrespond to ordered pairs that satisfy the equation It can be shown thatall the points above the line have coordinates that satisfy the inequality

Similarly, all points below the line have coordinates that satisfy the in-equality To see this, a few points have been selected on one side of the line.These points all satisfy as shown in the table below.x - y 6 1

x - y 7 1.x - y 6 1.

x - y = 1x - y = 1.

x - y = 1x - y = 1.x - y … 1,

112345

12345

2 3 4 512345

(0, 2)(1, 3)

(4, 2)

(3, 0)

x

y

112345

12345

2 3 4 512345(0, 1)

(1, 0)

x

y

x y 1

x y 1

x y 1

Point

Check (0, 2)True

Check True

Check True

Check True-4 6 1

-1 - 3 6 11-1, 32-2 6 1

-4 - 1-22 6 11-4, -22-3 6 1

-3 - 0 6 11-3, 02-2 6 1

0 - 2 6 1

x y<1

The graph of is in blue, and the graph of is in red.x - y 7 1x - y 6 1

The region above the line and the region below the line are called half-planes.Every line divides the plane (similar to a sheet of paper extending indefinitely in all di-rections) into two half-planes; the line is called the boundary.

How do we graph Recall that the inequality means

Thus, the graph of is the half-plane along with the boundary linex - y = 1.

x - y 6 1x - y … 1

x - y = 1 or x - y 6 1

x - y … 1x - y … 1?



E X A M P L E 1Graph: x + y 6 7

Solut ion First we graph the boundary line by graphing the equation We graph thisboundary as a dashed line because the inequality sign is and thus the points on theline are not solutions of the inequality x + y 6 7.

6 ,x + y = 7.

Graphing a Linear Inequality in Two VariablesStep 1. Graph the boundary line found by replacing the inequality sign with an

equal sign. If the inequality sign is or graph a dashed boundaryline (indicating that the points on the line are not solutions of the in-equality). If the inequality sign is or graph a solid boundary line(indicating that the points on the line are solutions of the inequality).

Step 2. Choose a point, not on the boundary line, as a test point. Substitutethe coordinates of this test point into the original inequality.

Step 3. If a true statement is obtained in Step 2, shade the half-plane thatcontains the test point. If a false statement is obtained, shade thehalf-plane that does not contain the test point.

… ,Ú

6 ,7


x - y 7 3

y

x

22468

10

2468

10

4 6 8 10246810

(0, 7)

(2, 5)

(7, 0)

(0, 0)

x

y

x y 7

Next, choose a test point, being careful not to choose a point on the boundary line. Wechoose (0, 0). Substitute the coordinates of (0, 0) into

Original inequality

Replace x with 0 and y with 0.True

Since the result is a true statement, (0, 0) is a solution of and every point inthe same half-plane as (0, 0) is also a solution. To indicate this, shade the entire half-plane containing (0, 0), as shown.

x + y 6 7,

0 6 7 0 + 0 6? 7

x + y 6 7

x + y 6 7.

22468

10

2468

10

4 6 8 10246810

(2, 5)

(0, 0)

x

y

x y 7

(0, 7)

(2, 5)

x y 7

TEACHING TIPFor Examples 1 and 2, ask studentswhat inequality describes the un-shaded region of each graph.



Solut ion We find the boundary line by graphing The boundary line is a dashed linesince the inequality symbol is We cannot use (0, 0) as a test point because it is apoint on the boundary line. We choose instead (0, 2).

Let and

True

Since the statement is true, we shade the half-plane that contains the test point (0, 2), asshown at the top of the next page.

0 6 4

y 2 .x 0 0 6? 2122 x 6 2y

6 .x = 2y .


y 6 3x

E X A M P L E 2Graph: 2x - y Ú 3

Solut ion Graph the boundary line by graphing Draw this line as a solid line sincethe inequality sign is and thus the points on the line are solutions of Once again, (0, 0) is a convenient test point since it is not on the boundary line.

Substitute 0 for x and 0 for y into the original inequality.

Let and False

Since the statement is false, no point in the half-plane containing (0, 0) is a solution.Shade the half-plane that does not contain (0, 0). Every point in the shaded half-planeand every point on the boundary line satisfies 2x - y Ú 3.

0 Ú 3y 0 .x 0 2102 - 0 Ú? 3

2x - y Ú 3

2x - y Ú 3.Ú ,2x - y = 3.


x - 4y … 4

112345

12345

2 3 4 512345

(0, 3)

x

y

2x y 3 2x y 3

(1q, 0 )

Graph: x 6 2y

H e l p f u l H i n tWhen graphing an inequality, make sure the test point is substituted into theoriginal inequality. For Example 2, we substituted the test point (0, 0) into theoriginal inequality not 2x - y = 3.2x - y Ú 3,

y

x

E X A M P L E 3

y

x


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CLASSROOM EXAMPLEGraph:

answer:

x 6 2


112345

12345

2 3 4 512345

(2, 1)

(0, 2)

(0, 0)x

y

x 2y

x 2y

Graph: 5x + 4y … 20

Solut ion We graph the solid boundary line and choose (0, 0) as the test point.

Let and

True

We shade the half-plane that contains (0, 0), as shown.

0 … 20

y 0 .x 0 5102 + 4102 …? 20

5x + 4y … 20

5x + 4y = 20

CLASSROOM EXAMPLEGraph:

answer:

3x + 2y Ú 12

6 7

67

1123

12345

2 3 4 5123

(4, 0)

(0, 5)

(0, 0)x

y

5x 4y 20

5x 4y 20

Graph: y 7 3

Solut ion We graph the dashed boundary line and choose (0, 0) as the test point. (Recallthat the graph of is a horizontal line with y-intercept 3.)

Let

False

We shade the half-plane that does not contain (0, 0), as shown on the top of the nextpage.

0 7 3

y 0 . 0 7? 3

y 7 3

y = 3y = 3

E X A M P L E 4

y

x

E X A M P L E 5

y

x


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112345

12345

2 3 4 512345(0, 0)

x

y

y 3

y 3

MENTAL MATHState whether the graph of each inequality includes its corresponding boundary line.

1. yes 2. no 3. yes 4. no

Decide whether (0, 0) is a solution of each given inequality.

5. yes 6. yes 7. no 8. no23

x +56

y 7 4x - y … -12x + 3y 6 10x + y 7 -5

x 7 0y Ú xx - y 7 -7y Ú x + 4





E X A M P L E 1

Graph the solution of the system: e 3x Ú y

x + 2y … 8

Solut ion We begin by graphing each inequality on the same set of axes. The graph of the solu-tion of the system is the region contained in the graphs of both inequalities. It is theirintersection.

First, graph The boundary line is the graph of Sketch a solidboundary line since the inequality means or The test point (1, 0)satisfies the inequality, so shade the half-plane that includes (1, 0).

3x = y .3x 7 y3x Ú y3x = y .3x Ú y .

3x y

3x y

112345

12345

2 3 4 512345 x

y

(1, 3)

(0, 0)

x 2y 8 x 2y 8

112345

12345

2 3 4 5 6 7 8 912345 x

y

Next, sketch a solid boundary line on the same set of axes. The testpoint (0, 0) satisfies the inequality so shade the half-plane that includes(0, 0). (For clarity, the graph of is shown on a separate set of axes.)

An ordered pair solution of the system must satisfy both inequalities. These solu-tions are points that lie in both shaded regions. The solution of the system is the purpleshaded region as seen on the top of the next page. This solution includes parts of bothboundary lines.

x + 2y … 8x + 2y … 8,

x + 2y = 8

TEACHING TIPSome students will have troublefinding the solution region evenwhen both inequalities of the systemhave been graphed correctly. Hereare two suggestions that may help.

1. Shade each inequality in thesystem with a different coloredpencil.

2. If shading with pencil lead, tryshading each inequality in thesystem at a different angle.

CLASSROOM EXAMPLEGraph the solution of the system:

answer:

ey - x Ú 4x + 3y … -2

y

x

O b j e c t i v e

1 Solve a system of linear inequalities.

1 In Section 4.5, we graphed linear inequalities in two variables. Just as two lin-ear equations make a system of linear equations, two linear inequalities make a systemof linear inequalities. Systems of inequalities are very important in a process called lin-ear programming. Many businesses use linear programming to find the most profitableway to use limited resources such as employees, machines, or buildings.

A solution of a system of linear inequalities is an ordered pair that satisfies eachinequality in the system. The set of all such ordered pairs is the solution set of the sys-tem. Graphing this set gives us a picture of the solution set. We can graph a system ofinequalities by graphing each inequality in the system and identifying the region ofoverlap.

4.6 S Y S T E M S O F L I N E A R I N E Q UA L I T I E S



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SYSTEMS OF L INEAR INEQUALIT IES SECTION 4.6 283

In linear programming, it is sometimes necessary to find the coordinates of thecorner point: the point at which the two boundary lines intersect. To find the point ofintersection, solve the related linear system

by the substitution method or the addition method. The lines intersect at thecorner point of the graph.

A87 , 247 B ,

e 3x = y

x + 2y = 8

Solutionregion

x 2y 8

3x y

112345

12345

2 3 4 512345 x

y

(¶, 247 )

Graphing the Solution of a System of Linear InequalitiesStep 1. Graph each inequality in the system on the same set of axes.

Step 2. The solutions of the system are the points common to the graphs ofall the inequalities in the system.

E X A M P L E 2

Graph the solution of the system: e x - y 6 2x + 2y 7 -1

Solut ion Graph both inequalities on the same set of axes. Both boundary lines are dashed linessince the inequality symbols are and The solution of the system is the regionshown by the purple shading. In this example, the boundary lines are not a part of thesolution.

7 .6

112345

12345

2 3 4 512345

Solutionregion

x

y

x 2y 1x y 2


answer:

e2x 6 y

5x + y 7 5y

x


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E X A M P L E 3

Graph the solution of the system: e -3x + 4y 6 12x Ú 2

Solut ion Graph both inequalities on the same set of axes.


answer:

e -2x + 5y 6 10y … 3

1123

123

67

45

2 3 4 512345

Solutionregion

x

y

3x 4y 12

x 2

Suppose you are a registered nurse.Today you are working at a healthfair providing free blood pressurescreenings. You measure an at-tendee’s blood pressure as 168/82(read as “168 over 82,” where thesystolic blood pressure is listed firstand the diastolic blood pressure islisted second). What would you rec-ommend that this health fair attendee do? Explain.

160

150

140

130

120

110

100

90

80

70

60

50

40

30

20

10

0

Dia

stol

ic (

mm

Hg)

Adult Blood Pressure

Systolic (mm Hg)0 20 40 60 80 100 120 140 160 180 200 220 240

Blood Pressure Category and Recommended Follow-up:Normal: recheck in 2 yearsHigh normal: recheck in 1 yearMild hypertension: confirm within 2 monthsModerate hypertension: see primary care physician within 1 monthSevere hypertension: see primary care physician within 1 weekVery severe hypertension: see primary care physician immediately

Source: National Institutes of Health and National Heart, Lung, and Blood Institute

y

x

The solution of the system is the purple shaded region, including a portion of the linex = 2.



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EXPONENTS SECTION 5.1 297

H e l p f u l H i n tBe careful when identifying the base of an exponential expression. Pay closeattention to the use of parentheses.

2 # 32 = 2 # 3 # 3 = 18-32 = -13 # 32 = -91-322 = 1-321-32 = 9The base is 3.The base is 3.The base is -3.

2 # 32-32(-322

5.1 E X P O N E N T S

O b j e c t i v e s

1 Evaluate exponential expressions.

2 Use the product rule for exponents.

3 Use the power rule for exponents.

4 Use the power rules for products and quotients.

5 Use the quotient rule for exponents, and define a number raised to the 0 power.

1 As we reviewed in Section 1.4, an exponent is a shorthand notation for repeat-ed factors. For example, can be written as The expression is called anexponential expression. It is also called the fifth power of 2, or we say that 2 is raised tothe fifth power.

The base of an exponential expression is the repeated factor. The exponent is thenumber of times that the base is used as a factor.

56 exponent

base 1-324 exponent

base

56 = 5 # 5 # 5 # 5 # 5 # 55

and 1-324 = 1-32 # 1-32 # 1-32 # 1-32('''')''''*

6 factors; each factor is 5 4 factors; each factor is -3

2525 .2 # 2 # 2 # 2 # 2TEACHING TIPAfter Example 1 ask students towrite each of the following expres-sions as an exponential expression.

Answers:1. 272. 73. 364.

5.

6. 75 3 # 52

a19b21

81

-62-3662 or 1-6227133

E X A M P L E 1Evaluate each expression.

a. b. c. d. e. f. g. 4 # 3210.523a12b4

-421-4223123

Solut ion a.

b. To raise 3 to the first power means to use 3 as a factor only once. Therefore,Also, when no exponent is shown, the exponent is assumed to be 1.

c. d.

e. f.

g.

Notice how similar is to in the example above.The difference betweenthe two is the parentheses. In the parentheses tell us that the base, or repeatedfactor, is In only 4 is the base.-42,-4.

1-422 ,1-422-42

4 # 32 = 4 # 9 = 36

10.523 = 10.5210.5210.52 = 0.125a12b4

=12

# 12

# 12

# 12

=116

-42 = -14 # 42 = -161-422 = 1-421-42 = 16

31 = 3.

23 = 2 # 2 # 2 = 8

CLASSROOM EXAMPLEEvaluate each expression.

a. b. c.d. e. f.answers:

a. 81 b. 7 c.d. e. f. 1804

9-8

-8

5 # 62A23 B2-23

1-2237134

TEACHING TIPMake sure that students under-stand the difference between and Reading the Helpful Hinton this page and working a fewmore examples should help.

Stress that to evaluate forexample, first identify the base ofthe exponent 2. The base is 5 soonly 5 is squared.

-52,

-42.1-422



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298 CHAPTER 5 EXPONENTS AND POLYNOMIALS

2 An exponent has the same meaning whether the base is a number or a variable.If x is a real number and n is a positive integer, then is the product of n factors, eachof which is x.

xn = x # x # x # x # x # Á # x('''')''''*

n factors of x

xn

E X A M P L E 2Evaluate each expression for the given value of x.

a. x is 5 b. x is -39

x2 ;2x3 ;

Solut ion a. b.

Exponential expressions can be multiplied, divided, added, subtracted, and them-selves raised to powers. By our definition of an exponent,

Also,

In both cases, notice that the result is exactly the same if the exponents are added.

This suggests the following rule.

54 # 53 = 54 + 3 = 57 and x2 # x3 = x2 + 3 = x5

= x5

= x # x # x # x # x

x2 # x3 = 1x # x2 # 1x # x # x2

54 # 53 = 15 # 5 # 5 # 525

# 15 # 5 # 523

4 factors of 5 3 factors of 5= 5 # 5 # 5 # 5 # 5 # 5 # 5

8

7 factors of 5= 57

= 250 =

99

= 1 = 2 # 125

=9

1-321-32 = 2 # 15 # 5 # 52 If x is -3,

9

x2 =9

1-322 If x is 5 , 2x3 = 2 # 1523CLASSROOM EXAMPLEEvaluate each expression for the givenvalue of x.

a. when x is 4

b. when x is

answers: a. 48 b. -2

-2x4

-8

3x2

Product Rule for ExponentsIf m and n are positive integers and a is a real number, then

am # an = am + n

For example,In other words, to multiply two exponential expressions with a common base,

keep the base and add the exponents. We call this simplifying the exponential expression.

35 # 37 = 35 + 7 = 312 .


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H e l p f u l H i n tDon’t forget that if noexponent is written, it isassumed to be 1.

E X A M P L E 3

Use the product rule to simplify.

a. b. c. d. e. f . a2 # b21-527 # 1-528y3 # y2 # y7y3 # yx4 # x642 # 45

Solut ion a.

b.

c.

d.

e.

f. Cannot be simplified because a and b are different bases.a2 # b2

1-527 # 1-528 = 1-527 + 8 = 1-5215

y3 # y2 # y7 = y3 + 2 + 7 = y12

= y4

= y3 + 1

y3 # y = y3 # y1

x4 # x6 = x4 + 6 = x10

42 # 45 = 42 + 5 = 47

CLASSROOM EXAMPLEUse the product rule to simplify.

a. b.c. d.e.answers:a. b. c.d. e. 1-3210s11

r6x1375

1-329 # 1-32s6 # s2 # s3r5 # r

x4 # x973 # 72

E X A M P L E 4

Use the product rule to simplify 12x221-3x52 .Solut ion

3 Exponential expressions can themselves be raised to powers. Let’s try to dis-cover a rule that simplifies an expression like By definition,

which can be simplified by the product rule for exponents.

1x223 = 1x221x221x22 = x2 + 2 + 2 = x6

1x223 = 1x221x221x223 factors of x2

1x223 .

Recall that means and means

Remove parentheses.

Group factors with common bases.

Simplify. = -6x7

= 2 # -3 # x2 # x5

12x221-3x52 = 2 # x2 # -3 # x5

-3 # x5 .-3x52 # x22x2

CLASSROOM EXAMPLEUse the product rule to simplify

answer: -12x1216x321-2x92 .

H e l p f u l H i n tThese examples will remind you of the difference between adding and multi-plying terms.

Addition

By the distributive property.

Cannot be combined.

Multiplication

By the product rule.

By the product rule. 17x214x22 = 7 # 4 # x # x2 = 28x1 + 2 = 28x3

15x3213x32 = 5 # 3 # x3 # x3 = 15x3 + 3 = 15x6

7x + 4x2 = 7x + 4x2

5x3 + 3x 3 = 15 + 32x3 = 8x3

5



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Notice that the result is exactly the same if we multiply the exponents.

The following property states this result.

1x223 = x2 # 3 = x6

Power Rule for ExponentsIf m and n are positive integers and a is a real number, then

1am2n = amn

For example,

To raise a power to a power, keep the base and multiply the exponents.

17225 = 72 # 5 = 710 .

E X A M P L E 5Use the power rule to simplify.

a. b. c. [1-523]71y8221x225

Solut ion a. b. c. [1-523]7 = 1-52211y822 = y8 # 2 = y161x225 = x2

# 5 = x10

CLASSROOM EXAMPLEUse the power rule to simplify.

a. b.answers: a. b. z18940

1z623194210H e l p f u l H i n tTake a moment to make sure that you understand when to apply the productrule and when to apply the power rule.

Product Rule Add Exponents Power Rule Multiply Exponents

1y622 = y6 # 2 = y12y6 # y2 = y6 + 2 = y8

1x527 = x5 # 7 = x35x5 # x7 = x5 + 7 = x12

::

4 When the base of an exponential expression is a product, the definition of still applies. To simplify for example,

means 3 factors of (xy).

Group factors with common bases.

Simplify.

Notice that to simplify the expression we raise each factor within the parenthe-ses to a power of 3.

In general, we have the following rule.

1xy23 = x3y3

1xy23 ,

= x3y3

= x # x # x # y # y # y

1xy23 1xy23 = 1xy21xy21xy21xy23 ,

xn

TEACHING TIPRemind students that although

The power of a product rule doesnot apply to the power of a sum.Have students square correctly.

1x + y22

1xy22 = x2y2 , 1x + y22 Z x2 + y2.

Power of a Product RuleIf n is a positive integer and a and b are real numbers, then

1ab2n = anbn

For example,In other words, to raise a product to a power, we raise each factor to the power.

13x25 = 35x5 .


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E X A M P L E 6Simplify each expression.

a. b. c. d. 1-5x2y3z22a13

mn3b212a231st24

Solut ion a. Use the power of a product rule.

b. Use the power of a product rule.

c.

d. Use the power of a product rule.

Use the power rule for exponents.

Let’s see what happens when we raise a quotient to a power. To simplify for example,

means 3 factors of

Multiply fractions.

Simplify.

Notice that to simplify the expression we raise both the numerator and the de-

nominator to a power of 3.

In general, we have the following.

axyb3

=x3

y3

axyb3

,

=x3

y3

=x # x # xy # y # y

a x

yb .ax

yb3

axyb3

= axyb ax

yb ax

yb

axyb3

,

= 25x4y6z2

1-5x2y3z22 = 1-522 # 1x222 # 1y322 # 1z122a1

3 mn3b2

= a13b2 # 1m22 # 1n322 =

19

m2n6

12a23 = 23 # a3 = 8a3

1st24 = s4 # t4 = s4t4

CLASSROOM EXAMPLESimplify each expression.

a. b. c.answers:

a. b. c. -8p12q6r381y4x7y7

1-2p4q2r2313y241xy27

Power of a Quotient RuleIf n is a positive integer and a and c are real numbers, then

aacbn

=an

cn , c Z 0

For example,

In other words, to raise a quotient to a power, we raise both the numerator andthe denominator to the power.

ay

7b4

=y4

74 .

E X A M P L E 7


a. b. a x3

3y5 b4am

nb7



a. Use the power of a quotient rule.

b. Use the power of a product or quotient rule.


5 Another pattern for simplifying exponential expressions involves quotients.

To simplify an expression like in which the numerator and the denominator

have a common base, we can apply the fundamental principle of fractions and dividethe numerator and the denominator by the common base factors. Assume for the re-mainder of this section that denominators are not 0.

Notice that the result is exactly the same if we subtract exponents of the common bases.

The quotient rule for exponents states this result in a general way.

x5

x3 = x5 - 3 = x2

= x2

= x # x

=x # x # x # x # x

x # x # x

x5

x3 =x # x # x # x # x

x # x # x

x5

x3 ,

=x12

81y20

a x3

3y5 b4

=1x324

34 # 1y524 , y Z 0

amnb7

=m7

n7 , n Z 0


Solut ion

CLASSROOM EXAMPLESimplify each expression.

a. b.

answers:

a. b.25x12

81y6

r6

s6

a 5x6

9y3b2a r

sb6

TEACHING TIPBefore introducing the quotientrule, have students simplify the

expressions and Then

have them write their own quotientrule.

x3

x2 .x9

x5,

x6

x,

Quotient Rule for ExponentsIf m and n are positive integers and a is a real number, then

as long as a is not 0.

am

an = am - n

For example,

In other words, to divide one exponential expression by another with a commonbase, keep the base and subtract exponents.

x6

x2 = x6 - 2 = x4 .

E X A M P L E 8Simplify each quotient.

a. b. c. d. e.2x5y2

xys2

t3

1-3251-322

47

43

x5

x2

Solut ion a. Use the quotient rule.

b. Use the quotient rule.47

43 = 47 - 3 = 44 = 256

x5

x2 = x5 - 2 = x3




c.

d. Cannot be simplified because s and t are different bases.

e. Begin by grouping common bases.

Use the quotient rule.

CONCEPT CHECKSuppose you are simplifying each expression. Tell whether you would add the exponents, subtract theexponents, multiply the exponents, divide the exponents, or none of these.

a. b. c. d.

Let’s now give meaning to an expression such as To do so, we will simplify in two ways and compare the results.

Apply the quotient rule.

Apply the fundamental principle for fractions.

Since and we define that as long as x is not 0.x0 = 1x3

x3 = 1,x3

x3 = x0

x3

x3 =x # x # xx # x # x

= 1

x3

x3 = x3 - 3 = x0

x3

x3x0 .

w45 # w9z16 + z8y15

y31x63221

= 2x4y1 or 2x4y

= 2 # 1x5 - 12 # 1y2 - 12 2x5y2

xy= 2 # x5

x1# y2

y1

s2

t3

1-3251-322 = 1-323 = -27

CLASSROOM EXAMPLESimplify each quotient.

a. b.

c. d.

answers:

a. b. 125

c. 16 d. 7a3b10

y4

7a4b11

ab

1-2214

1-2210

59

56

y7

y3


a. multiply b. subtractc. none of these d. add

Zero Exponentas long as a is not 0.a0 = 1,

In other words, any base raised to the 0 power is 1, as long as the base is not 0.

E X A M P L E 9Simplify the following expressions.

a. b. c. d. e. a 3100b0

-501-5201ab2030

Solut ion a.

b. Assume that neither a nor b is zero.

c.

d.

e.

In the next example, exponential expressions are simplified using two or more ofthe exponent rules presented in this section.

a 3100b0

= 1

-50 = -1 # 50 = -1 # 1 = -1

1-520 = 1

1ab20 = a0 # b0 = 1 # 1 = 1

30 = 1

CLASSROOM EXAMPLESimplify the following expressions.

a. b.c. d.answers:a. 1 b. 1 c. 1 d. -1

-601-62012r 2s2080


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E X A M P L E 1 0Simplify the following.

a. b. c. d.1a2b23a3b2

12x25x3

1x324xx7a -5x2

y3 b2

Solut ion a. Use the power of a product or quotient rule; then use the power rule for exponents.

b.

c. Use the power of a product rule; then use the quotient rule.

d. Begin by applying the power of a product rule to the numerator.



= a3b1 or a3b

= a6 - 3b3 - 2

=a6b3

a3b2

1a2b23a3b2 =

1a223 # b3

a3 # b2

12x25x3 =

25 # x5

x3 = 25 # x5 - 3 = 32x2

1x324xx7 =

x12 # x

x7 =x12 + 1

x7 =x13

x7 = x13 - 7 = x6

a -5x2

y3 b2

=1-5221x2221y322 =

25x4

y6

CLASSROOM EXAMPLESimplify the following.

a. b.

answers:

a. b.16x20

z1627y14

a -2x5

z4b413y823

y10

Suppose you are an auto mechanicand amateur racing enthusiast. Youhave been modifying the engine inyour car and would like to enter alocal amateur race. The racing classesdepend on the size, or displacement,of the engine.

Engine displacement can be calculated using the formula

the engine displacement in cubic centimeters (cc)

the bore or engine cylinder diameter in centimeters b = d =

d =p

4 b2sc. In the formula,

BRENTWOOD AMATEUR RACING CLUB

Racing DisplacementClass Limit

A up to 2000 ccB up to 2400 ccC up to 2650 ccD up to 3000 cc

the stroke or distance the piston travels up or down within the cylinder in centimeters

the number of cylinders the engine has.

You have made the following measurements on your modified engine: 8.4 cm bore, 7.6 cm stroke, and 6cylinders. In which racing class would you enter your car? Explain.

c = s =


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STUDY SKILLS REMINDERWhat should you do on the day of an exam?

On the day of an exam, try the following:

Allow yourself plenty of time to arrive at your classroom.

Read the directions on the test carefully.

Read each problem carefully as you take your test. Make surethat you answer the question asked.

Watch your time and pace yourself. Work the problems thatyou are most confident with first.

If you have time, check your work and answers.

Do not turn your test in early. If you have extra time, spend itdouble-checking your work.

Good luck!

MENTAL MATHState the bases and the exponents for each of the following expressions.

1. 2. 3. 4. 5.

6. 7. 8. 9. 10. 15x225x29 # 765 # 341-423-42-371-3265432

1. base: 3; exponent: 2 2. base: 5; exponent: 4 3. 4. base: 3; exponent: 7 5. base: 4; exponent: 26. 7. base: 5; exponent: 1; base: 3; exponent: 4 8. base: 9; exponent: 1; base: 7; exponent: 6base : -4; exponent : 3

base : -3; exponent : 6

9. base: 5; exponent: 1; base: x ; exponent: 2 10. base: 5x; exponent: 2


ADDING AND SUBTRACTING POLYNOMIALS SECTION 5.2 307

Expression Terms

5 5

7y37y3

9x4 , -7x , -19x4 - 7x - 1

4x2 , 3x4x2 + 3x

5.2 A D D I N G A N D S U B T R AC T I N G P O LY N O M I A L S

O b j e c t i v e s

1 Define monomial, binomial, trinomial, polynomial, and degree.

2 Find the value of a polynomial given replacement values for the variables.

3 Simplify a polynomial by combining like terms.

4 Add and subtract polynomials.

1 In this section, we introduce a special algebraic expression called a polynomial.Let’s first review some definitions presented in Section 2.1.

Recall that a term is a number or the product of a number and variables raised topowers. The terms of the expression are and 3x. The terms of the expres-sion are and -1.9x4 , -7x ,9x4 - 7x - 1

4x24x2 + 3x




The numerical coefficient of a term, or simply the coefficient, is the numerical factorof each term. If no numerical factor appears in the term, then the coefficient is understoodto be 1. If the term is a number only, it is called a constant term or simply a constant.

Term Coefficient

1

3

3 (constant) 3

-1-x2y

-4-4x

3x2

x5

For example,

is a polynomial. Notice that this polynomial is written in descending powers of x be-cause the powers of x decrease from left to right. (Recall that the term 1 can be thoughtof as )

On the other hand,

is not a polynomial because it contains an exponent, that is not a whole number.(We study negative exponents in Section 5 of this chapter.)

Some polynomials are given special names.

A monomial is a polynomial with exactly one term.A binomial is a polynomial with exactly two terms.A trinomial is a polynomial with exactly three terms.

The following are examples of monomials, binomials, and trinomials. Each ofthese examples is also a polynomial.

-5,

x-5 + 2x - 3

1x0 .

x5 - 3x3 + 2x2 - 5x + 1

PolynomialA polynomial in x is a finite sum of terms of the form where a is a realnumber and n is a whole number.

axn ,

POLYNOMIALS

Monomials Binomials Trinomials None of These

4 x6 + x4 - x3 + 1-q4 + q3 - 2q4x2 - 7

-y5 + y4 - 3y3 - y2 + yx5 + 7x2 - x3p + 2-3z

5x3 - 6x2 + 3x - 6x2 + 4xy + y2x + yax2

Each term of a polynomial has a degree.

Degree of a TermThe degree of a term is the sum of the exponents on the variables containedin the term.



E X A M P L E 1Find the degree of each term.

a. b. c. 25x3yz-3x2

Solut ion a. The exponent on x is 2, so the degree of the term is 2.

b. can be written as The degree of the term is the sum of its exponents,so the degree is or 5.

c. The constant, 2, can be written as (since ).The degree of 2 or is 0.

From the preceding, we can say that the degree of a constant is 0.Each polynomial also has a degree.

2x0x0 = 12x0

3 + 1 + 15x3y1z1 .5x3yz

CLASSROOM EXAMPLEFind the degree of each term.

a. b. c. 5answers: a. 3 b. 8 c. 0

2x5yz2-7y3

E X A M P L E 2Find the degree of each polynomial and tell whether the polynomial is a monomial, bi-nomial, trinomial, or none of these.

a. b. c. 7x + 3x3 + 2x2 - 115x - 10-2t2 + 3t + 6

Solut ion a. The degree of the trinomial is 2, the greatest degree of any of itsterms.

b. The degree of the binomial or is 1.

c. The degree of the polynomial is 3.7x + 3x3 + 2x2 - 1

15x1 - 1015x - 10

-2t2 + 3t + 6

CLASSROOM EXAMPLEFind the degree of each polynomial andtell whether the polynomial is a monomi-al, binomial, trinomial, or none of these.a.

b.c.answers:a. binomial, 1 b. none of these, 6c. trinomial, 2

10x2 - 6x - 6

9x - 3x6 + 5x4 + 2

-6x + 14

E X A M P L E 3Complete the table for the polynomial

Use the table to give the degree of the polynomial.

7x2y - 6xy + x2 - 3y + 7

Solut ion

2 Polynomials have different values depending on replacement values for thevariables.

The degree of the polynomial is 3.

CLASSROOM EXAMPLEComplete the table for the polynomial

answer:Term Numerical Degree of

Coefficient Term4

4ab 4 22

a 1 10-5-5

-2-2b2

-3-3a2b2

-3a2b2 + 4ab - 2b2 + a - 5.

Degree of a PolynomialThe degree of a polynomial is the greatest degree of any term of the polynomial.

Numerical DegreeTerm Coefficient of Term

7 32

1 21

7 7 0-3-3y

x2-6-6xy

7x2y




E X A M P L E 4

Find the value of the polynomial when x = -2.3x2 - 2x + 1

Solut ion Replace x with and simplify.

Many physical phenomena can be modeled by polynomials.

= 17

= 12 + 4 + 1

= 3142 + 4 + 1

3x2 - 2x + 1 = 31-222 - 21-22 + 1

-2

CLASSROOM EXAMPLEFind the value of the polynomial

when

answer: -25x = -1.6x2 + 11x - 20

TEACHING TIPFor Example 5, encourage students to think about how the given equationrelates to the problem. Here are some questions you could pose to get thestudents thinking.What does 1821 represent in the polynomial? What poly-nomial would you use if the building were only 950 feet tall? Do you thinkthis polynomial will give a good estimate of the height of the object for allvalues of t? (No, once the object hits the ground the polynomial does not apply.)

E X A M P L E 5

FINDING THE HEIGHT OF A DROPPED OBJECT

The CN Tower in Toronto, Ontario, is 1821 feet tall and is the world’s tallest self-sup-porting structure. An object is dropped from the Skypod of the Tower which is at 1150feet. Neglecting air resistance, the height of the object at time t seconds is given by thepolynomial Find the height of the object when second and whent = 7 seconds.

t = 1-16t2 + 1150.

Solut ion To find each height, we evaluate the polynomial when and when

Replace t with 1.

The height of the object at 1 second is 1134 feet.

Replace t with 7.

The height of the object at 7 seconds is 366 feet.

3 Polynomials with like terms can be simplified by combining the like terms. Re-call that like terms are terms that contain exactly the same variables raised to exactlythe same powers.

= 366

= -784 + 1150

= -161492 + 1150

-16t2 + 1150 = -161722 + 1150

= 1134

= -16 + 1150

= -16112 + 1150

-16t2 + 1150 = -161122 + 1150

t = 7.t = 1

1134 ft

t 1

366 ft

t 7

CLASSROOM EXAMPLEFind the height of the object in Example5 when seconds and when seconds.answer: 1006 ft; 574 ft

t = 6t = 3

Like Terms Unlike Terms

3x, 3y

y, 2y

6st2 , 4s2t12

a2b , -a2b

-2x2 , -5x

5x2 , -7x2

Only like terms can be combined. We combine like terms by applying the distributiveproperty.



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E X A M P L E 6Simplify each polynomial by combining any like terms.

a. b. c. d.25

x4 + 23

x3 - x2 + 110

x4 - 16

x311x2 + 5 + 2x2 - 7x + 3x2-3x + 7x

Solut ion a.

b. These terms cannot be combinedbecause x and are not like terms.

c.

Combine like terms.

d.

CONCEPT CHECKWhen combining like terms in the expression which of the following is the proper result?

a. b. c. d. -11x4-11x-8x2 - 3x-11x2

5x - 8x2 - 8x ,

=12

x4 +12

x3 - x2

=510

x4 +36

x3 - x2

= a 410

+110bx4 + a4

6-

16bx3 - x2

= a25

+110bx4 + a2

3-

16bx3 - x2

25

x4 +23

x3 - x2 +110

x4 -16

x3

= 13x2 - 2

11x2 + 5 + 2x2 - 7 = 11x2 + 2x2 + 5 - 7

3x2x + 3x2

-3x + 7x = 1-3 + 72x = 4x

CLASSROOM EXAMPLESimplify each polynomial by combiningany like terms.

a.b.c.d.

answers:

a. b.c.d. -

314 x3 + 1

8 x + 2

23x2 - 7x - 15

8x34y2 - 6

27 x3 - 1

4 x + 2 - 12 x3 + 3

8 x

23x2 - 6x - x - 15

7x3 + x3

14y2 + 3 - 10y2 - 9

Answer to Concept Check:b

E X A M P L E 7Combine like terms to simplify.

-9x2 + 3xy - 5y2 + 7xy

Solut ion = -9x2 + 10xy - 5y2

-9x2 + 3xy - 5y2 + 7xy = -9x2 + 13 + 72xy - 5y2

H e l p f u l H i n tThis term can be written as 10xy or 10yx.

CLASSROOM EXAMPLECombine like terms to simplify.a.b.

answers:a.b. 3x2y2 + y2 + 6x2

10ab - 6a2 + 8b2

+ x2 - y2 + 5x27x2y2 + 2y2 - 4y2x211ab - 6a2 - ba + 8b2

E X A M P L E 8Write a polynomial that describes the total area of the squares and rectangles shownbelow. Then simplify the polynomial.

Solut ion

Area: x # 2x+4 # x+3 # 3+3 # x+x # x

x 2xxx 3

3 4 xx 3

Recall that the areaof a rectangle islength times width.

q



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Combine like terms.

4 We now practice adding and subtracting polynomials.

= 3x2 + 7x + 9 = x2 + 3x + 9 + 4x + 2x2

Adding PolynomialsTo add polynomials, combine all like terms.

E X A M P L E 9

Add and 1-2x2 + x + 32 .1-2x2 + 5x - 12

CLASSROOM EXAMPLEWrite a polynomial that describes thetotal area. Then simplify.

answer: 2x2 + 13x + 20

x

x x

5x

5 4x

8x

Solut ion

= -4x2 + 6x + 2

= 1-2x2 - 2x22 + 15x + 1x2 + 1-1 + 32 1-2x2 + 5x - 12 + 1-2x2 + x + 32 = -2x2 + 5x - 1 - 2x2 + x + 3

CLASSROOM EXAMPLEAdd and

answer: -x2 - x1-6x2 + x - 12 .15x2 - 2x + 12

E X A M P L E 1 0

Add: 14x3 - 6x2 + 2x + 72 + 15x2 - 2x2 .Solut ion

Polynomials can be added vertically if we line up like terms underneath oneanother.

= 4x3 - x2 + 7

= 4x3 + 1-6x2 + 5x22 + 12x - 2x2 + 7

14x3 - 6x2 + 2x + 72 + 15x2 - 2x2 = 4x3 - 6x2 + 2x + 7 + 5x2 - 2x

CLASSROOM EXAMPLEAdd

answer: 3x5 - 4x3 - 1

+ 13x3 - 2x2 .13x5 - 7x3 + 2x - 12

E X A M P L E 1 1

Add using the vertical format.17y3 - 2y2 + 72 and 16y2 + 12Solut ion Vertically line up like terms and add.

To subtract one polynomial from another, recall the definition of subtraction.To sub-tract a number, we add its opposite: To subtract a polynomial, we alsoadd its opposite. Just as is the opposite of b, is the opposite of 1x2 + 52 .-1x2 + 52-b

a - b = a + 1-b2 .

7y3 - 2y2 + 7 6y2 + 17y3 + 4y2 + 8

CLASSROOM EXAMPLEAdd and using the vertical format.answer: 9y2 - 2y + 8

14y + 3219y2 - 6y + 52

E X A M P L E 1 2Subtract: 15x - 32 - 12x - 112 .

Solut ion From the definition of subtraction, we have

Add the opposite.

Apply the distributive property. = 15x - 32 + 1-2x + 112 15x - 32 - 12x - 112 = 15x - 32 + [-12x - 112]



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ADDING AND SUBTRACTING POLYNOMIALS SECTION 5.2 313 H e l p f u l H i n t

Notice the sign of eachterm is changed.

CLASSROOM EXAMPLESubtract answer: 5x + 8

19x + 52 - 14x - 32 .

E X A M P L E 1 3

Subtract: 12x3 + 8x2 - 6x2 - 12x3 - x2 + 12 .

Subtracting PolynomialsTo subtract two polynomials, change the signs of the terms of the polynomialbeing subtracted and then add.

= 3x + 8

= 15x - 2x2 + 1-3 + 112

Solut ion First, change the sign of each term of the second polynomial and then add.

Combine like terms. = 9x2 - 6x - 1

= 2x3 - 2x3 + 8x2 + x2 - 6x - 1

12x3 + 8x2 - 6x2 - 12x3 - x2 + 12 = 12x3 + 8x2 - 6x2 + 1-2x3 + x2 - 12

E X A M P L E 1 4

Subtract from using the vertical format.1-3y2 - 2y + 11215y2 + 2y - 62

Solut ion Arrange the polynomials in vertical format, lining up like terms.

-3y2 - 2y + 11 -3y2 - 2y + 11-15y2 + 2y - 62 -5y2 - 2y + 6

-8y2 - 4y + 17

E X A M P L E 1 5

CLASSROOM EXAMPLESubtract

answer: 8x3 - 11x2 + 12- 1-4x3 + x2 - 112 .

14x3 - 10x2 + 12

CLASSROOM EXAMPLESubtract from

using the vertical format.answer: -4y2 + y + 512y2 - 2y + 72

16y2 - 3y + 22

TEACHING TIPBefore Example 14, ask students tocorrectly translate the following“Subtract 8 from 10.” “Subtract afrom b.”

Subtract from the sum of 18z + 112 and 19z - 22 .15z - 72Solut ion Notice that is to be subtracted from a sum. The translation is

Remove grouping symbols.

Group like terms.

Combine like terms. = 12z + 16

= 8z + 9z - 5z + 11 - 2 + 7

= 8z + 11 + 9z - 2 - 5z + 7

[18z + 112 + 19z - 22] - 15z - 7215z - 72

CLASSROOM EXAMPLESubtract from the sum of

and answer: 13x - 9

112x - 52 .14x - 3213x + 12

E X A M P L E 1 6Add or subtract as indicated.

a.

b. 19a2b2 + 6ab - 3ab22 - 15b2a + 2ab - 3 - 9b2213x2 - 6xy + 5y22 + 1-2x2 + 8xy - y22

Solut ion a.

Combine like terms. = x2 + 2xy + 4y2

= 3x2 - 6xy + 5y2 - 2x2 + 8xy - y2

13x2 - 6xy + 5y22 + 1-2x2 + 8xy - y22

CLASSROOM EXAMPLEAdd or subtract as indicated.a.

b.

answers:a.b. 6x2y2 - 4 - 9x2y + 8xy2 - y2

5a2 - 2ab + 13b2

1-x2y2 + 7 - 8xy2 + 2y2215x2y2 + 3 - 9x2y + y22 -1-3a2 + ab - 7b2212a2 - ab + 6b22 -


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b. Change the sign of each termof the polynomial beingsubtracted.

Combine like terms. = 9a2b2 + 4ab - 8ab2 + 3 + 9b2

= 9a2b2 + 6ab - 3ab2 - 5b2a - 2ab + 3 + 9b2

19a2b2 + 6ab - 3ab22 - 15b2a + 2ab - 3 - 9b22

MENTAL MATHSimplify by combining like terms if possible.

1. 2. 3. 4.

5. 7x 6. 6z 7. 8. 8p3 + 3p28p3 + 3p25m2 + 2m5m2 + 2m7z - zx + 6x

2y521y5 - 19y57y34y3 + 3y313m56m5 + 7m5-14y-9y - 5y

4. 4; trinomial 5. 6; trinomial 6. 5; trinomial 13.–16. answers may vary 20. 22.23. 24. 126 ft; answers may vary 28. 47. 1x2 + 7x + 42 ft3k3 + 11-146 ft ; the object has reached the ground

1a2 -6; 1b2 -11a2 -4; 1b2 -3


MULTIPLYING POLYNOMIALS SECTION 5.3 317

O b j e c t i v e s

1 Use the distributive property to multiply polynomials.

2 Multiply polynomials vertically.

1 To multiply polynomials, we apply our knowledge of the rules and definitionsof exponents.

To multiply two monomials such as and use the associative andcommutative properties and regroup. Remember that to multiply exponential expres-sions with a common base we add exponents.

1-5x321-2x42 = 1-521-221x321x42 = 10x7

1-2x42 ,1-5x32

5.3 M U LT I P LY I N G P O LY N O M I A L S

E X A M P L E SMultiply.

1. Use the commutative and associative properties.Multiply.

2.

3.

To multiply polynomials that are not monomials, use the distributive property.

= 12x6

= 1-1221-121x5 # x2 1-12x521-x2 = 1-12x521-1x2

= -14x7

-7x2 # 2x5 = 1-7 # 221x2 # x52 = 24x2

6x # 4x = 16 # 421x # x2CLASSROOM EXAMPLEMultiply.

a. b.answers: a. b. 9y556y2

1-9y421-y27y # 8y

E X A M P L E 4Use the distributive property to find each product.

a. b. -3x215x2 + 6x - 125x12x3 + 62Solut ion a. Use the distributive property.

Multiply.

b.


Multiply.

We also use the distributive property to multiply two binomials. To multiplyby distribute the factor first.

Distribute

Apply distributive property a second time.Multiply.

Combine like terms.This idea can be expanded so that we can multiply any two polynomials.

= x2 + 4x + 3

= x2 + x + 3x + 3

= x1x2 + x112 + 31x2 + 31121x 32 . 1x + 321x + 12 = x1x + 12 + 31x + 12

1x + 321x + 12 ,1x + 32 = -15x4 - 18x3 + 3x2

= 1-3x2215x22 + 1-3x2216x2 + 1-3x221-12-3x215x2 + 6x - 12

= 10x4 + 30x

5x12x3 + 62 = 5x12x32 + 5x162CLASSROOM EXAMPLEUse the distributive property to find eachproduct.a.b.answers:a.b. -6x5 + 2x4 - 4x3

56x5 + 8x

-2x313x2 - x + 228x17x4 + 12

TEACHING TIPExample 4(b) can be illustratedwith an area diagram. Note thatthe result of the multiplication iswritten inside the rectangles.

5x2 +6x -1 -3x2 -15x4 -18x3 +3x2




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To Multiply Two PolynomialsMultiply each term of the first polynomial by each term of the second poly-nomial, and then combine like terms.

E X A M P L E 5

Multiply 13x + 2212x - 52 .Solut ion Multiply each term of the first binomial by each term of the second.

Multiply.

Combine like terms. = 6x2 - 11x - 10

= 6x2 - 15x + 4x - 10

13x + 2212x - 52 = 3x12x2 + 3x1-52 + 212x2 + 21-52

TEACHING TIPExample 5 can be illustrated withan area diagram.

3x 2 2x

-5 6x2

-15x 4x

-10

CLASSROOM EXAMPLEMultiply answer: 12x2 - x - 20

14x + 5213x - 42 .

E X A M P L E 6

Multiply 12x - y22 .

Solut ion Recall that so Multiply each term of thefirst polynomial by each term of the second.

Multiply.

Combine like terms. = 4x2 - 4xy + y2

= 4x2 - 2xy - 2xy + y2

12x - y212x - y2 = 2x12x2 + 2x1-y2 + 1-y212x2 + 1-y21-y212x - y22 = 12x - y212x - y2 .a2 = a # a ,

CLASSROOM EXAMPLEMultiply answer: 9x2 - 12xy + 4y2

13x - 2y22 .

E X A M P L E 7

Multiply by 13t2 - 4t + 22 .1t + 22Solut ion Multiply each term of the first polynomial by each term of the second.

Combine like terms. = 3t3 + 2t2 - 6t + 4

= 3t3 - 4t2 + 2t + 6t2 - 8t + 4

1t + 2213t2 - 4t + 22 = t13t22 + t1-4t2 + t122 + 213t22 + 21-4t2 + 2122CLASSROOM EXAMPLEMultiply .answer: 2x3 + x2 - 11x + 12

1x + 3212x2 - 5x + 42

E X A M P L E 8

Multiply 13a + b23 .

Solut ion Write as

= 27a3 + 27a2b + 9ab2 + b3

= 27a3 + 18a2b + 3ab2 + 9a2b + 6ab2 + b3

= 19a2 + 6ab + b223a + 19a2 + 6ab + b22b = 19a2 + 6ab + b2213a + b2

13a + b213a + b213a + b2 = 19a2 + 3ab + 3ab + b2213a + b213a + b213a + b213a + b2 .13a + b23

CLASSROOM EXAMPLEMultiply .answer: 8x3 - 12x2y + 6xy2 - y3

12x - y23


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MULTIPLYING POLYNOMIALS SECTION 5.3 319

Multiply Use a vertical format.13y2 - 4y + 121y + 22 .Solut ion Step 1. Multiply 2 by each term of the top polynomial. Write the first partial product

below the line.

Partial product

Step 2. Multiply y by each term of the top polynomial. Write this partial product un-derneath the previous one, being careful to line up like terms.

Partial productPartial product

Step 3. Combine like terms of the partial products.

Combine like terms.

Thus,

When multiplying vertically, be careful if a power is missing, you may want toleave space in the partial products and take care that like terms are lined up.

1y + 2213y2 - 4y + 12 = 3y3 + 2y2 - 7y + 2.

3y2 - 4y + 1* y + 2

6y2 - 8y + 23y3 - 4y2 + y 3y3 + 2y2 - 7y + 2

3y2 - 4y + 1* y + 2

6y2 - 8y + 23y3 - 4y2 + y

3y2 - 4y + 1* y + 2

6y2 - 8y + 2

CLASSROOM EXAMPLEMultiply vertically

answer:3y4 - 12y3 + 16y2 - 4y + 5

13y2 + 121y2 - 4y + 52 .

E X A M P L E 9

TEACHING TIPBefore doing Example 9, you maywish to review vertical multiplica-tion using the following example.

Try to use similar wording whenexplaining this example and Example 9.

134* 25

670268 3350

E X A M P L E 1 0

Multiply Use a vertical format.12x3 - 3x + 421x2 + 12 .Solut ion

Leave space for missing powers of x.

Combine like terms.

2x3 - 3x + 4* x2 + 1

2x3 - 3x + 42x5 - 3x3 + 4x2 2x5 - x3 + 4x2 - 3x + 4

CLASSROOM EXAMPLEFind the product of and

using the vertical format.answer:12x4 + 21x3 - 17x2 - 4x + 2

13x2 + 6x - 2214x2 - x - 12

MENTAL MATHFind the following products mentally. See Examples 1 through 3.

1. 5x(2y) 10xy 2. 7a(4b) 28ab 3. 4.

5. 6. 7. 8.

Simplify, if possible.9. 10. 11. 12. 11. cannot simplifya3a5

a2a2 + a5a101a225a7a2 # a5

32x8-8x1-4x72-27x5-9x3 # 3x215a45a213a2218x36x13x22z5z # z4x7x2 # x5

2 Another convenient method for multiplying polynomials is to use a verticalformat similar to the format used to multiply real numbers. We demonstrate thismethod by multiplying by 1y + 22 .13y2 - 4y + 12



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5.4 S P E C I A L P R O D U C T S

O b j e c t i v e s

1 Multiply two binomials using the FOIL method.

2 Square a binomial.

3 Multiply the sum and difference of two terms.

1 In this section, we multiply binomials using special products. First, a specialorder for multiplying binomials called the FOIL order or method is introduced. Thismethod is demonstrated by multiplying by

F stands for the product of the First terms.F

O stands for the product of the Outer terms.O

I stands for the product of the Inner terms.I

L stands for the product of the Last terms.L

F O I L

Combine like terms.

CONCEPT CHECK:Multiply using methods from the last section. Show that the product is still6x2 + 17x + 5.

13x + 1212x + 52

= 6x2 + 17x + 5

13x + 1212x + 52 = 6x2 + 15x + 2x + 5

13x + 1213x + 52112152 = 5

13x + 1212x + 5211212x2 = 2x

13x + 1212x + 5213x2152 = 15x

13x + 1212x + 5213x212x2 = 6x2

12x + 52 .13x + 12

TEACHING TIPPoint out that the special productsin this section are shortcuts for mul-tiplying binomials.They can all beworked out by the method in theprevious section in which each termin the first binomial is multiplied byevery term in the second binomial.

Multiply by the FOIL method.1x - 321x + 42Solut ion

F O I L

Combine like terms. = x2 + x - 12 = x2 + 4x - 3x - 12

(x - 3) ( x + 4) = (x)(x) + (x)(4) + (-3)(x) + (-3)(4)CLASSROOM EXAMPLEMultiply answer: x2 + 2x - 35

1x + 721x - 52 .

Multiply by the FOIL method.15x - 721x - 22

E X A M P L E 1

FF

IIOO

LL

E X A M P L E 2

Solut ionF O I L

Combine like terms. = 5x2 - 17x + 14

= 5x2 - 10x - 7x + 14

(5x - 7)(x - 2) = 5x(x) + 5x(-2) + (-7)(x) + (-7)(-2)

FF

II

OO

LL

CLASSROOM EXAMPLEMultiply answer: 6x2 - 25x + 4

16x - 121x - 42 .

Concept Check Answer:Multiply and simplify: 3x(2x + 5) + 1(2x + 5)




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SPECIAL PRODUCTS SECTION 5.4 323

E X A M P L E 3Multiply 21y + 6212y - 12 .

Solut ionF O I L

Now use the distributive property.

2 Now, try squaring a binomial using the FOIL method.

= 4y2 + 22y - 12

= 212y2 + 11y - 62 21y + 6212y - 12 = 212y2 - 1y + 12y - 62

CLASSROOM EXAMPLEMultiply answer: 9x2 - 33x - 12

31x - 4213x + 12 .

E X A M P L E 4

Multiply 13y + 122 .

Solut ionF O I L

= 9y2 + 6y + 1

= 9y2 + 3y + 3y + 1 = 13y213y2 + 13y2112 + 113y2 + 1112

13y + 122 = 13y + 1213y + 12

Notice the pattern that appears in Example 4.

is the first term of the binomialsquared.

6y is 2 times the product of both termsof the binomial.

1 is the second term of the binomial squared.

This pattern leads to the following, which can be used when squaring a binomial. Wecall these special products.

1122 1.

12213y2112 6y .

13y22 9y2 .9y213y + 122 = 9y2 + 6y + 1

CLASSROOM EXAMPLEMultiply answer: 4x2 + 20x + 25

12x + 522 .

TEACHING TIPConsider having students discoverpatterns for squaring binomialsthemselves. Before doing Exam-ple 4, you may want to have stu-dents multiply Thenhave them multiply Askif they notice any relationship be-tween the problem and its solution.

14x - 322 .14x + 322 .

Squaring a BinomialA binomial squared is equal to the square of the first term plus or minustwice the product of both terms plus the square of the second term.

1a - b22 = a2 - 2ab + b2

1a + b22 = a2 + 2ab + b2

This product can be visualized geometrically.

The area of the large square is side side.

The area of the large square is also the sum of theareas of the smaller rectangles.

Thus, 1a + b22 = a2 + 2ab + b2 .

Area = a2 + ab + ab + b2 = a2 + 2ab + b2

Area = 1a + b21a + b2 = 1a + b22#

a

a

b

b

b2

a2

ab

ab

a b

a b

TEACHING TIPNow is probably a good time toonce again remind students that

Point out thepartitioned square illustration inthis section. It may help students ifthey visually see the product.

1x + y22 Z x2 + y2 .



E X A M P L E 5Use a special product to square each binomial.

a. b. c. d. 1x2 - 7y2212x + 5221p - q221t + 222

Solut ion

a.

b.

c.

d. 17y22 = x4 - 14x2y + 49y2+21x2217y2-1x222=1x2 - 7y2252 = 4x2 + 20x + 25+212x2152+12x22=12x + 522q2 = p2 - 2pq + q2+21p21q2-p2=1p - q2222 = t2 + 4t + 4+21t2122+t2=1t + 222

øøTTΩ

secondterm

squaredplus

twice theproduct ofthe terms

plusor

minus

firstterm

squaredCLASSROOM EXAMPLEMultiply.

a. b.answers:a.

b. x4 - 6x2y + 9y2

y2 + 6y + 9

1x2 - 3y221y + 322

H e l p f u l H i n tNotice that

The middle term 2ab is missing.

Likewise,

1a - b22 = 1a - b21a - b2 = a2 - 2ab + b2

1a - b22 Z a2 - b2

1a + b22 = 1a + b21a + b2 = a2 + 2ab + b2

1a + b22 Z a2 + b2

Another special product is the product of the sum and difference of the same twoterms, such as Finding this product by the FOIL method, we see apattern emerge.

F O I L

Notice that the middle two terms subtract out. This is because the Outer product is theopposite of the Inner product. Only the difference of squares remains.

3

= x2 - y2

(x + y) (x - y) = x2 - xy + xy - y2

1x + y21x - y2 .

FF

II

OO

L

Multiplying the Sum and Difference of Two TermsThe product of the sum and difference of two terms is the square of the first term minus the square of the second term.

1a + b21a - b2 = a2 - b2



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SPECIAL PRODUCTS SECTION 5.4 325

E X A M P L E 6Use a special product to multiply.

a. b. c.

d. e. 13x2 - 5y213x2 + 5y212p - q212p + q2ax -

14b ax +

14b16t + 7216t - 7241x + 421x - 42

Solut ion

a.

b.

c.

d.

e. 13x2 - 5y213x2 + 5y2 = 13x222 - 15y22 = 9x4 - 25y2

12p - q212p + q2 = 12p22 - q2 = 4p2 - q2

ax -14b ax +

14b = x2 - a1

4b2

= x2 -116

72 = 36t2 - 49-16t + 7216t - 72 = 16t22422 = 41x2 - 162 = 4x2 - 64-41x + 421x - 42 = 41x2

øTT

secondterm

squaredminus

firstterm

squared

CONCEPT CHECKMatch each expression on the left to the equivalent expression or expressions in the list below.

a. b. c. d. e.

Let’s now practice multiplying polynomials in general. If possible, use a specialproduct.

a2 + 2ab + b2a2 - 2ab + b2a2 + b2a2 - b21a + b21a + b21a + b21a - b21a + b22

CLASSROOM EXAMPLEMultiply.a.b.c.answers:a. b.c. 4x4 - 36y2

x2 - 19x2 - 49

12x2 - 6y212x2 + 6y2Ax - 1

3 B Ax + 13 B

1x + 721x - 72

TEACHING TIPPoint out to students that the abili-ty to recognize the patterns theydiscovered in this section will helpthem factor binomials in the nextchapter.

Concept Check Answer:a or e, b

E X A M P L E 7Multiply.

a. b. c.d. e. 1a - 321a2 + 2a - 121y4 + 2213y2 - 12

1y - 0.621y + 0.6217x + 4221x - 5213x + 42

Solut ion a. FOIL.

b. Squaring a binomial.

c. Multiplying the sum and difference of 2 terms.

d.

e. I’ve inserted this product as a reminder that since it is not a binomial times a bino-mial, the FOIL order may not be used.

Multiplying each termof the binomial byeach term of thetrinomial.

= a3 - a2 - 7a + 3 = a3 + 2a2 - a - 3a2 - 6a + 3

1a - 321a2 + 2a - 12 = a1a2 + 2a - 12 - 31a2 + 2a - 12

1y4 + 2213y2 - 12 = 3y6 - y4 + 6y2 - 2

1y - 0.621y + 0.62 = y2 - 10.622 = y2 - 0.36

= 49x2 + 56x + 16

17x + 422 = 17x22 + 217x2142 + 42

= 3x2 - 11x - 20

1x - 5213x + 42 = 3x2 + 4x - 15x - 20


a.b.

answers:

a.b. 5x6 - 11x4 + 5x2 - 11

49x2 - 14x + 1

1x4 + 1215x2 - 11217x - 122



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MENTAL MATHAnswer each exercise true or false.

1. false 2. For the product of the first terms is true

3. false 4. The product is a polynomial of degree 5. false1x - 121x3 + 3x - 121x + 421x - 42 = x2 + 16

2x2 .1x + 6212x - 121x + 422 = x2 + 16



5.5 N E G AT I V E E X P O N E N T S A N D S C I E N T I F I C N OTAT I O N

O b j e c t i v e s

1 Evaluate numbers raised to negative integer powers.

2 Use all the rules and definitions for exponents to simplify exponential expressions.

3 Write numbers in scientific notation.

4 Convert numbers from scientific notation to standard form.

1 Our work with exponential expressions so far has been limited to exponentsthat are positive integers or 0. Here we expand to give meaning to an expression like

Suppose that we wish to simplify the expression If we use the quotient rulefor exponents, we subtract exponents:

x2

x5 = x2 - 5 = x-3 , x Z 0

x2

x5 .

x-3 .



NEGATIVE EXPONENTS AND SCIENTIF IC NOTATION SECTION 5.5 329

But what does mean? Let’s simplify using the definition of

Divide numerator and denominator by common factors by applying the fundamental principle for fractions.

If the quotient rule is to hold true for negative exponents, then must equal

From this example, we state the definition for negative exponents.

1

x3 .x-3

=1

x3

=x # x

x # x # x # x # x

x2

x5 =x # x

x # x # x # x # x

xn .x2

x5x-3

Negative ExponentsIf a is a real number other than 0 and n is an integer, then

a-n =1an

TEACHING TIPA common student mistake is thefollowing: Remind stu-dents that a negative exponent hasnothing to do with the sign of thesimplified expression.

2-3 = -8.

For example,

In other words, another way to write is to take its reciprocal and change the sign ofits exponent.

a-n

x-3 =1

x3 .

E X A M P L E 1Simplify by writing each expression with positive exponents only.

a. b. c. d. e. f.1

7-2

1

y-41-22-42-1 + 4-12x-33-2

Solut ion a. Use the definition of negative exponents.

b. Use the definition of negative exponents.2x-3 = 2 # 1

x3 =2

x3

3-2 =1

32 =19


a. b. c.answers:

a. b. c. 815

7

x41

125

5-1 + 3-17x-45-3

H e l p f u l H i n tDon’t forget that since there are no parentheses, only x is the base for the ex-ponent -3.

c.

d.

e. f.1

7-2 =11

72

=72

1 or 49

1

y-4 =11

y4

= y4

1-22-4 =1

1-224 =1

1-221-221-221-22 =1

16

2-1 + 4-1 =12

+14

=24

+14

=34


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H e l p f u l H i n tFrom Example 1, we see that

Also, notice that a negative exponent does not affect the sign of its base. Thekey word to remember when working with negative exponents is reciprocal.

1

5-2 =52

1 or 25

y-7

x-11 =x11

y7

x-2 =1

x2 2-3 =1

23 =18

1

y-4 =y4

1 or y4 .

E X A M P L E 2Simplify each expression. Write results using positive exponents only.

a. b. c. d.5-3

2-5

p-4

q-9

1

3-4

1

x-3

TEACHING TIPConsider asking students if theysee another approach to Example4(a). For example.

=278

=33

23

a23b-3

= a32b3

Solut ion a. b. c. d.5-3

2-5 =25

53 =32125

p-4

q-9 =q9

p4

1

3-4 =34

1= 81

1

x-3 =x3

1= x3


a. b. c.

answers:

a. b. 32 c.y2

x5y7

x-5

y-2

1

2-5

1

y-7

E X A M P L E 3Simplify each expression. Write answers with positive exponents.

a. b. c.x-5

x7

3

x-4

y

y-2

Solut ion a. Remember that

b. c.

2 All the previously stated rules for exponents apply for negative exponentsalso. Here is a summary of the rules and definitions for exponents.

x-5

x7 = x-5 - 7 = x-12 =1

x12

3

x-4 = 3 # 1

x-4 = 3 # x4 or 3x4

am

an am-n .y

y-2 =y1

y-2 = y1 -1-22 = y3


a. b.

answers:

a. b.7

z6x6

7z-2

z4

x

x-5

Summary of Exponent RulesIf m and n are integers and a, b, and c are real numbers, then:

Product rule for exponents:

Power rule for exponents:

Power of a product:

Power of a quotient:

Quotient rule for exponents:

Zero exponent:

Negative exponent: a-n =1an , a Z 0

a0 = 1, a Z 0

am

an = am - n , a Z 0

aacbn

=an

cn , c Z 0

1ab2n = anbn

1am2n = am # nam # an = am + n


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Simplify the following expressions. Write each result using positive exponents only.

a. b. c.

d. e. f. a -2x3y

xy-1 b31y-3z62-64-1x-3y

4-3x2y-6

a3a2

bb-31x324x

x7a23b-3

E X A M P L E 4

Solut ion a.

b. Use the power rule.

c. Raise each factor in the numerator

and the denominator to the power.

Use the power rule.

Use the negative exponent rule.

Write as 27.

d.

e.

f.

3 Both very large and very small numbers frequently occur in many fields ofscience. For example, the distance between the sun and the planet Pluto is approxi-mately 5,906,000,000 kilometers, and the mass of a proton is approximately0.00000000000000000000000165 gram. It can be tedious to write these numbers in thisstandard decimal notation, so scientific notation is used as a convenient shorthand forexpressing very large and very small numbers.

a -2x3y

xy-1 b3

=1-223x9y3

x3y-3 =-8x9y3

x3y-3 = -8x9 - 3y3 -1-32 = -8x6y6

1y-3z62-6 = y18 # z-36 =y18

z36

4-1x-3y

4-3x2y-6 = 4-1 -1-32x-3 - 2y1 -1-62 = 42x-5y7 =42y7

x5 =16y7

x5

33 =b3

27a6

=b3

33a6

=3-3a-6

b-3

3

a3a2

bb-3

=3-31a22-3

b-3

1x324xx7 =

x12 # x

x7 =x12 + 1

x7 =x13

x7 = x13 - 7 = x6

a23b-3

=2-3

3-3 =33

23 =278


a. b.

c.

answers:

a. b. c.1

36x3y3

y2

81x6

a20

b35

13x-2y2-2

4x7y

a9x3

yb-21a-4b72-5

SunPluto5,906,000,000

kilometers

proton

Mass of proton is approximately0.000 000 000 000 000 000 000 001 65 gram

Scientific NotationA positive number is written in scientific notation if it is written as the prod-uct of a number a, where and an integer power r of 10:

a * 10r

1 … a 6 10,



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The numbers below are written in scientific notation. The sign for multiplica-tion is used as part of the notation.

(Distance between the sun and Pluto)

(Mass of a proton)

The following steps are useful when writing numbers in scientific notation.

1 * 10-3 8.1 * 10-5 1.65 * 10-24

5

4

2.03 * 102 7.362 * 107 5.906 * 109

*

To Write a Number in Scientific NotationStep 1. Move the decimal point in the original number to the left or right

so that the new number has a value between 1 and 10.

Step 2. Count the number of decimal places the decimal point is moved inStep 1. If the original number is 10 or greater, the count is positive.If the original number is less than 1, the count is negative.

Step 3. Multiply the new number in Step 1 by 10 raised to an exponentequal to the count found in Step 2.

E X A M P L E 5

Write each number in scientific notation.

a. 367,000,000 b. 0.000003 c. 20,520,000,000 d. 0.00085

Solut ion a. Step 1. Move the decimal point until the number is between 1 and 10.367,000,000

8 places

Step 2. The decimal point is moved 8 places, and the original number is 10 orgreater, so the count is positive 8.

Step 3.

b. Step 1. Move the decimal point until the number is between 1 and 10.0.000003

6 placesStep 2. The decimal point is moved 6 places, and the original number is less than

1, so the count is

Step 3.

c.

d.

4 A number written in scientific notation can be rewritten in standard form. Forexample, to write in standard form, recall that

Notice that the exponent on the 10 is positive 3, and we moved the decimal point 3places to the right.

To write in standard form, recall that

7.29 * 10-3 = 7.29a 11000

b =7.291000

= 0.00729

10-3 =1

103 =1

1000.7.29 * 10-3

8.63 * 103 = 8.63110002 = 8630

103 = 1000.8.63 * 103

0.00085 = 8.5 * 10-4

20,520,000,000 = 2.052 * 1010

0.000003 = 3.0 * 10-6

-6.

367,000,000 = 3.67 * 108.

TEACHING TIPShow students that Example 5(a),367,000,000, can be rewritten inother ways, such as

These prod-ucts are not called scientific nota-tion because 367 and 36.7 are notbetween 1 and 10.

367 * 106, 36.7 * 107.

CLASSROOM EXAMPLEWrite each number in scientific notation.

a. 9,060,000,000

b. 0.00017

answers:

a. b. 1.7 * 10-49.06 * 109



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The exponent on the 10 is negative 3, and we moved the decimal to the left 3 places.In general, to write a scientific notation number in standard form, move the deci-

mal point the same number of places as the exponent on 10. If the exponent is positive,move the decimal point to the right; if the exponent is negative, move the decimal pointto the left.

E X A M P L E 6Write each number in standard notation, without exponents.

a. b. c. d. 3.007 * 10-58.4 * 1077.358 * 10-31.02 * 105

Solut ion a. Move the decimal point 5 places to the right.

b. Move the decimal point 3 places to the left.

c. 7 places to the right

d. 5 places to the left

CONCEPT CHECKWhich number in each pair is larger?

a. or b. or c. or

Performing operations on numbers written in scientific notation makes use of therules and definitions for exponents.

6.3 * 10-55.6 * 10-42.7 * 1049.2 * 10-22.1 * 1057.8 * 103

3.007 * 10-5 = 0.00003007

8.4 * 107 = 84,000,000.

7.358 * 10-3 = 0.007358.

1.02 * 105 = 102,000.

CLASSROOM EXAMPLEWrite the numbers in standard notation.

a. b.answers:a. 0.0003062 b. 6,002,000

6.002 * 1063.062 * 10-4


a. b.c. 5.6 * 10-4

2.7 * 1042.1 * 105

E X A M P L E 7Perform each indicated operation. Write each result in standard decimal notation.

a. b.12 * 102

6 * 10-318 * 10-6217 * 1032Solut ion a.

b.12 * 102

6 * 10-3 =126

* 102 -1-32 = 2 * 105 = 200,000

= 0.056

= 56 * 10-3

18 * 10-6217 * 1032 = 18 # 72 * 110-6 # 1032

CLASSROOM EXAMPLEPerform each indicated operation. Writeeach result in standard decimal notation.

a.

b.

answers:a. 0.36 b. 40,000,000

8 * 104

2 * 10-3

19 * 107214 * 10-92

Scientific Notation

To enter a number written in scientific notation on a scientific calculator, locate the

scientific notation key, which may be marked or To enter

press The display should read 3.1 07 . 7 . EE 3.1 3.1 * 107, EXP . EE




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Enter each number written in scientific notation on your calculator.

1. 2.

3. 4.

Multiply each of the following on your calculator. Notice the form of the result.

5. 6.

Multiply each of the following on your calculator.Write the product in scientific notation.

7. 8. 18.76 * 10-4211.237 * 109213.26 * 106212.5 * 10132

230,000 * 10003,000,000 * 5,000,000

-9.9811 * 10-26.6 * 10-9-4.8 * 10145.31 * 103

Suppose you are a paralegal for a law firm. You are investigating the facts of amineral rights case. Drilling on property adjacent to the client’s has struck anatural gas reserve.The client believes that a portion of this reserve lies withinher own property boundaries and she is, therefore, entitled to a portion of theproceeds from selling the natural gas. As part of your investigation, you con-tact two different experts, who fax you the following estimates for the size ofthe natural gas reserve:

Expert A Expert B

Estimate of entire reserve: Estimate of entire reserve:cubic feet cubic feet

Estimate of size of reserve on Estimate of size of reserve onclient’s property client’s property:

cubic feet cubic feet2.68 * 1061.84 * 107

6.7 * 1064.6 * 107

How, if at all, would you use these estimates in the case? Explain.

MENTAL MATHState each expression using positive exponents only.

1. 2. 3. 4. 5. 6. 16y716

y-74y34

y-3x31

x-3y61

y-6

3

x33x-35

x25x-2


DIVIS ION OF POLYNOMIALS SECTION 5.6 337

O b j e c t i v e s

1 Divide a polynomial by a monomial.

2 Use long division to divide a polynomial by another polynomial.

1 Now that we know how to add, subtract, and multiply polynomials, we practicedividing polynomials.

To divide a polynomial by a monomial, recall addition of fractions. Fractions thathave a common denominator are added by adding the numerators:

If we read this equation from right to left and let a, b, and c be monomials, wehave the following:

c Z 0,

ac

+bc

=a + b

c

5.6 D I V I S I O N O F P O LY N O M I A L S

Dividing a Polynomial By a MonomialDivide each term of the polynomial by the monomial.

a + bc

=ac

+bc

, c Z 0

Throughout this section, we assume that denominators are not 0.

E X A M P L E 1

Divide by 2m.6m2 + 2m

Solut ion We begin by writing the quotient in fraction form. Then we divide each term of thepolynomial by the monomial 2m.

Simplify. = 3m + 1

6m2 + 2m

2m=

6m2

2m+

2m

2m

6m2 + 2mCLASSROOM EXAMPLEDivide: by answer: 5x + 1

5x2 .25x3 + 5x2

TEACHING TIPAsk students whether the answerto Example 1 would be true for allvalues of m. Have them verify thatit is not true for m = 0.

Check We know that if then must equal

Thus, to check, we multiply.

The quotient checks.3m + 1

2m13m + 12 = 2m13m2 + 2m112 = 6m2 + 2m

6m2 + 2m .2m # 13m + 126m2 + 2m

2m= 3m + 1,

E X A M P L E 2

Divide 9x5 - 12x2 + 3x

3x2 .

Solut ionDivide each term by

Simplify. = 3x3 - 4 +1x

3x2 . 9x5 - 12x2 + 3x

3x2 =9x5

3x2 -12x2

3x2 +3x

3x2



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The quotient is 281 R 7, which can be written as

Recall that division can be checked by multiplication.To check a division problem suchas this one, we see that

Now we demonstrate long division of polynomials.

13 # 281 + 7 = 3660

; remainder; divisor

281 713

281133660

26 106 104

20 13

7

Notice that the quotient is not a polynomial because of the term This expression is

alled a rational expression—we will study rational expressions further in Chapter 7.Al-though the quotient of two polynomials is not always a polynomial, we may still checkby multiplying.

1x

.CLASSROOM EXAMPLE

Divide

answer: 6x5 + 2 -1x

30x7 + 10x2 - 5x

5x2.

Check

= 9x5 - 12x2 + 3x

3x2a3x3 - 4 +1xb = 3x213x32 - 3x2142 + 3x2a 1

xb

E X A M P L E 3

Divide 8x2y2 - 16xy + 2x

4xy.

Solut ionDivide each term by 4xy.

Simplify. = 2xy - 4 +1

2y

8x2y2 - 16xy + 2x

4xy=

8x2y2

4xy-

16xy

4xy+

2x

4xy

CLASSROOM EXAMPLE

Divide

answer:

4x2y2 - 6 +2x

12x3y3 - 18xy + 6y

3xy.

Check

CONCEPT CHECK

In which of the following is simplified correctly?

a. b. x c.

2 To divide a polynomial by a polynomial other than a monomial, we use aprocess known as long division. Polynomial long division is similar to number long di-vision, so we review long division by dividing 13 into 3660.

x + 1x

5+ 1

x + 55

= 8x2y2 - 16xy + 2x

4xya2xy - 4 +1

2yb = 4xy12xy2 - 4xy142 + 4xya 1

2yb

Subtract and bring down the next digit in the dividend.

Subtract and bring down the next digit in the dividend.

Subtract. There are no more digits to bring down, so the remainder is 7.1 # 13 13

8 # 13 104

2 # 13 26T

∂

H e l p f u l H i n tRecall that 3660 is called the dividend.

Concept Check Answer:a


DIVIS ION OF POLYNOMIALS SECTION 5.6 339

Divide by using long division.x + 3x2 + 7x + 12

Solut ion

Now we repeat this process.

The quotient is x + 4.

x + 4 x + 3x2 + 7x + 12

-x2 +- 3x

4x + 12-

4x +- 120

To subtract, change the signs of these terms and add.

x x + 3x2 + 7x + 12

-x2 +- 3x

4x + 12

To subtract, change the signs of these terms and add.

E X A M P L E 4

CLASSROOM EXAMPLEDivide by answer: x + 7

x + 5.x2 + 12x + 35

How many times does x divide Multiply: Subtract and bring down the next term.

x1x 32 .x2?

x2

x x .

How many times does x divide

Multiply:

Subtract. The remainder is 0.

41x 32 .

4x? 4x

x 4 .

Check We check by multiplying.

or

The quotient checks.

x2 + 7x + 12=0+1x + 42#1x + 32 TTTTdividend=remainder+quotient#divisor

E X A M P L E 5

Divide by using long division.3x - 16x2 + 10x - 5

Solut ion

Thus divided by is with a remainder of Thiscan be written as

; remainder; divisor

6x2 + 10x - 53x - 1

= 2x + 4 +-1

3x - 1

-1.12x + 4213x - 1216x2 + 10x - 52

T

2x + 43x - 16x2 + 10x - 5

-6x2 -+ 2x

12x - 5-

12x -+ 4

-1

so 2x is a term of the quotient.

Multiply

Subtract and bring down the next term.

Subtract. The remainder is 1 .

12x

3x 4 , 413x 12

2x13x 12.6x2

3x 2x ,

CLASSROOM EXAMPLEDivide by answer:

3x + 5 -2

2x - 1

2x - 1.6x2 + 7x - 7

Check To check, we multiply Then we add the remainder, to thisproduct.

The quotient checks.

In Example 5, the degree of the divisor, is 1 and the degree of the re-mainder, is 0. The division process is continued until the degree of the remainderpolynomial is less than the degree of the divisor polynomial.

-1,3x - 1,

= 6x2 + 10x - 5

13x - 1212x + 42 + 1-12 = 16x2 + 12x - 2x - 42 - 1

-1,13x - 1212x + 42 .

T


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MENTAL MATH

E X A M P L E 6

Divide 4x2 + 7 + 8x3

2x + 3.

Solut ion Before we begin the division process, we rewrite as Notice that we have written the polynomial in descending order and have

represented the missing x term by 0x.

Thus,

Remainder.

4x2 + 7 + 8x3

2x + 3= 4x2 - 4x + 6 +

-112x + 3

.

4x2 - 4x + 62x + 38x3 + 4x2 + 0x + 7

8x3 +- 12x2 -8x2 + 0x

-+ 8x2 -+ 12x 12x + 7

-12x +- 18

-11

+ 0x + 7.8x3 + 4x24x2 + 7 + 8x3

CLASSROOM EXAMPLE

Divide

answer:

3x2 - 2x + 1 +3

3x + 2

5 - x + 9x3

3x + 2.

E X A M P L E 7

Divide 2x4 - x3 + 3x2 + x - 1

x2 + 1.

Solut ion Before dividing, rewrite the divisor polynomial as The 0xterm represents the missing term in the divisor.

Thus,

Remainder.

2x4 - x3 + 3x2 + x - 1

x2 + 1= 2x2 - x + 1 +

2x - 2

x2 + 1.

2x2 - x + 1x2 + 0x + 12x4 - x3 + 3x2 + x - 1

-2x4 +- 0x3 +- 2x2

-x3 + x2 + x -+ x3 -+ 0x2 -+ x

x2 + 2x - 1-

x2 +- 0x +- 12x - 2

x11x2 + 0x + 12 .1x2 + 12

CLASSROOM EXAMPLE

Divide

answer: x2 + x + 1

x3 - 1x - 1

.

Simplify each expression mentally.

1. 2. y 3. 4. 5.

6. 7. 8. 9. k2k7

k5k3k5

k2p5p8

p3k2k7

k5

k3k5

k2p5p8

p3a2a3

a

y2

ya2a6

a4


O b j e c t i v e s

1 Find the greatest common factor of a list of integers.

2 Find the greatest common factor of a list of terms.

3 Factor out the greatest common factor from a polynomial.

4 Factor a polynomial by grouping.

When an integer is written as the product of two or more other integers, each of theseintegers is called a factor of the product.This is true for polynomials, also.When a poly-nomial is written as the product of two or more other polynomials, each of these poly-nomials is called a factor of the product.

The process of writing a polynomial as a product is called factoring thepolynomial.




Notice that factoring is the reverse process of multiplying.

x2 + 5x + 6 = 1x + 221x + 32

1x + 221x + 32 = x2 + 5x + 6

x2 # x3 = x5

2 # 3 = 6

354 CHAPTER 6 FACTORING POLYNOMIALS

1 The GCF of a list of integers is the largest integer that is a factor of all the inte-gers in the list. For example, the GCF of 12 and 20 is 4 because 4 is the largest integerthat is a factor of both 12 and 20. With large integers, the GCF may not be easily foundby inspection. When this happens, use the following steps.

Recall from Section 1.3 that a prime number is a whole number other than 1, whoseonly factors are 1 and itself.

Concept Check Answer:The result would bebecause factoring is the re-

verse process of multiplying.21x - 422x - 8;

Finding the GCF of a List of IntegersStep 1. Write each number as a product of prime numbers.

Step 2. Identify the common prime factors.

Step 3. The product of all common prime factors found in step 2 is the greatestcommon factor. If there are no common prime factors, the greatestcommon factor is 1.

6.1 T H E G R E AT E S T C O M M O N FAC TO R A N D FAC TO R I N G B Y G R O U P I N G

CONCEPT CHECKMultiply:

What do you think the result of factoring would be? Why?2x - 8

21x - 42

factoring

multiplying

The first step in factoring a polynomial is to see whether the terms of the polynomial havea common factor. If there is one, we can write the polynomial as a product by factoringout the common factor.We will usually factor out the greatest common factor (GCF).



THE GREATEST COMMON FACTOR AND FACTORING BY GROUPING SECTION 6.1 355

Find the GCF of each list of numbers.

a. 28 and 40 b. 55 and 21 c. 15, 18, and 66

Solut ion a. Write each number as a product of primes.

There are two common factors, each of which is 2, so the GCF is

b.

There are no common prime factors; thus, the GCF is 1.

c.

The only prime factor common to all three numbers is 3, so the GCF is

2 The greatest common factor of a list of variables raised to powers is found in asimilar way. For example, the GCF of and is because each term contains afactor of and no higher power of x is a factor of each term.

There are two common factors, each of which is x, so the or From thisexample, we see that the GCF of a list of common variables raised to powers is thevariable raised to the smallest exponent in the list.

x2 .GCF = x # x

x5 = x # x # x # x # x

x3 = x # x # x

x2 = x # x

x2x2x5x2 , x3 ,

GCF = 3

66 = 2 # 3 # 11 18 = 2 # 3 # 3 = 2 # 32 15 = 3 # 5

21 = 3 # 7 55 = 5 # 11

GCF = 2 # 2 = 4

40 = 2 # 2 # 2 # 5 = 23 # 5 28 = 2 # 2 # 7 = 22 # 7

E X A M P L E 2Find the GCF of each list of terms.

a. and b. y, and y7y4 ,x5x3 , x7 ,

E X A M P L E 1

Solut ion a. The GCF is since 3 is the smallest exponent to which x is raised.

b. The GCF is or y, since 1 is the smallest exponent on y.

In general, the greatest common factor (GCF) of a list of terms is the product ofthe GCF of the numerical coefficients and the GCF of the variable factors.

y1

x3 ,

Find the GCF of each list of terms.

a. and b. and c. and a6b2a3b2 , a5b ,y58y2 , y3 ,-8x6x2 , 10x3 ,

E X A M P L E 3

Solut ion a. The GCF of the numerical coefficients 6, 10, and is 2.The GCF of variable factors and x is x.Thus, the GCF of and is 2x.-8x6x2 , 10x3 ,

x2 , x3 ,-8

CLASSROOM EXAMPLEFind the GCF of each list of terms.

a. and b.

answer: a. b. xy4

x and x10y8y4 , y5 ,

CLASSROOM EXAMPLEFind the GCF of each list of numbers.

a. 45 and 75 b. 32 and 33

c. 14, 24, and 60answer: a. 15 b. 1 c. 2




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b. The GCF of the numerical coefficients 8, 1, and 1 is 1.The GCF of variable factors and is Thus, the GCF of and is or

c. The GCF of and is

The GCF of b, and is b. Thus, the GCF of the terms is

3 The first step in factoring a polynomial is to find the GCF of its terms. Once wedo so, we can write the polynomial as a product by factoring out the GCF.

The polynomial for example, contains two terms: 8x and 14.The GCF ofthese terms is 2. We factor out 2 from each term by writing each term as a product of 2and the term’s remaining factors.

Using the distributive property, we can write

Thus, a factored form of is 214x + 72 .8x + 14

= 214x + 72 8x + 14 = 2 # 4x + 2 # 7

8x + 14 = 2 # 4x + 2 # 7

8x + 14,

a3b .b2b2 ,

a3 .a6a3 , a5 ,

y2 .1y2y58y2 , y3 ,y2 .y5y2 , y3 ,

E X A M P L E 4Factor each polynomial by factoring out the GCF.

a. b. y5 - y76t + 18

Solut ion a. The GCF of terms 6t and 18 is 6.


Our work can be checked by multiplying 6 and

b. The GCF of and is Thus,

= y511 - y22 y5 - y7 = 1y521 - 1y52y2

y5 .y7y5

61t + 32 = 6 # t + 6 # 3 = 6t + 18, the original polynomial .

1t + 32 . = 61t + 32

6t + 18 = 6 # t + 6 # 3CLASSROOM EXAMPLEFactor.a. b.answer:

a. b. z21z - 12717x - 52z3 - z249x - 35

Concept Check Answer:c

H e l p f u l H i n tDon’t forget the 1.

H e l p f u l H i n tA factored form of is not

Although the terms have been factored (written as a product), thepolynomial has not been factored (written as a product). A factoredform of is the product 214x + 72 .8x + 14

8x + 14

2 # 4x + 2 # 7

8x + 14

CONCEPT CHECKWhich of the following is/are factored form(s) of

a. 7 b. c. d. 71t + 21271t + 327 # t + 7 # 3

7t + 21?

CLASSROOM EXAMPLEFind the GCF of each list of terms.

a. and 6x

b.answer: a. 3x b. ab2

11ab3 , a2b3, and ab2

-9x2 , 15x4 ,



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E X A M P L E 5Factor -9a5 + 18a2 - 3a .

Solut ion

In Example 5 we could have chosen to factor out a instead of 3a. If we factorout a we have

= -3a13a4 - 6a + 12 -9a5 + 18a2 - 3a = 1-3a213a42 + 1-3a21-6a2 + 1-3a2112

-3a ,-3a

= 3a1-3a4 + 6a - 12 -9a5 + 18a2 - 3a = 13a21-3a42 + 13a216a2 + 13a21-12

E X A M P L E 6

Factor 25x4z + 15x3z + 5x2z .

Solut ion

CLASSROOM EXAMPLE

Factor answer: 9xy312y2 + y + 12

18xy5 + 9xy4 + 9xy3 .

TEACHING TIPA common mistake is for studentsto factor as

Before workingExample 7, make sure that stu-dents understand the concept offactoring out a common factor.

1x + 32215 + y2.51x + 32 + y1x + 32

The greatest common factor is

25x4z + 15x3z + 5x2z = 5x2z15x2 + 3x + 125x2z .

Factor 51x + 32 + y1x + 32 .Solut ion The binomial is the greatest common factor. Use the distributive property to

factor out

51x + 32 + y1x + 32 = 1x + 3215 + y21x + 32 .1x + 32

H e l p f u l H i n tDon’t forget the -1.

H e l p f u l H i n tNotice the changes in signs when factoring out -3a .

E X A M P L E 7

E X A M P L E 8

Factor 3m2n1a + b2 - 1a + b2 .

H e l p f u l H i n tBe careful when the GCF of the terms is the same as one of the terms in the poly-nomial.The greatest common factor of the terms of is 2x.Whenfactoring out 2x from the terms of don’t forget a term of 1.

Check by multiplying.

2x14x - 3x2 + 12 = 8x2 - 6x3 + 2x

= 2x14x - 3x2 + 12 8x2 - 6x3 + 2x = 2x14x2 - 2x13x22 + 2x112

8x2 - 6x3 + 2x ,8x2 - 6x3 + 2x

CLASSROOM EXAMPLE

Factor answer:

or-2b16b2 - 5b + 222b1-6b2 + 5b - 22

-12b3 + 10b2 - 4b .

Solut ion The greatest common factor is

3m2n1a + b2 - 11a + b2 = 1a + b213m2n - 121a + b2 .

CLASSROOM EXAMPLE

Factor

answer: 1y - 121x - 132x1y - 12 - 131y - 12 .

CLASSROOM EXAMPLE

Factor

answer: 12m + n215x2y3 - 12(2m + n) - (2m - n).

5x2y3


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CLASSROOM EXAMPLE

Factor by grouping.

answer: 15a + 2b21a 125a2 + 2ab - 5a - 2b


Factoring a Four-term Polynomial by GroupingStep 1. Arrange the terms so that the first two terms have a common factor and

the last two terms have a common factor.

Step 2. For each pair of terms, use the distributive property to factor out thepair’s greatest common factor.

Step 3. If there is now a common binomial factor, factor it out.

Step 4. If there is no common binomial factor in step 3, begin again, rearrang-ing the terms differently. If no rearrangement leads to a common bino-mial factor, the polynomial cannot be factored.

CLASSROOM EXAMPLE

Factor answer: 1b + 421a + 72

ab + 4a + 7b + 28.

E X A M P L E 1 0

Factor by grouping.3x2 + 4xy - 3x - 4y

Solut ion The first two terms have a common factor x. Factor from the last two terms so thatthe common binomial factor of appears.

Next, factor out the common factor

= 13x + 4y21x - 1213x + 4y2 .

3x2 + 4xy - 3x - 4y = x13x + 4y2 - 113x + 4y213x + 4y2 -1

E X A M P L E 9Factor by grouping. Check by multiplying.xy + 2x + 3y + 6

Solut ion The GCF of the first two terms is x, and the GCF of the last two terms is 3.

xy + 2x + 3y + 6 = x1y + 22 + 31y + 22('''')''''*

Next, factor out the common binomial factor of

To check, multiply by

Thus, the factored form of is 1y + 221x + 32 .xy + 2x + 3y + 6

1y + 221x + 32 = xy + 2x + 3y + 6, the original polynomial .

1x + 32 .1y + 22x1y + 22 + 31y + 22 = 1y + 221x + 32

1y + 22 .

H e l p f u l H i n tNotice that this is not a factored form of the original polynomial. It is a sum,not a product.

4 Once the GCF is factored out, we can often continue to factor the polynomial,using a variety of techniques. We discuss here a technique for factoring polynomialscalled grouping.



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E X A M P L E 1 1

Factor 4ax - 4ab - 2bx + 2b2 .

Solut ion First, factor out the common factor 2 from all four terms.

Factor out 2 from all four terms.Factor out common factors from each pair of terms.Factor out the common binomial.

Notice that we factored out instead of b from the second pair of terms so that thebinomial factor of each pair is the same.

-b

= 21x - b212a - b2 = 2[2a1x - b2 - b1x - b2] = 212ax - 2ab - bx + b22

4ax - 4ab - 2bx + 2b2

CLASSROOM EXAMPLE

Factor

answer: 51x + y213z - y215xz + 15yz - 5xy - 5y2 .

Factoring out a greatest common factor first makes factoring by any method eas-ier, as we see in the next example.

H e l p f u l H i n tOne more reminder: When factoring a polynomial, make sure the polynomial iswritten as a product. For example, it is true that

but is not a factored form of the original polyno-mial since it is a sum (difference), not a product. The factored form of

is 13x + 4y21x - 12 .3x2 + 4xy - 3x - 4y

x13x + 4y2 - 113x + 4y23x2 + 4xy - 3x - 4y = x13x + 4y2 - 113x + 4y2

MENTAL MATHFind the prime factorization of the following integers.

1. 14 2. 15 3. 10 4. 70

Find the GCF of the following pairs of integers.

5. 6, 15 3 6. 20, 15 5 7. 3, 18 3 8. 14, 35 7

2 # 5 # 72 # 53 # 52 # 7

STUDY SKILLS REMINDERHow Well Do You Know Your Textbook?

See if you can answer the questions below.

1. What does the icon mean?



4. Where can you find a review for each chapter? What answersto this review can be found in the back of your text?

5. Each chapter contains an overview of the chapter along withexamples. What is this feature called?

6. Does this text contain any solutions to exercises? If so, where?


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FACTORING TRINOMIALS OF THE FORM x2 + bx + c SECTION 6.2 361

6.2 FAC TO R I N G T R I N O M I A L S O F T H E F O R M x 2 + b x + c

O b j e c t i v e s

1 Factor trinomials of the form

2 Factor out the greatest common factor and then factor a trinomial of the form

1 In this section, we factor trinomials of the form such as

Notice that for these trinomials, the coefficient of the squared variable is 1.

Recall that factoring means to write as a product and that factoring and multi-plying are reverse processes. Using the FOIL method of multiplying binomials,

x2 + 4x + 3, x2 - 8x + 15, x2 + 4x - 12, r2 - r - 42

x2 + bx + c ,

x 2 + bx + c .

x 2 + bx + c .




Factoring a Trinomial of the Form To factor a trinomial of the form look for two numbers whoseproduct is c and whose sum is b. The factored form of is

T—– product is c —————

T1x + one number21x + other number2q—– sum is b ————q

x2 + bx + cx2 + bx + c ,

x

2 bx c

we have thatF O I L

Thus, a factored form of is Notice that the product of the first terms of the binomials is the first

term of the trinomial. Also, the product of the last two terms of the binomials isthe third term of the trinomial. The sum of these same terms is

the coefficient of the middle term, x, of the trinomial.

The product of these numbers is 3.

The sum of these numbers is 4.

Many trinomials, such as the one above, factor into two binomials. To factorlet’s assume that it factors into two binomials and begin by writing two

pairs of parentheses. The first term of the trinomial is so we use x and x as the firstterms of the binomial factors.

To determine the last term of each binomial factor, we look for two integers whoseproduct is 10 and whose sum is 7. Since our numbers must have a positive product anda positive sum, we list pairs of positive integer factors of 10 only.

Positive Factors of 10 Sum of Factors

1, 10

2, 5

The correct pair of numbers is 2 and 5 because their product is 10 and their sum is 7.Now we can fill in the last terms of the binomial factors.

To see if we have factored correctly, multiply.

Combine like terms. = x2 + 7x + 10

1x + 221x + 52 = x2 + 5x + 2x + 10

x2 + 7x + 10 = 1x + 221x + 52

2 + 5 = 7

1 + 10 = 11

x2 + 7x + 10 = 1x + 21x + 2x2 ,

x2 + 7x + 10,

T— – – — –——

T x2 + 4x + 3 = 1x + 321x + 12

q—— –——q

3 + 1 = 4,3 # 1 = 3,

x # x = x2 ,1x + 321x + 12 .x2 + 4x + 3

= x2 + 4x + 3

1x + 321x + 12 = x2 + 1x + 3x + 3

H e l p f u l H i n tSince multiplication is commutative, the factored form of can bewritten as either or 1x + 521x + 22 .1x + 221x + 52 x2 + 7x + 10



E X A M P L E 1

Factor x2 + 7x + 12.

Solut ion Begin by writing the first terms of the binomial factors.

Next, look for two numbers whose product is 12 and whose sum is 7. Since our numbersmust have a positive product and a positive sum, we look at positive pairs of factors of12 only.

Positive Factors of 12 Sum of Factors

1, 122, 63, 4

The correct pair of numbers is 3 and 4 because their product is 12 and their sum is 7.Use these numbers as the last terms of the binomial factors.

To check, multiply by 1x + 42 .1x + 32x2 + 7x + 12 = 1x + 321x + 42

3 + 4 = 72 + 6 = 8

1 + 12 = 13

1x + 21x + 2CLASSROOM EXAMPLE

Factor

answer: 1x + 421x + 52x2 + 9x + 20.

Factor x2 - 8x + 15.

E X A M P L E 2

Solut ion Begin by writing the first terms of the binomials.

Now look for two numbers whose product is 15 and whose sum is Since our numbersmust have a positive product and a negative sum, we look at negative factors of 15 only.

Negative Factors of 15 Sum of Factors

The correct pair of numbers is and because their product is 15 and their sum isThen

x2 - 8x + 15 = 1x - 321x - 52-8.

-5-3

-3 + 1-52 = -8-3, -5-1 + 1-152 = -16-1, -15

-8.

1x + 21x + 2CLASSROOM EXAMPLE

Factor

answer: 1x - 221x - 112x2 - 13x + 22.

E X A M P L E 3

Factor x2 + 4x - 12.

Solut ionLook for two numbers whose product is and whose sum is 4. Since our numbershave a negative product, their signs must be different.

Factors of Sum of Factors

3 + 1-42 = -13, -4-3 + 4 = 1-3, 4

2 + 1-62 = -42, -6-2 + 6 = 4-2, 6

1 + 1-122 = -111, -12-1 + 12 = 11-1, 12

12

-12x2 + 4x - 12 = 1x + 21x + 2

CLASSROOM EXAMPLE

Factor answer: 1x + 921x - 42

x2 + 5x - 36.



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E X A M P L E 4

Factor r2 - r - 42.

Solut ion Because the variable in this trinomial is r, the first term of each binomial factor is r.

Find two numbers whose product is and whose sum is the numerical coeffi-cient of r. The numbers are 6 and Therefore,

r2 - r - 42 = 1r + 621r - 72-7.

-1,-42

r2 - r - 42 = 1r + 21r + 2CLASSROOM EXAMPLE

Factor

answer: 1q + 521q - 82q2 - 3q - 40.

E X A M P L E 5

Factor a2 + 2a + 10.

Solut ion Look for two numbers whose product is 10 and whose sum is 2. Neither 1 and 10 nor 2and 5 give the required sum, 2. We conclude that is not factorable withintegers. The polynomial is called a prime polynomial.a2 + 2a + 10

a2 + 2a + 10

CLASSROOM EXAMPLE

Factor

answer: prime polynomial

y2 + 6y + 15.

E X A M P L E 6Factor x2 + 5xy + 6y2 .

Solut ion

Look for two terms whose product is and whose sum is 5y, the coefficient of x in themiddle term of the trinomial. The terms are 2y and 3y because and

Therefore,

The following sign patterns may be useful when factoring trinomials.

x2 + 5xy + 6y2 = 1x + 2y21x + 3y22y + 3y = 5y .

2y # 3y = 6y26y2

x2 + 5xy + 6y2 = 1x + 21x + 2CLASSROOM EXAMPLE

Factor

answer: 1a - 3b21a - 10b2a2 - 13ab + 30b2 .

The correct pair of numbers is and 6 since their product is and their sum is 4.Hence

x2 + 4x - 12 = 1x - 221x + 62

-12-2

H e l p f u l H i n t — S i g n P a t t e r n sA positive constant in a trinomial tells us to look for two numbers with thesame sign. The sign of the coefficient of the middle term tells us whether thesigns are both positive or both negative.

both samepositive sign

both samenegative sign

x2 - 10x + 16 = 1x - 221x - 82TT

x2 + 10x + 16 = 1x + 221x + 82TT


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2 Remember that the first step in factoring any polynomial is to factor out thegreatest common factor (if there is one other than 1 or ).-1

A negative constant in a trinomial tells us to look for two numbers with oppositesigns.

opposite oppositesigns signs

x2 + 6x - 16 = 1x + 821x - 22 x2 - 6x - 16 = 1x - 821x + 22TT

E X A M P L E 7

Factor 3m2 - 24m - 60.

Solut ion

CLASSROOM EXAMPLEFactor

a.

b.

answer: a.b. 5x31x + 121x - 62

x1x + 421x - 125x5 - 25x4 - 30x3

x3 + 3x2 - 4x

First factor out the greatest common factor, 3, from each term.

Next, factor by looking for two factors of whose sum is Thefactors are and 2.

Remember to write the common factor 3 as part of the answer. Check by multiplying.

= 3m2 - 24m - 60

31m + 221m - 102 = 31m2 - 8m - 202

3m2 - 24m - 60 = 31m + 221m - 102-10

-8.-20m2 - 8m - 20

3m2 - 24m - 60 = 31m2 - 8m - 202

H e l p f u l H i n tWhen factoring a polynomial, remember that factored out common factors arepart of the final factored form. For example,

Thus, factored completely is 51x + 221x - 52 .5x2 - 15x - 50

= 51x + 221x - 52 5x2 - 15x - 50 = 51x2 - 3x - 102

MENTAL MATHComplete the following.

1. 2. 3.

4. 5. 6. + 42x2 + 10x + 24 = 1x + 621x+ 22x2 + 4x + 4 = 1x + 221x- 112x2 - 13x + 22 = 1x - 221x- 32x2 - 7x + 12 = 1x - 421x+ 72x2 + 12x + 35 = 1x + 521x+ 52x2 + 9x + 20 = 1x + 421x



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FACTORING TRINOMIALS OF THE FORM ax2 + bx + c SECTION 6.3 367

6.3 FAC TO R I N G T R I N O M I A L S O F T H E F O R M ax2 bx c

æ¡ Incorrect middle term

O b j e c t i v e s

1 Factor trinomials of the form

2 Factor out a GCF before factoring a trinomial of the form

3 Factor perfect square trinomials.

4 Factor trinomials of the form by grouping.

1 In this section, we factor trinomials of the form such as

Notice that the coefficient of the squared variable in these trinomials is a number otherthan 1. We will factor these trinomials using a trial-and-check method based on FOILand our work in the last section.

To begin, let’s review the relationship between the numerical coefficients of thetrinomial and the numerical coefficients of its factored form. For example, since

the factored form of is

Notice that 2x and x are factors of the first term of the trinomial. Also, 6 and 1 arefactors of 6, the last term of the trinomial, as shown:

Also notice that 13x, the middle term, is the sum of the following products:

Middle term

Let’s use this pattern to factor First, we find factors of Since all nu-merical coefficients in this trinomial are positive, we will use factors with positive nu-merical coefficients only. Thus, the factors of are 5x and x. Let’s try these factors asfirst terms of the binomials. Thus far, we have

Next, we need to find positive factors of 2. Positive factors of 2 are 1 and 2. Nowwe try possible combinations of these factors as second terms of the binomials until weobtain a middle term of 7x.

15x + 121x"

+ 22 = 5x2 + 11x + 2 1x 5

+10x 11x

5x2 + 7x + 2 = 15x + 21x + 2

5x2

5x2 .5x2 + 7x + 2.

2x2 + 13x + 6 = 12x + 121x"

+ 62 1x

5

+12x 13x

2x2 + 13x + 6 = (2x + 1)(x + 6)2x # x

1 # 6

2x2 ,

2x2 + 13x + 6 = 12x + 121x + 622x2 + 13x + 612x + 121x + 62 = 2x2 + 13x + 6,

3x2 + 11x + 6, 8x2 - 22x + 5, 2x2 + 13x - 7

ax2 + bx + c ,

ax2 + bx + c

ax2 + bx + c .

ax2 + bx + c .




Let’s try switching factors 2 and 1.

Thus the factored form of is To check, we multi-ply and The product is 5x2 + 7x + 2.1x + 12 .15x + 22 15x + 221x + 12 .5x2 + 7x + 2

15x + 221x"

+ 12 = 5x2 + 7x + 2 2x 5

+ 5x

7x

E X A M P L E 1Factor .3x2 + 11x + 6

Solut ion Since all numerical coefficients are positive, we use factors with positive numerical co-efficients. We first find factors of

Factors of

If factorable, the trinomial will be of the form

Next we factor 6.

Factors of

Now we try combinations of factors of 6 until a middle term of 11x is obtained. Let’s try1 and 6 first.

Now let’s next try 6 and 1.

Before multiplying, notice that the terms of the factor have a common factor of3. The terms of the original trinomial have no common factor otherthan 1, so the terms of the factored form of can contain no common fac-tor other than 1. This means that is not a factored form.

Next let’s try 2 and 3 as last terms.

Thus the factored form of is 13x + 221x + 32 .3x2 + 11x + 6

13x + 221x"

+ 32 = 3x2 + 11x + 6

2x 5

+ 9x

11x

13x + 621x + 123x2 + 11x + 63x2 + 11x + 6

3x + 6

13x + 621x + 12

13x + 121x"

+ 62 = 3x2 + 19x + 6 1x 5

+18x

19x

6 : 6 = 1 # 6, 6 = 2 # 3

3x2 + 11x + 6 = 13x + 21x + 2

3x2 : 3x2 = 3x # x

3x2 .CLASSROOM EXAMPLE

Factor

answer: 12x + 5212x + 124x2 + 12x + 5.

æ

¡ Correct middle term

æ

¡ Incorrect middle term

æ

Correct middle term¡

H e l p f u l H i n tIf the terms of a trinomial have no common factor (other than 1), then theterms of neither of its binomial factors will contain a common factor (otherthan 1).



For example,

3x2 + 11x + 6 13x + 321x + 225

no common factor

CONCEPT CHECKDo the terms of have a common factor? Without multiplying, decide which of the fol-lowing factored forms could not be a factored form of

a. b. c. d. 13x + 921x + 2213x + 621x + 3213x + 221x + 9213x + 1821x + 123x2 + 29x + 18.

3x2 + 29x + 18

Concept Check Answer:no; a, c, d

E X A M P L E 2Factor .8x2 - 22x + 5

Solut ion

æ

æ

Incorrect middle term¡

Incorrect middle term¡

æ


CLASSROOM EXAMPLEFactor.a.b.answer:a.

b. 12x - 321x - 4213x - 1212x - 122x2 - 11x + 126x2 - 5x + 1

E X A M P L E 3Factor: 2x2 + 13x - 7

Solut ion Factors of

Factors of

We try possible combinations of these factors:

-7 : -7 = -1 # 7, -7 = 1 # -7

2x2 : 2x2 = 2x # x

Common factor of 3, so cannot be the factorization of 3x2 11x 6

qq

Factors of

We’ll try 8x and x.

Since the middle term, has a negative numerical coefficient, we factor 5 intonegative factors.

Let’s try and

Now let’s try and

Don’t give up yet! We can still try other factors of Let’s try 4x and 2x with and

The factored form of is 14x - 1212x - 52 .8x2 - 22x + 5

14x - 1212x"

- 52 = 8x2 - 22x + 5 -2x 5

+1-20x2 -22x

-5.-18x2 .

18x - 521x"

- 12 = 8x2 - 13x + 5 -5x 5

+1-8x2-13x

-1.-5

18x - 121x"

- 52 = 8x2 - 41x + 5 -1x 5

+1-40x2 -41x

-5.-1Factors of 5 : 5 = -1 # -5

-22x ,

8x2 - 22x + 5 = 18x + 21x + 2

8x2 : 8x2 = 8x # x , 8x2 = 4x # 2x


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CLASSROOM EXAMPLEFactor answer:15x + 2217x - 22

35x2 + 4x - 4.

E X A M P L E 4Factor .10x2 - 13xy - 3y2

Solut ion Factors of Factors of

We try some combinations of these factors:

Correct middle term

The factored form of is

2 Don’t forget that the best first step in factoring any polynomial is to look for acommon factor to factor out.

12x - 3y215x + y2 .10x2 - 13xy - 3y2

12x - 3y215x + y2 = 10x2 - 13xy - 3y2 15x + 3y212x - y2 = 10x2 + xy - 3y2 1x + 3y2110x - y2 = 10x2 + 29xy - 3y2 110x - 3y21x + y2 = 10x2 + 7xy - 3y2

-3y2 : -3y2 = -3y # y , -3y2 = 3y # -y

10x2 : 10x2 = 10x # x , 10x2 = 2x # 5x

CLASSROOM EXAMPLEFactor answer: 17a + 2b212a - b2

14a2 - 3ab - 2b2 .

E X A M P L E 5Factor 24x4 + 40x3 + 6x2 .

Solut ion Notice that all three terms have a common factor of First, factor out

Next, factor Factors of

Since all terms in the trinomial have positive numerical coefficients, factor 3 using pos-itive factors only.

Factors of 3:

We try some combinations of the factors.

Correct middle term

The factored form of is 2x212x + 3216x + 12 .24x4 + 40x3 + 6x2

2x212x + 3216x + 12 = 2x2112x2 + 20x + 32 2x2112x + 121x + 32 = 2x2112x2 + 37x + 32 2x214x + 3213x + 12 = 2x2112x2 + 13x + 32

3 = 1 # 3

12x2 : 12x2 = 6x # 2x , 12x2 = 4x # 3x , 12x2 = 12x # x

12x2 + 20x + 3.

24x4 + 40x3 + 6x2 = 2x2112x2 + 20x + 322x2 .2x2 .

CLASSROOM EXAMPLE

Factor

answer: 3x12y - 121y + 626xy2 + 33xy - 18x .

E X A M P L E 6

Factor 4x2 - 12x + 9.

Solut ion Factors of Since the middle term has a negative numerical coefficient, factor 9 into negativefactors only.

-12x

4x2 : 4x2 = 2x # 2x , 4x2 = 4x # x

Incorrect middle term

Correct middle term

The factored form of is 12x - 121x + 72 .2x2 + 13x - 7

12x - 121x + 72 = 2x2 + 13x - 7

12x + 121x - 72 = 2x2 - 13x - 7

H e l p f u l H i n tDon’t forget to include the common factor in the factored form.


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CLASSROOM EXAMPLE

Factor

answer: 13n - 1229n2 - 6n + 1.

Factors of 9:

The correct combination is

Thus, which can also be written as

3 Notice in Example 6 that The trinomialis called a perfect square trinomial since it is the square of the binomi-

al

In the last chapter, we learned a shortcut special product for squaring a binomial,recognizing that

The trinomial is a perfect square trinomial, since it is the square ofthe binomial We can use this pattern to help us factor perfect square trinomials.To use this pattern, we must first be able to recognize a perfect square trinomial. A tri-nomial is a perfect square when its first term is the square of some expression a, its lastterm is the square of some expression b, and its middle term is twice the product of theexpressions a and b.When a trinomial fits this description, its factored form is 1a + b22 .

a + b .a2 + 2ab + b2

1a + b22 = a2 + 2ab + b2

2x - 3.4x2 - 12x + 9

4x2 - 12x + 9 = 12x - 322 .

12x - 322 .4x2 - 12x + 9 = 12x - 3212x - 32 ,

12x - 3212x"

- 32 = 4x2 - 12x + 9 -6x 5

+1-6x2 -12x

9 = -3 # -3, 9 = -1 # -9

æ


Perfect Square Trinomials

a2 - 2ab + b2 = 1a - b22 a2 + 2ab + b2 = 1a + b22

E X A M P L E 7Factor x2 + 12x + 36.

Solut ion This trinomial is a perfect square trinomial since:

1. The first term is the square of x:2. The last term is the square of 6:3. The middle term is twice the product of x and 6:Thus, x2 + 12x + 36 = 1x + 622 .

12x = 2 # x # 6.36 = 1622 .x2 = 1x22 .

CLASSROOM EXAMPLE

Factor

answer: 1x + 1022x2 + 20x + 100.

Factor 25x2 + 25xy + 4y2 .

Solut ion

CLASSROOM EXAMPLEFactor answer: 13r + s)(3r + 7s2

9r2 + 22rs + 7s2 .

Determine whether or not this trinomial is a perfect square by considering the samethree questions.

1. Is the first term a square? Yes,2. Is the last term a square? Yes,3. Is the middle term twice the product of 5x and 2y? No. not 25xy.Therefore, is not a perfect square trinomial. It is factorable, though.Using earlier techniques, we find that 25x2 + 25xy + 4y2 = 15x + 4y215x + y2 .25x2 + 25xy + 4y2

2 # 5x # 2y = 20xy ,4y2 = 12y22 .25x2 = 15x22 .

E X A M P L E 8


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Factoring Trinomials of the Form by GroupingStep 1. Find two numbers whose product is and whose sum is b.

Step 2. Write the middle term, bx, using the factors found in Step 1.

Step 3. Factor by grouping.

a # c

ax2 bx c

E X A M P L E 9Factor 4m2 - 4m + 1.

Solut ion This is a perfect square trinomial since and

4 If we extend our work from Section 6.1, grouping can also be used to factor tri-nomials of the form To use this method, write the trinomial as a four-term polynomial. For example, to factor using grouping, find twonumbers whose product is and whose sum is 11. Since we want a positiveproduct and a positive sum, we consider positive pairs of factors of 24 only.

Factors of 24 Sum of Factors

1, 24

2, 123, 8

The factors are 3 and 8. Use these factors to write the middle term 11x as Re-place 11x with in the original trinomial and factor by grouping.

In general, we have the following:

= 12x + 321x + 42 = x12x + 32 + 412x + 32 = 12x2 + 3x2 + 18x + 122

2x2 + 11x + 12 = 2x2 + 3x + 8x + 12

3x + 8x3x + 8x .

3 + 8 = 112 + 12 = 141 + 24 = 25

2 # 12 = 242x2 + 11x + 12

ax2 + bx + c .

4m2 - 4m + 1 = 12m - 1224m = 2 # 2m # 1.4m2 = 12m22 , 1 = 1122 ,

CLASSROOM EXAMPLE

Factor

answer: 15x - 22225x2 - 20x + 4.

E X A M P L E 1 0Factor by grouping.8x2 - 14x + 5

Solut ion This trinomial is of the form with and Step 1. Find two numbers whose product is or and whose sum is b or

The numbers are and Step 2. Write as so that

Step 3. Factor by grouping.

= 12x - 1214x - 52 8x2 - 4x - 10x + 5 = 4x12x - 12 - 512x - 12

8x2 - 14x + 5 = 8x2 - 4x - 10x + 5

-4x - 10x-14x

-10.-4-14.8 # 5 = 40,a # c

c = 5.a = 8, b = -14,ax2 + bx + c

CLASSROOM EXAMPLEFactor by grouping.answer: 1x + 4213x + 22

3x2 + 14x + 8

H e l p f u l H i n tA perfect square trinomial that is not recognized as such can be factored byother methods.



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E X A M P L E 1 1Factor by grouping.3x2 - x - 10

Solut ion

STUDY SKILLS REMINDERAre You Satisfied with Your Performance in ThisCourse Thus Far?If not, ask yourself the following questions:

Am I attending all class periods and arriving on time?

Am I working and checking my homework assignments?

Am I getting help when I need it?

In addition to my instructor, am I using the supplements tothis text that could help me? For example, the tutorial videolessons? MathPro, the tutorial software?

Am I satisfied with my performance on quizzes and tests?

If you answered no to any of these questions, read or rereadSection 1.1 for suggestions in these areas. Also, you may wantto contact your instructor for additional feedback.

In and Step 1. Find two numbers whose product is or and whose sum is

b or The numbers are and 5.Step 2.

Step 3.

= 1x - 2213x + 52 = 3x1x - 22 + 51x - 223x2 - x - 10 = 3x2 - 6x + 5x - 10

-6-1.31-102 = -30a # c

c = -10.3x2 - x - 10, a = 3, b = -1,

CLASSROOM EXAMPLE

Factor by grouping.

answer: 12x + 1213x - 526x2 - 7x - 5

E X A M P L E 1 2Factor by grouping.4x2 + 11x - 3

Solut ion In and Step 1. Find two numbers whose product is or and whose sum is b

or 11. The numbers are and 12.Step 2.

= 14x - 121x + 32 = x14x - 12 + 314x - 12

4x2 + 11x - 3 = 4x2 - 1x + 12x - 3-1

41-32 = -12a # c

c = -3.4x2 + 11x - 3, a = 4, b = 11,

CLASSROOM EXAMPLE

Factor by grouping.

answer: 215x - 1213x - 2230x2 - 26x + 4

State whether or not each trinomial is a perfect trinomial square.

1. yes 2. yes 3. no

4. no 5. yes 6. yesy2 - 16y + 649y2 + 6y + 1x2 - 4x + 2

y2 + 2y + 49x2 - 12x + 4x2 + 14x + 49

MENTAL MATH


FACTORING BINOMIALS SECTION 6.4 375

6.4 FAC TO R I N G B I N O M I A L S

O b j e c t i v e s

1 Factor the difference of two squares.

2 Factor the sum or difference of two cubes.

1 When learning to multiply binomials in Chapter 5, we studied a special prod-uct, the product of the sum and difference of two terms, a and b:

For example, the product of and is

The binomial is called a difference of squares. In this section, we use the patternfor the product of a sum and difference to factor the binomial difference of squares.

To use this pattern to help us factor, we must be able to recognize a difference ofsquares. A binomial is a difference of squares when it is the difference of the square ofsome expression a and the square of some expression b.

x2 - 9

1x + 321x - 32 = x2 - 9

x - 3x + 3

1a + b21a - b2 = a2 - b2

Difference of Two Squares

a2 - b2 = 1a + b21a - b2

E X A M P L E 1

Factor x2 - 25.

Solut ion is the difference of two squares since Therefore,

Multiply to check.

x2 - 25 = x2 - 52 = 1x + 521x - 52x2 - 25 = x2 - 52.x2 - 25

CLASSROOM EXAMPLEFactor .answer: 1a + 421a - 42

a2 - 16




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Write as and as

x4 - y6 = 1x222 - 1y322 = 1x2 + y321x2 - y321y322 .y61x222x4

CLASSROOM EXAMPLEFactor .answer: 1p2 + 921p + 321p - 32

p4 - 81

E X A M P L E 4

Factor 9x2 - 36.

Solut ion Remember when factoring always to check first for common factors. If there are com-mon factors, factor out the GCF and then factor the resulting polynomial.

Factor out the GCF 9.

In this example, if we forget to factor out the GCF first, we still have the difference oftwo squares.

This binomial has not been factored completely since both terms of both binomial fac-tors have a common factor of 3.

Then

Factoring is easier if the GCF is factored out first before using other methods.

9x2 - 36 = 13x + 6213x - 62 = 31x + 2231x - 22 = 91x + 221x - 22

3x + 6 = 31x + 22 and 3x - 6 = 31x - 22

9x2 - 36 = 13x22 - 1622 = 13x + 6213x - 62

= 91x + 221x - 22 = 91x2 - 222

9x2 - 36 = 91x2 - 42CLASSROOM EXAMPLEFactor.a. b.answer:a.

b. -13x + 10213x - 102314x2 + 1212x + 1212x - 12

-9x2 + 10048x4 - 3

E X A M P L E 5

Factor x2 + 4.

Solut ion The binomial is the sum of squares since we can write as Wemight try to factor using or But when multiplying tocheck, neither factoring is correct.

In both cases, the product is a trinomial, not the required binomial. In fact, is aprime polynomial.

x2 + 4

1x - 221x - 22 = x2 - 4x + 4 1x + 221x + 22 = x2 + 4x + 4

1x - 221x - 22 .1x + 221x + 22 x2 + 22.x2 + 4x2 + 4

CLASSROOM EXAMPLEFactor.a.

b.

answer:a. prime polynomial

b. Ac + 35 B Ac - 3

5 B

c2 -9

25

s2 + 9


E X A M P L E 2Factor each difference of squares.

a. b. 25a2 - 9b24x2 - 1

Solut ion a.

b. 25a2 - 9b2 = 15a22 - 13b22 = 15a + 3b215a - 3b24x2 - 1 = 12x22 - 12 = 12x + 1212x - 12

CLASSROOM EXAMPLEFactor each difference of squares.a. b.answer:a.

b. 14a + 7b214a - 7b215x + 1215x - 12

16a2 - 49b225x2 - 1

E X A M P L E 3

Factor x4 - y6 .

Solut ion


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H e l p f u l H i n tAfter the greatest common factor has been removed, the sum of two squarescannot be factored further using real numbers.

2 Although the sum of two squares usually does not factor, the sum or differ-ence of two cubes can be factored and reveals factoring patterns. The pattern for thesum of cubes is illustrated by multiplying the binomial and the trinomial

Sum of cubes.

The pattern for the difference of two cubes is illustrated by multiplying the binomialby the trinomial The result is

Difference of cubes.1x - y21x2 + xy + y22 = x3 - y3

x2 + xy + y2 .x - y

1x + y21x2 - xy + y22 = x3 + y3

x2 - xy + y2 x + y x2y - xy2 + y3

x3 - x2y + xy2 x3 + y3

x2 - xy + y2 .x + y

Sum or Difference of Two Cubes

a3 - b3 = 1a - b21a2 + ab + b22 a3 + b3 = 1a + b21a2 - ab + b22

E X A M P L E 6

Factor x3 + 8.

Solut ion First, write the binomial in the form

Write in the form .

If we replace a with x and b with 2 in the formula above, we have

= 1x + 221x2 - 2x + 42 x3 + 23 = 1x + 22[x2 - 1x2122 + 22]

a3 b3x3 + 8 = x3 + 23

a3 + b3 .

H e l p f u l H i n tWhen factoring sums or differences of cubes, notice the sign patterns.

same sign

opposite signs always positivesame sign

opposite signs always positive

T—— — –————Tx3 - y3 = 1x - y21x2 + xy + y22

q

T— —— — –———Tx3 + y3 = 1x + y21x2 - xy + y22

CLASSROOM EXAMPLE

Factor

answer: 1x + 321x2 - 3x + 92x3 + 27.



E X A M P L E 7

Factor y3 - 27.

Solut ion Write in the form

= 1y - 321y2 + 3y + 92 = 1y - 32[y2 + 1y2132 + 32]

a3 b3. y3 - 27 = y3 - 33

CLASSROOM EXAMPLE

Factor

answer: 1z - 221z2 + 2z + 42z3 - 8.

E X A M P L E 8

Factor 64x3 + 1.

Solut ion

= 14x + 12116x2 - 4x + 12 = 14x + 12[14x22 - 14x2112 + 12]

64x3 + 1 = 14x23 + 13

CLASSROOM EXAMPLE

Factor

answer: 15a - 12125a2 + 5a + 12125a3 - 1.

E X A M P L E 9

Factor 54a3 - 16b3 .

Solut ion Remember to factor out common factors first before using other factoring methods.

Factor out the GCF 2.

Difference of two cubes.

= 213a - 2b219a2 + 6ab + 4b22 = 213a - 2b2[13a22 + 13a212b2 + 12b22]= 2[13a23 - 12b23]

54a3 - 16b3 = 2127a3 - 8b32CLASSROOM EXAMPLE

Factor

answer:212x + 5y214x2 - 10xy + 25y22

16x3 + 250y3 .

A graphing calculator is a convenient tool for evaluating an expression at a given replacement value. For example, let’s evaluate when To do so, storethe value 2 in the variable x and then enter and evaluate the algebraic expression.

The value of when is You may want to use this method for eval-uating expressions as you explore the following.

We can use a graphing calculator to explore factoring patterns numerically.Use your calculator to evaluate and for eachvalue of x given in the table. What do you observe?

1x - 122x2 - 2x + 1, x2 - 2x - 1,

-8.x = 2x2 - 6x

2 : X2

X2 - 6X-8

x = 2.x2 - 6x




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Notice in each case that Because for each x in the table thevalue of and the value of are the same, we might guess that

We can verify our guess algebraically with multiplication:

1x - 121x - 12 = x2 - x - x + 1 = x2 - 2x + 1

x2 - 2x + 1 = 1x - 122 .1x - 122x2 - 2x + 1

x2 - 2x - 1 Z 1x - 122 .

x = 0

x = -12.1

x = 2.7

x = -3

x = 5

1x - 122x2 - 2x - 1x2 - 2x + 1

MENTAL MATHWrite each number as a square.

1. 1 2. 25 3. 81 4. 64 5. 9 6. 100

Write each number as a cube.

7. 1 8. 64 9. 8 10. 27 33234313

1023282925212

2. 4. 6. 7. 8.2 2

914 + 3x214 - 3x2111 + 10x2111 - 10x217a + 4217a - 421x + 1021x - 1021y + 921y - 92

17. 18. 20. 23.24. 25. 26. 14x - 12116x2 + 4x + 1212a + 1214a2 - 2a + 121b - 221b2 + 2b + 42

1a + 321a2 - 3a + 921x7 + y221x7 - y2217b2 + 1217b2 - 1217a2 + 4217a2 - 42


SOLVING QUADRATIC EQUATIONS BY FACTORING SECTION 6.5 385

6.5 S O LV I N G Q UA D R AT I C E Q UAT I O N S B Y FAC TO R I N G

O b j e c t i v e s

1 Define quadratic equation.

2 Solve quadratic equations by factoring.

3 Solve equations with degree greater than 2 by factoring.

31. 34. 13y + 5212y - 3215 - 2x214 + x222

99. 105. answers may vary1x + 221x - 221x + 7268.69.70.

71.

73.

86.83. 1x - 1521x - 82

2y13y + 521y - 3213x + 521x - 1217t - 1212t - 121x - y217 + y215 + x21x + y21y + 321y - 321x2 + 32

79. 80. 90. 13ab + 2219a2b2 - 6ab + 4213 - x213 + x212 - x212 + x213 - x213 + x211 - x211 + x2

91.97.106. yes; 91x2 + 9y221a2 + 221a + 222xy11 + 6x211 - 6x2




256 feet

Quadratic EquationA quadratic equation is one that can be written in the form

where a, b, and c are real numbers, and a Z 0.ax2 + bx + c = 0,

Zero Factor TheoremIf a and b are real numbers and if then or b = 0.a = 0ab = 0,

1 Linear equations, while versatile, are not versatile enough to model manyreal-life phenomena. For example, let’s suppose an object is dropped from the top of a256-foot cliff and we want to know how long before the object strikes the ground. Theanswer to this question is found by solving the equation (See Exam-ple 1 in the next section.) This equation is called a quadratic equation because it con-tains a variable with an exponent of 2, and no other variable in the equation containsan exponent greater than 2. In this section, we solve quadratic equations by factoring.

-16t2 + 256 = 0.

Notice that the degree of the polynomial is 2. Here are a few moreexamples of quadratic equations.

Quadratic Equations

The form is called the standard form of a quadratic equation. Thequadratic equations and are in standard form.One side of the equation is 0 and the other side is a polynomial of degree 2 written indescending powers of the variable.

2 Some quadratic equations can be solved by making use of factoring and thezero factor theorem.

-16t2 + 256 = 03x2 + 5x + 6 = 0ax2 + bx + c = 0

3x2 + 5x + 6 = 0 x2 = 9 y2 + y = 1

ax2 + bx + c

This theorem states that if the product of two numbers is 0 then at least one of thenumbers must be 0.

E X A M P L E 1Solve 1x - 321x + 12 = 0.

Solut ion If this equation is to be a true statement, then either the factor must be 0 or thefactor must be 0. In other words, either

x - 3 = 0 or x + 1 = 0

x + 1x - 3


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If we solve these two linear equations, we have

Thus, 3 and are both solutions of the equation To check, wereplace x with 3 in the original equation. Then we replace x with in the originalequation.

-11x - 321x + 12 = 0.-1

x = 3 or x = -1

CheckReplace x with 3. Replace x with True True

The solutions are 3 and or we say that the solution set is 5-1, 36 .-1,

1-42102 = 0 0142 = 01 . 1-1 - 321-1 + 12 0 13 - 3213 + 12 0

1x - 321x + 12 = 0 1x - 321x + 12 = 0

CLASSROOM EXAMPLE

Solve

answer: 7, -2

1x - 721x + 22 = 0.

H e l p f u l H i n tThe zero factor property says that if a product is 0, then a factor is 0.

If then or If then or If then or

Use this property only when the product is 0. For example, if we donot know the value of a or b. The values may be or or any other two numbers whose product is 8.

a = 8, b = 1,a = 2, b = 4a # b = 8,

2x - 3 = 0.x + 7 = 01x + 7212x - 32 = 0,x + 5 = 0.x = 0x1x + 52 = 0,

b = 0.a = 0a # b = 0,

E X A M P L E 2

Solve x2 - 9x = -20.

Solut ion First, write the equation in standard form; then factor.

Write in standard form by adding 20 to both sides.Factor.

Next, use the zero factor theorem and set each factor equal to 0.

Set each factor equal to 0.Solve.

Check the solutions by replacing x with each value in the original equation. The solu-tions are 4 and 5.

The following steps may be used to solve a quadratic equation by factoring.

x = 4 or x = 5 x - 4 = 0 or x - 5 = 0

1x - 421x - 52 = 0 x2 - 9x + 20 = 0

x2 - 9x = -20CLASSROOM EXAMPLE

Solve

answer: 12, 2

x2 - 14x = -24.

Solving Quadratic Equations by FactoringStep 1. Write the equation in standard form:

Step 2. Factor the quadratic completely.

Step 3. Set each factor containing a variable equal to 0.

Step 4. Solve the resulting equations.

Step 5. Check each solution in the original equation.

ax2 + bx + c = 0.

Since it is not always possible to factor a quadratic polynomial, not all quadraticequations can be solved by factoring. Other methods of solving quadratic equationsare presented in Chapter 9.



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E X A M P L E 3Solve x12x - 72 = 4.

Solut ion

CLASSROOM EXAMPLE

Solve answer: 5, -1

x1x - 42 = 5.

First, write the equation in standard form; then factor.

Multiply.Write in standard form.Factor.Set each factor equal to zero.Solve.

Check both solutions and 4.- 12

x = - 12

2x = -1 or x = 4 2x + 1 = 0 or x - 4 = 0

12x + 121x - 42 = 0 2x2 - 7x - 4 = 0

2x2 - 7x = 4 x12x - 72 = 4

E X A M P L E 4Solve -2x2 - 4x + 30 = 0.

Solut ion The equation is in standard form so we begin by factoring out a common factor of

Factor out Factor the quadratic.

Next, set each factor containing a variable equal to 0.

Set each factor containing a variable equal to 0.

or Solve.Note that the factor is a constant term containing no variables and can never equal0. The solutions are and 3.

3 Some equations involving polynomials of degree higher than 2 may also besolved by factoring and then applying the zero factor theorem.

-5-2

x = 3x = -5

x + 5 = 0 or x - 3 = 0

-21x + 521x - 32 = 02 . -21x2 + 2x - 152 = 0

-2x2 - 4x + 30 = 0

-2.

CLASSROOM EXAMPLE

Solve answer: -2, 6

-5x2 + 20x + 60 = 0.

H e l p f u l H i n tTo solve the equation above do not set each factor equal to 4.Remember that to apply the zero factor property, one side of the equation mustbe 0 and the other side of the equation must be in factored form.

x12x - 72 = 4,

E X A M P L E 5Solve 3x3 - 12x = 0.

Solut ion Factor the left side of the equation. Begin by factoring out the common factor of 3x.

Factor out the GCF 3x.Factor a difference of squares.x2 4 , 3x1x + 221x - 22 = 0

3x1x2 - 42 = 0 3x3 - 12x = 0CLASSROOM EXAMPLE

Solve answer: 0, 3 , -3

2x3 - 18x = 0.



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Begin by factoring out the GCF 2x.

Factor out the GCF 2x.

Factor the quadratic.

Set each factor containing a variable equal to 0.Solve.

Check by replacing x with each solution in the cubic equation. The solutions are 0,and 5.

In Chapter 3, we graphed linear equations in two variables, such as Recall that to find the x-intercept of the graph of a linear equation, let and solvefor x. This is also how to find the x-intercepts of the graph of a quadratic equation intwo variables, such as y = x2 - 5x + 4.

y = 0y = 5x - 6.

-3,

x = 0 or x = 5 or x = -3

2x = 0 or x - 5 = 0 or x + 3 = 0

2x1x - 521x + 32 = 0

2x1x2 - 2x - 152 = 0

2x3 - 4x2 - 30x = 0


Set each factor equal to 0.Solve.

Thus, the equation has three solutions: and 2.To check, replace xwith each solution in the original equation.

Let Let Let

Substituting or 2 into the original equation results each time in a true equation.The solutions are and 2.0, -2,

0, -2, 0 = 0 0 = 0

3182 - 24 0 31-82 + 24 0 0 = 0 31223 - 12122 0 31-223 - 121-22 0 31023 - 12102 0

x 2 .x 2 .x 0 .

0 , -2,3x3 - 12x = 0

x = 0 or x = -2 or x = 2

3x = 0 or x + 2 = 0 or x - 2 = 0

E X A M P L E 6Solve 15x - 1212x2 + 15x + 182 = 0.

Solut ion

Factor the trinomial.

Set each factor equal to 0.

Solve.

The solutions are and Check by replacing x with each solution in the orig-

inal equation. The solutions are and 15

.-6, - 32

,

-6.15

, - 32

,

x =15 or x = -

32

5x = 1 or 2x = -3 or x = -6

5x - 1 = 0 or 2x + 3 = 0 or x + 6 = 0

15x - 1212x + 321x + 62 = 0

15x - 1212x2 + 15x + 182 = 0

CLASSROOM EXAMPLE

Solve

answer: -3, - 13

, 7

1x + 3213x2 - 20x - 72 = 0.

E X A M P L E 7

Solve 2x3 - 4x2 - 30x = 0.

Solut ion

CLASSROOM EXAMPLE

Solve

answer: 0, 4 , -1

3x3 - 9x2 - 12x = 0.


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112345

12345

2 3 4 512345

(1, 0) (4, 0)x

y

Let and solve for x.

Let Factor.Set each factor equal to 0.Solve.

The x-intercepts of the graph of are (1, 0) and (4, 0).The graph of is shown in the margin.In general, a quadratic equation in two variables is one that can be written in the

form where The graph of such an equation is called aparabola and will open up or down depending on the value of a.

Notice that the x-intercepts of the graph of are the real num-ber solutions of Also, the real number solutions of

are the x-intercepts of the graph of We studymore about graphs of quadratic equations in two variables in Chapter 9.

y = ax2 + bx + c .0 = ax2 + bx + c0 = ax2 + bx + c .

y = ax2 + bx + c

a Z 0.y = ax2 + bx + c

y = x2 - 5x + 4y = x2 - 5x + 4

x = 1 or x = 4 x - 1 = 0 or x - 4 = 0

0 = 1x - 121x - 42y 0 . 0 = x2 - 5x + 4

y = x2 - 5x + 4

y = 0


Find the x-intercepts of the graph of y = x2 - 5x + 4.

Solut ion

CLASSROOM EXAMPLE

Find the x-intercepts of the graph of

answer: 15, 02A - 32 , 0 B

y = 2x2 - 7x - 15.

A grapher may be used to find solutions of a quadratic equation whether the relat-ed quadratic polynomial is factorable or not. For example, let’s use a grapher to ap-proximate the solutions of To do so, graph Recall that the x-intercepts of this graph are the solutions ofNotice that the graph appears to have an x-intercept between and and one be-tween 0 and 1. Many graphers contain a TRACE feature. This feature activates a graphcursor that can be used to trace along a graph while the corresponding x- and y-coordi-nates are shown on the screen. Use the TRACE feature to confirm that x-intercepts liebetween and and also 0 and 1. To approximate the x-intercepts to the nearesttenth, use a ROOT or a ZOOM feature on your grapher or redefine the viewing win-dow. (A ROOT feature calculates the x-intercept.A ZOOM feature magnifies the view-ing window around a specific location such as the graph cursor.) If we redefine thewindow to [0, 1] on the x-axis and on the y-axis, the following graph is generated.By using the TRACE feature, we can conclude that one x-intercept is approximately0.6 to the nearest tenth. By repeating these steps for the other x-intercept, we findthat it is approximately -4.6.

[-1, 1]

-4-5

-4-50 = x2 + 4x - 3.

y1 = x2 + 4x - 3.0 = x2 + 4x - 3.


1

1

1

x .63829787 y .03938431

0

y x2 4x 3 10

10

1010

E X A M P L E 8

Graph of y ax2 bx cx-intercepts are solutions of 0 ax2 bx c

no solution 2 solutions1 solution 2 solutions

x

y y y y

x x x


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Use a grapher to approximate the real number solutions to the nearest tenth. If an equation has no real number solution, state so.

1.

3.

5. -x2 + x + 5 = 0

2x2 + x + 2 = 0

3x2 - 4x - 6 = 0 2.

4.

6. 10x2 + 6x - 3 = 0

-4x2 - 5x - 4 = 0

x2 - x - 9 = 0

Solve each equation by inspection.

1. 3, 7 2. 5, 2 3.

4. 5. 6. 1, -21x - 121x + 22 = 0-1, 31x + 121x - 32 = 0-2, -31x + 221x + 32 = 0

-8, -61x + 821x + 62 = 01a - 521a - 22 = 01a - 321a - 72 = 0

MENTAL MATH

2. 4. 0, 7 6. 8. 9. 24. 0, 7 45. 47. 49. 56. no real solutionx - 6 x + 1 = 0-3, -2


QUADRATIC EQUATIONS AND PROBLEM SOLVING SECTION 6.6 393

6.6 Q UA D R AT I C E Q UAT I O N S A N D P R O B L E M S O LV I N G

O b j e c t i v e

1 Solve problems that can be modeled by quadratic equations.

1 Some problems may be modeled by quadratic equations. To solve these prob-lems, we use the same problem-solving steps that were introduced in Section 2.5.Whensolving these problems, keep in mind that a solution of an equation that models a prob-lem may not be a solution to the problem. For example, a person’s age or the length ofa rectangle is always a positive number. Discard solutions that do not make sense as so-lutions of the problem.

E X A M P L E 1

FINDING THE LENGTH OF TIME

For a TV commercial, a piece of luggage is dropped from a cliff 256 feet above theground to show the durability of the luggage. Neglecting air resistance, the height h infeet of the luggage above the ground after t seconds is given by the quadratic equation

Find how long it takes for the luggage to hit the ground.h = -16t2 + 256

256 feet

Solut ion 1. UNDERSTAND. Read and reread the problem.Then draw a picture of the problem.

CLASSROOM EXAMPLEAn object is dropped from the roof of a144-foot-tall building. Neglecting air re-sistance, the height h in feet of the objectabove ground after t seconds is given bythe quadratic equation

Find how long it takes the object to hitthe ground.answer: 3 seconds

h = -16t2 + 144


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E X A M P L E 2

FINDING A NUMBER

The square of a number plus three times the number is 70. Find the number.

Solut ion

CLASSROOM EXAMPLEThe square of a number minus twice thenumber is 63. Find the number.answer: 9 and -7

The equation models the height of the falling luggage at time t. Fa-miliarize yourself with this equation by finding the height of the luggage at

and When the height of the suitcase is

When the height of the suitcase is

2. TRANSLATE. To find how long it takes the luggage to hit the ground, we want toknow the value of t for which the height

3. SOLVE. We solve the quadratic equation by factoring.

4. INTERPRET. Since the time t cannot be negative, the proposed solution is 4 seconds.Check: Verify that the height of the luggage when t is 4 seconds is 0.When

State: The solution checks and the luggage hits the ground 4 seconds after it isdropped.

t = 4 seconds, h = -161422 + 256 = -256 + 256 = 0 feet .

t = 4 t = -4 t - 4 = 0 or t + 4 = 0

0 = -161t - 421t + 42 0 = -161t2 - 162 0 = -16t2 + 256

0 = -16t2 + 256

h = 0.

h = -161222 + 256 = 192 feet .

t = 2 seconds,

h = -161122 + 256 = 240 feet .

t = 1 second,t = 2 seconds.t = 1 second

h = -16t2 + 256

1. UNDERSTAND. Read and reread the problem. Suppose that the number is 5. Thesquare of 5 is or 25. Three times 5 is 15. Then not 70, so the num-ber must be greater than 5. Remember, the purpose of proposing a number, such as5, is to better understand the problem. Now that we do, we will let number.

2. TRANSLATE.

3. SOLVE.

Subtract 70 from both sides.Factor.Set each factor equal to 0.Solve.

4. INTERPRET.

Check: The square of is or 100. Three times is orThen the correct sum, so checks.-10100 + 1-302 = 70,-30.

31-102-101-1022 ,-10

x = -10 x = 7 x + 10 = 0 or x - 7 = 0

1x + 1021x - 72 = 0 x2 + 3x - 70 = 0

x2 + 3x = 70

70=3x+x2TTTTT

70isthree timesthe number

plusthe square of a

number

x = the

25 + 15 = 40,52


1. UNDERSTAND. Read and reread the problem. Since we are finding the length ofthe base and the height, we let

and since the height is 2 meters less than twice the base,

An illustration is shown to the left.2. TRANSLATE. We are given that the area of the triangle is 30 square meters, so we

use the formula for area of a triangle.

3. SOLVE. Now we solve the quadratic equation.

Multiply.Write in standard form.Factor.Set each factor equal to 0.

4. INTERPRET. Since x represents the length of the base, we discard the solution The base of a triangle cannot be negative. The base is then 6 meters and the heightis

Check: To check this problem, we recall that or

The required area

State: The base of the triangular sail is 6 meters and the height is 10 meters.

The next example makes use of the Pythagorean theorem and consecutive inte-gers. Before we review this theorem, recall that a right triangle is a triangle that con-tains a 90° or right angle. The hypotenuse of a right triangle is the side opposite theright angle and is the longest side of the triangle. The legs of a right triangle are theother sides of the triangle.

12

1621102 = 30

12

base # height = area,

2162 - 2 = 10 meters .

-5.

x = 6 x = -5 x - 6 = 0 or x + 5 = 0

1x - 621x + 52 = 0 x2 - x - 30 = 0

30 = x2 - x

30 =12

x12x - 22

12x - 22#x#12

=30

TTTT

height#base#12

=area oftriangle

2x - 2 = the height

x = the length of the base

FINDING THE BASE AND HEIGHT OF A SAIL

The square of 7 is or 49. Three times 7 is 3(7), or 21. Then thecorrect sum, so 7 checks.State: There are two numbers. They are and 7.-10

49 + 21 = 70,72


Height 2x 2

Base x

E X A M P L E 3

The height of a triangular sail is 2 meters less than twice the length of the base. If thesail has an area of 30 square meters, find the length of its base and the height.

Solut ion

CLASSROOM EXAMPLEThe length of a rectangle is 5 feet morethan its width. The area of the rectangleis 176 square feet. Find the length andthe width of the rectangle.answer: length: 16 ft; width: 11 ft



Study the following diagrams for a review of consecutive integers.

Consecutive integers:

If x is the first integer:

Consecutive even integers:

If x is the first even integer:

Consecutive odd integers:

If x is the first odd integer: x , x + 2, x + 4

x , x + 2, x + 4

x , x + 1, x + 2


42

12 1410

42

11 139

6 units

4 units8 units

x

x 2

x 4

Pythagorean TheoremIn a right triangle, the sum of the squares of the lengths of the two legs isequal to the square of the length of the hypotenuse.

1leg22 + 1leg22 = 1hypotenuse22 or a2 + b2 = c2

leg b

leg a

hypotenuse c

H e l p f u l H i n tIf you use this formula,don’t forget that c repre-sents the length of thehypotenuse.

E X A M P L E 4

FINDING THE DIMENSIONS OF A TRIANGLE

Find the lengths of the sides of a right triangle if the lengths can be expressed as threeconsecutive even integers.

Solut ion 1. UNDERSTAND. Read and reread the problem. Let’s suppose that the length ofone leg of the right triangle is 4 units. Then the other leg is the next even integer, or6 units, and the hypotenuse of the triangle is the next even integer, or 8 units. Re-member that the hypotenuse is the longest side. Let’s see if a triangle with sides ofthese lengths forms a right triangle. To do this, we check to see whether thePythagorean theorem holds true.

FalseOur proposed numbers do not check, but we now have a better understanding of theproblem.

We let and be three consecutive even integers. Since these in-tegers represent lengths of the sides of a right triangle, we have

x + 4 = hypotenuse 1longest side2 x + 2 = other leg x = one leg

x + 4x , x + 2,

52 = 64 16 + 36 64 42 + 62 82

21

6 75

H e l p f u l H i n tThis 2 means that evennumbers are 2 units be-tween each other.

H e l p f u l H i n tThis 2 means that oddnumbers are 2 units be-tween each other.



CLASSROOM EXAMPLESolve.a. Find two consecutive odd integers

whose product is 23 more than theirsum.

b. The length of one leg of a right trian-gle is 7 meters less than the length ofthe other leg. The length of the hy-potenuse is 13 meters. Find thelengths of the legs.

answer: a. 5 and 7 or and b. 5m, 12m

-3-5

2. TRANSLATE. By the Pythagorean theorem, we have that

3. SOLVE. Now we solve the equation.

Multiply.Combine like terms.Write in standard form.Factor.Set each factor equal to 0.

4. INTERPRET. We discard since length cannot be negative. If thenand

Check: Verify that or or

State: The sides of the right triangle have lengths 6 units, 8 units, and 10 units.

100 = 36 + 64.102 = 62 + 82,1hypotenuse22 = 1leg22 + 1leg22 ,

x + 4 = 10.x + 2 = 8x = 6,x = -2

x = 6 x = -2 x - 6 = 0 or x + 2 = 0

1x - 621x + 22 = 0 x2 - 4x - 12 = 0 x2 + 8x + 16 = 2x2 + 4x + 4 x2 + 8x + 16 = x2 + x2 + 4x + 4

1x + 422 = x2 + 1x + 222 1x + 422 = 1x22 + 1x + 222

1hypotenuse22 = 1leg22 + 1leg22

8 units

6 units 10 units

Suppose you are a land-scaper. You are landscaping apublic park and have just put in aflower bed measuring 8 feet by12 feet. You would also like tosurround the bed with a decora-tive floral border consisting of

low-growing, spreading plants. Each plant will cover approximately 1square foot when mature, and you have 224 plants to use. How wide ofa strip of ground should you prepare around the flower bed for the bor-der? Explain.

2x 8

x

x x

x

2x 1212 ft

8 ft

1 foot

1 foot

Grows to cover 1 sq ft


SIMPLIFYING RATIONAL EXPRESSIONS SECTION 7.1 411

Rational Expression

A rational expression is an expression that can be written in the form

where P and Q are polynomials and Q does not equal 0.

P

Q,

7.1 S I M P L I F Y I N G R AT I O N A L E X P R E S S I O N S

Rational Expressions

Rational expressions have different values depending on what value replaces thevariable. Next, we review the standard order of operations by finding values of ration-al expressions for given replacement values of the variable.

3y3

8 -4p

p3 + 2p + 1 5x2 - 3x + 2

3x + 7

E X A M P L E 1

Find the value of for the given replacement values.

a. b. x = -2x = 5

x + 42x - 3

.

Solut ion a. Replace each x in the expression with 5 and then simplify.

b. Replace each x in the expression with and then simplify.

For a negative fraction such as recall from Chapter 1 that

In general, for any fraction

2-7

=-27

= - 27

2- 7 ,

x + 42x - 3

=-2 + 4

21-22 - 3=

2-7

or - 27

-2

x + 42x - 3

=5 + 4

2152 - 3=

910 - 3

=97

CLASSROOM EXAMPLESFind the value of for the given replacement values.

b. 37

3

x - 35x + 1

O b j e c t i v e s

1 Find the value of a rational expression given a replacement number.

2 Identify values for which a rational expression is undefined.

3 Write rational expressions in lowest terms.

1 As we reviewed in Chapter 1, a rational number is a number that can be writ-ten as a quotient of integers. A rational expression is also a quotient; it is a quotient ofpolynomials.

-a

b=

a

-b= -

a

b, b Z 0




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412 CHAPTER 7 RATIONAL EXPRESSIONS

This is also true for rational expressions. For example,Notice the parentheses.

2 In the definition of rational expression on the previous page, notice that wewrote Q is not 0 for the denominator Q. This is because the denominator of a rationalexpression must not equal 0 since division by 0 is not defined. This means we must becareful when replacing the variable in a rational expression by a number. For example,suppose we replace x with 5 in the rational expression The expression becomes

But division by 0 is undefined. Therefore, in this rational expression we can allow x tobe any real number except 5. A rational expression is undefined for values that makethe denominator 0.

2 + x

x - 5=

2 + 55 - 5

=70

2 + xx - 5 .

-1x + 22x

=x + 2

-x= -

x + 2x

T

E X A M P L E 2Are there any values for x for which each rational expression is undefined?

a. b. c. d.2

x2 + 1

x3 - 6x2 - 10x

3x2 + 2

x2 - 3x + 2

x

x - 3

Solut ion To find values for which a rational expression is undefined, find values that make thedenominator 0.

a. The denominator of is 0 when or when Thus, when the expression is undefined.

b. Set the denominator equal to zero.

Factor.

Set each factor equal to zero.

Solve.

Thus, when or the denominator is 0. So the rational ex-

pression is undefined when or when

c. The denominator of is never zero, so there are no values of x for

which this expression is undefined.

d. No matter which real number x is replaced by, the denominator does notequal 0, so there are no real numbers for which this expression is undefined.

3 A fraction is said to be written in lowest terms or simplest form when the nu-merator and denominator have no common factors other than 1 (or ). For example,the fraction is in lowest terms since the numerator and denominator have no com-mon factors other than 1 (or ).

The process of writing a rational expression in lowest terms or simplest form iscalled simplifying a rational expression.The following fundamental principle of ration-al expressions is used to simplify a rational expression.

-1

710

-1

x2 + 1

x3 - 6x2 - 10x

3

x = 1.x = 2x2 + 2

x2 - 3x + 2

x2 - 3x + 2x = 1,x = 2

x = 2 or x = 1

x - 2 = 0 or x - 1 = 0

1x - 221x - 12 = 0

x2 - 3x + 2 = 0

xx - 3

x = 3,x = 3.x - 3 = 0xx - 3

CLASSROOM EXAMPLEAre there any values for x for which eachrational expression is undefined?

a. b.

c.

answers: a.b. or c. nox - 1x = -4

x = -2

x2 - 3x + 25

x - 3

x2 + 5x + 4

x

x + 2

$%&



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Factor the numerator and denominator. Then apply the fundamental principle.

21a2b

3a5b=

7 # 3 # a2 # b

3 # a3 # a2 # b=

7

a3

Simplifying a rational expression is similar to simplifying a fraction. To simplifythe fraction we factor the numerator and the denominator, look for common factorsin both, and then use the fundamental principle.

To simplify the rational expression we also factor the numerator

and denominator, look for common factors in both, and then use the fundamental prin-ciple of rational expressions.

This means that the rational expression has the same value as the rational

expression for all values of x except 2 and (Remember that when x is 2, thedenominator of both rational expressions is 0 and when x is the original rationalexpression has a denominator of 0.)

As we simplify rational expressions, we will assume that the simplified rationalexpression is equal to the original rational expression for all real numbers except thosefor which either denominator is 0. The following steps may be used to simplify rationalexpressions.

-3,-3.x - 3

x - 2

x2 - 9

x2 + x - 6

x2 - 9

x2 + x - 6=1x - 321x + 321x - 221x + 32 =

x - 3x - 2

x2 - 9

x2 + x - 6,

1520

=3 # 5

2 # 2 # 5=

32 # 2

=34

1520 ,

Simplifying a Rational ExpressionStep 1. Completely factor the numerator and denominator.

Step 2. Apply the fundamental principle of rational expressions to divide outcommon factors.

TEACHING TIPIt may be useful for students to usethe TABLE feature on a graphing

calculator to evaluate

and for various values of x toverify that the expressions are thesame for all values except and2. Then ask students which expres-sion is easier to evaluate. Finally,point out that one reason for sim-plifying rational expressions beforeevaluating them is to simplify yourwork.

-3

x - 3x - 2

x2 - 9

x2 + x - 6

E X A M P L E 3

Write in simplest form.21a2b

3a5b

Solut ion

Fundamental Principle of Rational ExpressionsIf P, Q, and R are polynomials, and Q and R are not 0,

PR

QR=

P

Q

CLASSROOM EXAMPLE

Write in simplest terms.

answer:3x

5y6

15x2y

25xy7.


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Factor the numerator and denominator and apply the fundamental principle.

x2 + 8x + 7

x2 - 4x - 5=1x + 721x + 121x - 521x + 12 =

x + 7x - 5

Factor the numerator and denominator, if possible, and then apply the fundamentalprinciple.

5x - 5

x3 - x2 =51x - 12x21x - 12 =

5

x2

Factor the numerator and denominator. Then apply the fundamental principle.

6a - 3315

=312a - 112

3 # 5=

2a - 115

E X A M P L E 4

Simplify:6a - 33

15

Solut ion

CLASSROOM EXAMPLE

Simplify:

answer:2a - 1

3

14a - 721

E X A M P L E 5

Simplify:5x - 5

x3 - x2

Solut ion

CLASSROOM EXAMPLE

Simplify:

answer:x3

5

x4 + x3

5x + 5

E X A M P L E 6

Simplify:x2 + 8x + 7

x2 - 4x - 5

Solut ion

CLASSROOM EXAMPLE

Simplify:

answer:x + 9x - 1

x2 + 11x + 18

x2 + x - 2

H e l p f u l H i n tWhen simplifying a rational expression, the fundamental principle applies tocommon factors, not common terms.

Common factors. These can be Common terms. Fundamental principle divided out. does not apply. This is in simplest form.

x + 2x

x # 1x + 22x # x

=x + 2

x

CONCEPT CHECKRecall that the fundamental principle applies to common factors only. Which of the following are nottrue? Explain why.

a. b. c. d.2x + 3

2= x + 3

3772

=32

2x + 102

= x + 53 - 13 + 5

= - 15

TEACHING TIPGive students a concrete examplethat illustrates why can besimplified to 2, and can-not be simplified to 2. For example,evaluate each expression for and then Have students no-tice that the first expression evalu-ates to 2 for both values of x butthe second expression does not.

x = 5.x = 1

(x + 2)/x2x/x

E X A M P L E 7

Simplify each rational expression.

a. b.x - y

y - x

x + y

y + xConcept Check Answers:a, c, d




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Suppose you are a forensic lab techni-cian. You have been asked to try to iden-tify a piece of metal found at a crimescene. You know that one way to analyzethe piece of metal is to find its density(mass per unit volume), using the formula

After weighing the metal, you find its mass as 36.2 grams.

You have also determined the volume of the metal to be 4.5 milliliters. Use thisinformation to decide which type of metal this piece is most likely made of, andexplain your reasoning.What other characteristics might help the identification?

density =mass

volume.

Densities

Metal Density (g/ml)

Aluminum 2.7Iron 7.8Lead 11.5Silver 10.5

Solut ion

CLASSROOM EXAMPLESimplify:

a. b.

answers: a. 1 b. -1

a - b

b - a

a + b

b + a

a. The expression can be simplified by using the commutative property of addi-tion to rewrite the denominator as

b. The expression can be simplified by recognizing that and are op-posites. In other words, Proceed as follows:

x - y

y - x=

1 # 1x - y21-121x - y2 =

1-1

= -1

y - x = -11x - y2 .x - yy - xx - y

y - x

x + y

y + x=

x + y

x + y= 1

x + y .y + x

x + yy + x

E X A M P L E 8

Simplify:4 - x2

3x2 - 5x - 2

Solut ion Factor.

Write as

Simplify. =1-1212 + x2

3x + 1 or

-2 - x

3x + 1

11x 22 .2 x =1-121x - 2212 + x21x - 2213x + 12

4 - x2

3x2 - 5x - 2=12 - x212 + x21x - 2213x + 12

CLASSROOM EXAMPLE

Simplify:

answer:

or -x + 5

x-

x - 5x

x2 - 10x + 25

5x - x2

STUDY SKILLS REMINDERHow Are You Doing?

If you haven’t done so yet, take a few moments and think abouthow you are doing in this course. Are you working toward yourgoal of successfully completing this course? Is your performanceon homework, quizzes, and tests satisfactory? If not, you mightwant to see your instructor to see if he/she has any suggestions on



how you can improve your performance. Let me once again re-mind you that, in addition to your instructor, there are manyplaces to get help with your mathematics course. A few sugges-tions are below.

This text has an accompanying video lesson for every sectionin this text.

The back of this book contains answers to odd-numbered ex-ercises and selected solutions.

MathPro is available with this text. It is a tutorial software pro-gram with lessons corresponding to each section in the text.

There is a student solutions manual available that containsworked-out solutions to odd-numbered exercises as well as so-lutions to every exercise in the Integrated Reviews, ChapterReviews, Chapter Tests, and Cumulative Reviews.

Don’t forget to check with your instructor for other local re-sources available to you, such as a tutoring center.

MENTAL MATHFind any real numbers for which each rational expression is undefined. See Example 2.

1. 2. 3. 4.

Decide which rational expression can be simplified. (Do not actually simplify.)

5. no 6. yes 7. yes 8. nox + 2x + 8

5 - x

x - 53 + x

x + 3x

x + 7

x = 5, x = 6x + 2

1x - 521x - 62x = 0, x = 1x2 + 4x - 2

x1x - 12x = 3x2 - 5x

x - 3x = 0

x + 5x


MULTIPLYING AND DIV IDING RATIONAL EXPRESSIONS SECTION 7.2 419

Multiplying Rational ExpressionsLet P, Q, R, and S be polynomials. Then

as long as and S Z 0.Q Z 0

P

Q# R

S=

PR

QS

O b j e c t i v e s

1 Multiply rational expressions.

2 Divide rational expressions.

1 Just as simplifying rational expressions is similar to simplifying number frac-tions, multiplying and dividing rational expressions is similar to multiplying and divid-ing number fractions.To find the product of fractions and rational expressions, multiplythe numerators and multiply the denominators.

35

# 14

=3 # 15 # 4

=320 and x

y + 1# x + 3y - 1

=x1x + 32

1y + 121y - 12

TEACHING TIPContinually remind studentsthroughout the next few sectionsthat operations on rational expres-sions are the same as on fractions.

E X A M P L E 1Multiply.

a. b.-7x2

5y# 3y5

14x2

25x

2# 1

y3

7.2 M U LT I P LY I N G A N D D I V I D I N G R AT I O N A L E X P R E S S I O N S



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Solut ion To multiply rational expressions, multiply the numerators and then multiply the de-nominators of both expressions. Then simplify if possible.

a.

The expression is in simplest form.

b. Multiply.

The expression is not in simplest form, so we factor the numerator and

the denominator and apply the fundamental principle.

When multiplying rational expressions, it is usually best to factor each numeratorand denominator. This will help us when we apply the fundamental principle to writethe product in lowest terms.

= - 3y4

10

=-1 # 7 # 3 # x2 # y # y4

5 # 2 # 7 # x2 # y

-7x2 # 3y5

5y # 14x2

-7x2

5y# 3y5

14x2 =-7x2 # 3y5

5y # 14x2

25x

2y3

25x

2# 1

y3 =25x # 1

2 # y3 =25x

2y3


a. b.

answers: a. b. - 2a2

9b

16y

3x2

-5a3

3b3# 2b2

15a

16y

3# 1

x2

E X A M P L E 2

Multiply:x2 + x

3x# 65x + 5

Solut ion Factor numerators and denominators.

Multiply.

Simplify.

The following steps may be used to multiply rational expressions.

=25

=x1x + 12 # 2 # 3

3x # 51x + 12

x2 + x

3x# 65x + 5

=x1x + 12

3x# 2 # 351x + 12

CLASSROOM EXAMPLE

Multiply:

answer:12

x - 1

6x + 67

# 14

x2 - 1

Multiplying Rational ExpressionsStep 1. Completely factor numerators and denominators.

Step 2. Multiply numerators and multiply denominators.

Step 3. Simplify or write the product in lowest terms by applying the funda-mental principle to all common factors.

CONCEPT CHECK

Which of the following is a true statement?

a. b. c. d.x

7 #

x + 54

=2x + 5

283x

# 12

=3

2x

2x

# 5x

=10x

13

# 12

=15





H e l p f u l H i n tDon’t forget how to find reciprocals. The reciprocal of

a

b is

ba

, a Z 0, b Z 0.

Dividing Rational ExpressionsLet P, Q, R, and S be polynomials. Then,

as long as and R Z 0.Q Z 0, S Z 0,

P

Q,

R

S=

P

Q# S

R=

PS

QR

For example, to divide by multiply by

32

,78

=32

# 87

=3 # 4 # 2

2 # 7=

127

87 .3

278 ,3

2

Factor.

Multiply.

Apply the fundamentalprinciple.

Next, recall that and are opposites so that

Write as

Apply the fundamental principle.

2 We can divide by a rational expression in the same way we divide by a fraction.To divide by a fraction, multiply by its reciprocal.

=-31x + 125x12x - 32 or -

31x + 125x12x - 32

111 x2 .x 1 =31x + 121-1211 - x2

5x11 - x212x - 32

x - 1 = -111 - x2 .1 - xx - 1

=31x + 121x - 12

5x11 - x212x - 32

=31x + 1212x + 321x - 12

5x11 - x212x - 3212x + 32

3x + 3

5x - 5x2# 2x2 + x - 3

4x2 - 9=

31x + 125x11 - x2 # 12x + 321x - 12

12x - 3212x + 32

E X A M P L E 3

Multiply:3x + 3

5x - 5x2# 2x2 + x - 3

4x2 - 9

CLASSROOM EXAMPLE

Multiply:

answer:41x + 22

7x13x - 12

4x + 8

7x2 - 14x# 3x2 - 5x - 2

9x2 - 1

Solut ion



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Solut ionMultiply by the reciprocal of

Simplify. =3y9

160

=3x3y9

160x3

4x3

y2.

3x3y7

40,

4x3

y2 =3x3y7

40# y2

4x3

CLASSROOM EXAMPLE

Divide:

answer:7xy

3

7x2

6,

x

2y

E X A M P L E 5

Divide by 2x + 4

5

1x - 121x + 2210

Solut ion

The following may be used to divide by a rational expression.

=x - 1

4

=1x - 121x + 22 # 5

5 # 2 # 2 # 1x + 22

1x - 121x + 22

10,

2x + 45

=1x - 121x + 22

10# 52x + 4

CLASSROOM EXAMPLE

Divide:

answer:2x + 3

9

12x + 321x - 426

,3x - 12

2

TEACHING TIPRemind students not to look forcommon factors that may be divid-ed out until the operation is multiplication.

Dividing by a Rational ExpressionMultiply by its reciprocal.

E X A M P L E 6

Divide:6x + 2

x2 - 1,

3x2 + x

x - 1

Solut ion Multiply by the reciprocal.

Factor and multiply.

Simplify. =2

x1x + 12

=213x + 121x - 12

1x + 121x - 12 # x13x + 12

6x + 2

x2 - 1,

3x2 + x

x - 1=

6x + 2

x2 - 1# x - 1

3x2 + xCLASSROOM EXAMPLE

Divide:

answer:2

x21x - 22

10x + 4

x2 - 4,

5x3 + 2x2

x + 2

E X A M P L E 7

Divide:2x2 - 11x + 5

5x - 25,

4x - 210

Solut ionMultiply by the reciprocal.


Simplify. =11 or 1

=12x - 121x - 52 # 2 # 5

51x - 52 # 212x - 12

2x2 - 11x + 5

5x - 25,

4x - 210

=2x2 - 11x + 5

5x - 25# 104x - 2

CLASSROOM EXAMPLE

Divide:

answer: 1

3x2 - 10x + 87x - 14

,9x - 12

21

E X A M P L E 4

Divide:3x3y7

40,

4x3

y2

Multiply by the reciprocal of


Simplify.

2x 4

5.



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E X A M P L E 8

CONVERTING FROM SQUARE YARDS TO SQUARE FEET

The largest casino in the world is the Foxwoods Resort Casino in Ledyard, CT. Thegaming area for this casino is 21,444 square yards. Find the size of the gaming area insquare feet. (Source: The Guinness Book of Records)

CLASSROOM EXAMPLEThe largest building in the world is theBoeing Company’s Everett, Washington,factory complex where the 747, 767,and 777 jets are built. The volume of thisbuilding is 472,370,319 cubic feet. Con-vert this to cubic yards.answer: 17,495,197 cu yd

Now that we know how to multiply fractions and rational expressions, we can usethis knowledge to help us convert between units of measure. To do so, we will use unitfractions.A unit fraction is a fraction that equals 1. For example, since wehave the unit fractions

12 in.1 ft

= 1 and 1 ft12 in.

= 1

12 in. = 1 ft ,

Solut ion There are 9 square feet in 1 square yard.

Unit fraction

= 192,996 square feet

21,444 square yards = 21,444 sq. yd # 9 sq. ft

1 sq. yd

2

1 square yard 9 square feet

1 ydor 3 ft

1 yd or3 ft

H e l p f u l H i n tWhen converting a unit of measurement, if possible, write the unit fraction so thatthe numerator contains the units we are converting to and the denominator con-tains the original units. For example, suppose we want to convert 48 inches to feet.

Unit fraction

=4812

ft = 4 ft

; Units converting to; Original units

48 in. =48 in.

1 # 1 ft

12 in.

4

MENTAL MATHFind the following products. See Example 1.

1. 2. 3. 4. 5. 6.37

y

7# 3y

95

9x

# x

54x5

11z3

x5

11# 4

z3

5y2

7x2

57

# y2

x2

3x

4y

3x

4# 1y

2x

3y

2y

# x

3


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7.3 A D D I N G A N D S U B T R AC T I N G R AT I O N A L E X P R E S S I O N S W I T H C O M M O ND E N O M I N ATO R S A N D L E A S T C O M M O N D E N O M I N ATO R

O b j e c t i v e s

1 Add and subtract rational expressions with the same denominator.

2 Find the least common denominator of a list of rational expressions.

3 Write a rational expression as an equivalent expression whose denominator is given.

1 Like multiplication and division, addition and subtraction of rational expres-sions is similar to addition and subtraction of rational numbers. In this section, we addand subtract rational expressions with a common (or the same) denominator.

Add: Add:

Add the numerators and place the sum over the common denominator.

Simplify. Simplify. =12

x + 2 =

85

9

x + 2+

3x + 2

=9 + 3x + 2

65

+25

=6 + 2

5

9x + 2

+3

x + 265

+25

Adding and Subtracting Rational Expressions with Common Denominators

If and are rational expressions, then

To add or subtract rational expressions, add or subtract numerators and placethe sum or difference over the common denominator.

P

R+

Q

R=

P + Q

R and P

R-

Q

R=

P - Q

R

Q

R

P

R

E X A M P L E 1

Add:5m

2n+

m

2n

Solut ion

CLASSROOM EXAMPLE

Add:

answer:3x

y

8x

3y+

x

3y

Add the numerators.

Simplify the numerator by combining like terms.

Simplify by applying the fundamental principle. =3mn

=6m

2n

5m

2n+

m

2n=

5m + m

2n

TEACHING TIPRemind students throughout thissection that to add or subtract whenthe denominators are the same, addor subtract numerators and keepthe common denominator.



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ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH COMMON DENOMINATORS SECTION 7.3 427

2 To add and subtract fractions with unlike denominators, first find a least commondenominator (LCD), and then write all fractions as equivalent fractions with the LCD.

For example, suppose we add and The LCD of denominators 3 and 5 is 15,since 15 is the least common multiple (LCM) of 3 and 5.That is, 15 is the smallest num-ber that both 3 and 5 divide into evenly. Rewrite each fraction so that its denominatoris 15. (Notice how we apply the fundamental principle of rational expressions.)

We are multiplying by 1.

To add or subtract rational expressions with unlike denominators, we also first findan LCD and then write all rational expressions as equivalent expressions with the LCD.The least common denominator (LCD) of a list of rational expressions is a polynomial ofleast degree whose factors include all the factors of the denominators in the list.

qq

83

+25

=81523152 +

21325132 =

4015

+6

15=

40 + 615

=4615

25 .8

3

H e l p f u l H i n tNotice how the numerator has been subtracted in Example 3.

This sign applies to the So parentheses are inserted entire numerator of here to indicate this.

3x2 + 2x

x - 1-

10x - 5x - 1

=3x2 + 2x - 110x - 52

x - 1

TTT10x - 5.

-10x - 5

E X A M P L E 3

Subtract:3x2 + 2x

x - 1-

10x - 5x - 1

.

Solut ion

CLASSROOM EXAMPLE

Subtract:

answer: 2x - 3

2x2 + 5x

x + 2-

4x + 6x + 2

Subtract the numeratorsNotice the parentheses.


Combine like terms.

Factor.

Simplify. = 3x - 5

=1x - 1213x - 52

x - 1

=3x2 - 8x + 5

x - 1

=3x2 + 2x - 10x + 5

x - 1

3x2 + 2x

x - 1-

10x - 5x - 1

=13x2 + 2x2 - 110x - 52

x - 1

TEACHING TIPTo reinforce this concept, have stu-dents work a few extra examples:

2x + 3x + 6

-x - 3x + 6

a

7-

a - 37

3x

-2 + y

x

E X A M P L E 2

Subtract:2y

2y - 7-

72y - 7

Solut ion Subtract the numerators.

Simplify. =11 or 1

2y

2y - 7-

72y - 7

=2y - 7

2y - 7CLASSROOM EXAMPLE

Subtract:

answer: 1

2x

2x - 5-

52x - 5

H e l p f u l H i n tParentheses are insertedso that the entire numera-tor, is subtracted.10x - 5,



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Finding the Least Common Denominator (LCD)Step 1. Factor each denominator completely.

Step 2. The least common denominator (LCD) is the product of all unique fac-tors found in step 1, each raised to a power equal to the greatest num-ber of times that the factor appears in any one factored denominator.

E X A M P L E 4Find the LCD for each pair.

a. b.7

5x,

6

15x2

18

, 322

Solut ion a. Start by finding the prime factorization of each denominator.

Next, write the product of all the unique factors, each raised to a power equal to thegreatest number of times that the factor appears in any denominator.

The greatest number of times that the factor 2 appears is 3.The greatest number of times that the factor 11 appears is 1.

b. Factor each denominator.

The greatest number of times that the factor 5 appears is 1.The greatest number of times that the factor 3 appears is 1.The greatest number of times that the factor x appears is 2.

LCD = 31 # 51 # x2 = 15x2

5x = 5 # x and 15x2 = 3 # 5 # x2

LCD = 23 # 111 = 8 # 11 = 88

8 = 2 # 2 # 2 = 23 and 22 = 2 # 11CLASSROOM EXAMPLEFind the LCD for each pair.

a. b.

answers: a. 45 b. 24x5

5

6x3,

11

8x5

29

, 7

15

E X A M P L E 5Find the LCD of

a. b.3x

and 6

x + 47x

x + 2 and

5x2

x - 2

Solut ion a. The denominators and are completely factored already. The factorappears once and the factor appears once.

b. The denominators x and cannot be factored further. The factor x appearedonce and the factor appears once.

LCD = x1x + 42x + 4

x + 4

LCD = 1x + 221x - 22x - 2x + 2

x - 2x + 2

CLASSROOM EXAMPLEFind the LCD of

a. and

b. and

answers: a.b. y1y - 32

1a + 521a - 52

5y

y - 3-2y

7a

a - 53a

a + 5

E X A M P L E 6

Find the LCD of and2

1m + 522 .6m2

3m + 15

Solut ion Factor each denominator.

3m + 15 = 31m + 521m + 522 is already factored.


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ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH COMMON DENOMINATORS SECTION 7.3 429

E X A M P L E 9

Write as an equivalent fraction with the given denominator.

4b

9a=

27a2b

4b

9aCLASSROOM EXAMPLEWrite as an equivalent rational expres-sion with the given denominator.

answer:8x3y

20x2y2

2x

5y=

20x2y2

Solut ion Ask yourself: “What do we multiply 9a by to get ” The answer is 3ab, sinceMultiply the numerator and denominator by 3ab.

4b

9a=

4b13ab29a13ab2 =

12ab2

27a2b

9a13ab2 = 27a2b .27a2b?

CLASSROOM EXAMPLEFind the LCD of

answer: 31x - 4227x2

1x - 422 and 5x

3x - 12.


E X A M P L E 7

Find the LCD of andt + 5

t2 + 3t + 2.

t - 10

t2 - t - 6

Solut ion Start by factoring each denominator.

LCD = 1t - 321t + 221t + 12 t2 + 3t + 2 = 1t + 121t + 22 t2 - t - 6 = 1t - 321t + 22CLASSROOM EXAMPLE

Find the LCD of and

answer: 1y + 321y - 221y - 12y + 4

y2 - 3y + 2.

y + 5

y2 + 2y - 3

The greatest number of times that the factor 3 appears is 1.The greatest number of times that the factor appears in any onedenominator is 2.

CONCEPT CHECK

Choose the correct LCD of and

a. b. c. d. 5x1x + 1221x + 1231x + 122x + 1

5x + 1

.x

1x + 122

LCD = 31m + 522m + 5

E X A M P L E 8

Find the LCD of and10

2 - x.

2x - 2

Solut ion The denominators and are opposites. That is, Useor as the LCD.

3 Next we practice writing a rational expression as an equivalent rational ex-pression with a given denominator. To do this, we apply the fundamental principle,which says that or equivalently that This can be seen by recalling thatmultiplying an expression by 1 produces an equivalent expression. In other words,

P

Q=

P

Q# 1 =

P

Q# R

R=

PR

QR.

PQ = PR

QR .PRQR = P

Q ,

LCD = x - 2 or LCD = 2 - x

2 - xx - 22 - x = -11x - 22 .2 - xx - 2

CLASSROOM EXAMPLE

Find the LCD of and

answer: 1x + 42 or 14 - x29

4 - x.

6x - 4

TEACHING TIPBefore discussing Example 8, haveyour students find the LCD of

2x - 2 and 10

2 - x 1Example 822

17 - 2 and 102 - 17

215 and 10

- 15


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E X A M P L E 1 0Write the rational expression as an equivalent rational expression with the givendenominator.

5

x2 - 4= 1x - 221x + 221x - 42

Solut ion First, factor the denominator as

If we multiply the original denominator by the result is the newdenominator Thus, multiply the numerator and the denomi-nator by

Factoreddenominator

=5x - 20

1x - 221x + 221x - 42

53

5

x2 - 4=

51x - 221x + 22 =

51x - 421x - 221x + 221x - 42

x - 4.1x + 221x - 221x - 42 . x - 4,1x - 221x + 22

1x - 221x + 22 .x2 - 4

CLASSROOM EXAMPLEWrite as an equivalent rational expres-sion with the given denominator.

answer:3x - 9

1x + 521x - 521x - 32

3

x2 - 25= 1x + 521x - 521x - 32


1. 1 2. 3. 4.

5. 6. 7. 8.8x

712x

7-

4x

7

7 - 10y

575

-10y

5-

712

- 412

-3

1219

89

-79

5y

8

3y

8+

2y

87x

93x

9+

4x

9611

511

+1

1123

+13

MENTAL MATH


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O b j e c t i v e

1 Add and subtract rational expressions with unlike denominators.

1 In the previous section, we practiced all the skills we need to add and subtractrational expressions with unlike or different denominators. The steps are as follows:

7.4A D D I N G A N D S U B T R AC T I N G R AT I O N A L E X P R E S S I O N S W I T H U N L I K ED E N O M I N ATO R S

Adding or Subtracting Rational Expressions with Unlike DenominatorsStep 1. Find the LCD of the rational expressions.

Step 2. Rewrite each rational expression as an equivalent expression whosedenominator is the LCD found in Step 1.

Step 3. Add or subtract numerators and write the sum or difference over thecommon denominator.

Step 4. Simplify or write the rational expression in simplest form.



ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH UNLIKE DENOMINATORS SECTION 7.4 433

Subtract:6x

x2 - 4-

3x + 2

Solut ion Since the We write equivalentexpressions with the LCD as denominators.

Subtract numerators. Write the differenceover the common denominator.

Apply the distributive property in the numerator.

Combine like terms in the numerator.

Next we factor the numerator to see if this rational expression can be simplified.

Factor.

Apply the fundamental principle to simplify. =3

x - 2

=31x + 22

1x + 221x - 22

=3x + 6

1x + 221x - 22

=6x - 3x + 61x + 221x - 22

=6x - 31x - 221x + 221x - 22

6x

x2 - 4-

3x + 2

=6x

1x - 221x + 22 -31x - 22

1x + 221x - 22

LCD = 1x - 221x + 22 .x2 - 4 = 1x + 221x - 22 ,CLASSROOM EXAMPLE

Subtract:

answer:5

x - 3

10x

x2 - 9-

5x + 3

Perform each indicated operation.

a. b.3

10x2 +7

25x

a

4-

2a

8

Solut ion

CLASSROOM EXAMPLEPerform each operation.

a. b.

answers:

a. 0 b.25x + 44

40x2

58x

+11

10x2

y

5-

3y

15

a. First, we must find the LCD. Since and the Next wewrite each fraction as an equivalent fraction with the denominator 8, then we subtract.

b. Since and the We writeeach fraction as an equivalent fraction with a denominator of

Add numerators. Write the sum over the common denominator.

=15 + 14x

50x2

=15

50x2 +14x

50x2

3

10x2 +7

25x=

315210x2152 +

712x225x12x2

50x2 .LCD = 2 # 52 # x2 = 50x2 .25x = 5 # 5 # x ,10x2 = 2 # 5 # x # x

ƒ ————— – – — –——Ta

4-

2a

8=

a1224122 -

2a

8=

2a

8-

2a

8=

2a - 2a

8=

08

= 0

ƒ—————— ——q

LCD = 23 = 8.8 = 23,4 = 22

TEACHING TIPIn Example 1b, for instance, helpstudents understand that the ex-pression has the same

value as for all numbersexcept 0. To see this have studentsevaluate and the

simplified form for

They should get in both cases.43200

x = 2.15 + 14x50x2

310x2 + 7

25x

15 + 14x50x2 ,

310x2 + 7

25x

E X A M P L E 1

E X A M P L E 2


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Add:23t

+5

t + 1

Solut ion The LCD is We write each rational expression as an equivalent rational ex-pression with a denominator of

Add numerators. Write the sum over thecommon denominator.

Apply the distributive property in thenumerator.

Combine like terms in the numerator. =17t + 2

3t1t + 12

=2t + 2 + 15t

3t1t + 12

=21t + 12 + 513t2

3t1t + 12

23t

+5

t + 1=

21t + 123t1t + 12 +

513t21t + 1213t2

3t1t + 12 .3t1t + 12 .CLASSROOM EXAMPLE

Add:

answer:19x + 5

7x1x + 12

57x

+2

x + 1

E X A M P L E 3

TEACHING TIPContinue to remind students thatthey may not add or subtract nu-merators until the denominatorsare the same.

E X A M P L E 4

Subtract:7

x - 3-

93 - x

Solut ion To find a common denominator, we notice that and are opposites. That is,We write the denominator as and simplify.

Apply

Subtract numerators. Write the difference over the common denominator.

=16

x - 3

=7 - 1-92

x - 3

a

b

a

b. =

7x - 3

--9

x - 3

7

x - 3-

93 - x

=7

x - 3-

9-1x - 32

-1x - 323 - x3 - x = -1x - 32 . 3 - xx - 3

CLASSROOM EXAMPLE

Subtract:

answer:25

x - 6

10x - 6

-15

6 - x

TEACHING TIPIn Example 4, to verify for studentsthat and are oppo-sites, replace x with several valuesand have students notice the results. To verify that and

are equal, have studentsreplace x with several values andnotice the results.

-1x - 323 - x

3 - xx - 3

E X A M P L E 5

Add: 1 +m

m + 1

Solut ion Recall that 1 is the same as The LCD of and and is

Write 1 as

Multiply both the numerator and the =11m + 1211m + 12 +

m

m + 1

11 . 1 +

m

m + 1=

11

+m

m + 1

m + 1.m

m + 111

11

.

CLASSROOM EXAMPLE

Add:

answer:3x + 10

x + 5

2 +x

x + 5

denominator of by m 1 .11


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ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH UNLIKE DENOMINATORS SECTION 7.4 435


Combine like terms in the numerator. =2m + 1m + 1

=m + 1 + m

m + 1

Subtract:3

2x2 + x-

2x

6x + 3

Solut ion First, we factor the denominators.

The LCD is We write equivalent expressions with denominators of

Subtract numerators. Write the differenceover the common denominator.

=9 - 2x2

3x12x + 12

=3132

x12x + 12132 -2x1x2

312x + 121x2

3x12x + 12 . 3x12x + 12 .

3

2x2 + x-

2x

6x + 3=

3x12x + 12 -

2x

312x + 12CLASSROOM EXAMPLE

Subtract:

answer:16 - 3x2

4x13x + 22

4

3x2 + 2x-

3x

12x + 8

Add:2x

x2 + 2x + 1+

x

x2 - 1

Solut ion First we factor the denominators.

Now we write the rational expressions as equivalent expressions with denominators ofthe LCD.


Apply the distributive property in the numerator.

The numerator was factored as a last step to see if the rational expression could be sim-plified further. Since there are no factors common to the numerator and the denomi-nator, we can’t simplify further.

=3x2 - x

1x + 1221x - 12 or x13x - 121x + 1221x - 12

=2x2 - 2x + x2 + x

1x + 1221x - 12

=2x1x - 12 + x1x + 121x + 1221x - 12

=2x1x - 12

1x + 121x + 121x - 12 +x1x + 12

1x + 121x - 121x + 12

1x + 121x + 121x - 12 ,

2x

x2 + 2x + 1+

x

x2 - 1=

2x

1x + 121x + 12 +x

1x + 121x - 12CLASSROOM EXAMPLE

Add:

answer:x17x - 102

1x + 2221x - 22

6x

x2 + 4x + 4+

x

x2 - 4

E X A M P L E 6

E X A M P L E 7


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MENTAL MATHMatch each exercise with the first step needed to perform the operation. Do not actually perform the operation.

1. D 2. C 3. A 4. B

A. Multiply the first rational expression by the reciprocal of the second rational expression.B. Find the LCD. Write each expression as an equivalent expression with the LCD as denominator.C. Multiply numerators, then multiply denominators.D. Subtract numerators. Place the difference over a common denominator.

9x - 2

-x

x + 2x + 1

x,

x - 1x

2a

# 31a + 62

34

-y

4



7.5 S O LV I N G E Q UAT I O N S C O N TA I N I N G R AT I O N A L E X P R E S S I O N S

O b j e c t i v e s

1 Solve equations containing rational expressions.

2 Solve equations containing rational expressions for a specified variable.

1 In Chapter 2, we solved equations containing fractions. In this section, we contin-ue the work we began in Chapter 2 by solving equations containing rational expressions.

Examples of Equations Containing Rational Expressions

To solve equations such as these, use the multiplication property of equality to clearthe equation of fractions by multiplying both sides of the equation by the LCD.

x

5+

x + 29

= 8 and x + 19x - 5

=2

3x

TEACHING TIPYou may want to begin this sectionwith a review of solving linearequations, such as 3x + 16 = 1.

Solve:x

2+

83

=16

Solut ion The LCD of denominators 2, 3, and 6 is 6, so we multiply both sides of the equation by 6.


Multiply and simplify.Subtract 16 from both sides.Divide both sides by 3. x = -5

3x = -15 3 # x + 16 = 1

6ax

2b + 6a8

3b = 6a1

6b

6ax

2+

83b = 6a1

6b

E X A M P L E 1

CLASSROOM EXAMPLE

Solve:

answer: -3

x

4+

45

=120

H e l p f u l H i n tMake sure that eachterm is multiplied by theLCD, 6.

Check To check, we replace x with in the original equation.

Replace x with

True

This number checks, so the solution is -5.

16

=16

5 . -52

+83

16

x

2+

83

=16

-5



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SOLVING EQUATIONS CONTAINING RATIONAL EXPRESSIONS SECTION 7.5 439

Solut ion In this equation, 0 cannot be a solution because if x is 0, the rational expression is un-defined. The LCD is x, so we multiply both sides of the equation by x.


Simplify.

Now we write the quadratic equation in standard form and solve for x.

3x - 6 = x2 + 8x

x132 - xa 6xb = x # x + x # 8

xa3 -6xb = x1x + 82

6x

H e l p f u l H i n tMultiply each term by x.

Solve:t - 4

2-

t - 39

=518

Solut ion The LCD of denominators 2, 9, and 18 is 18, so we multiply both sides of the equationby 18.


Simplify.Use the distributive property.Combine like terms.

Solve for t. t = 5 7t = 35

7t - 30 = 5 9t - 36 - 2t + 6 = 5

91t - 42 - 21t - 32 = 5

18a t - 42b - 18a t - 3

9b = 18a 5

18b

18a t - 42

-t - 3

9b = 18a 5

18b

CLASSROOM EXAMPLE

Solve:

answer: -6

x + 23

-x - 1

5=

115

H e l p f u l H i n tMultiply each term by 18.

Check

Replace t with 5.

Simplify.

True

The solution is 5.

Recall from Section 7.1 that a rational expression is defined for all real numbersexcept those that make the denominator of the expression 0. This means that if anequation contains rational expressions with variables in the denominator, we must becertain that the proposed solution does not make the denominator 0. If replacing thevariable with the proposed solution makes the denominator 0, the rational expressionis undefined and this proposed solution must be rejected.

518

=518

12

-29

518

5 - 4

2-

5 - 39

518

t - 4

2-

t - 39

=518

Solve: 3 -6x

= x + 8

E X A M P L E 2

E X A M P L E 3

CLASSROOM EXAMPLE

Solve:

answer: -6, 1

2 +6x

= x + 7


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Check To check these solutions, we replace x in the original equation by and then by

If If

True True

Both and are solutions.

The following steps may be used to solve an equation containing rational ex-pressions.

-2-3

6 = 6 5 = 5

3 - 1-32 6 3 - 1-22 5

3 -6

-2 -2 + 8 3 -

6-3

-3 + 8

3 -6x

= x + 8 3 -6x

= x + 8

x = -2 :x = -3 :

-2.-3,

Factor.

or Set each factor equal to 0 and solve.

Notice that neither nor makes the denominator in the original equation equal to 0.-2-3

x = -2 x = -3

x + 2 = 0 x + 3 = 0

0 = 1x + 321x + 22 0 = x2 + 5x + 6

Solving an Equation Containing Rational ExpressionsStep 1. Multiply both sides of the equation by the LCD of all rational expres-

sions in the equation.

Step 2. Remove any grouping symbols and solve the resulting equation.

Step 3. Check the solution in the original equation.

E X A M P L E 4

Solve:4x

x2 - 25+

2x - 5

=1

x + 5

Solut ion The denominator factors as The LCD is thenso we multiply both sides of the equation by this LCD.

= 1x + 521x - 52 # 1x + 5

1x + 521x - 52 # 4x

x2 - 25+ 1x + 521x - 52 # 2

x - 5

= 1x + 521x - 52a 1x + 5

b

1x + 521x - 52a 4x

1x + 521x - 52 +2

x - 5b

1x + 521x - 52 , 1x + 521x - 52 .x2 - 25

CLASSROOM EXAMPLE

Solve:

answer: -1

2x + 3

+3

x - 3=

-2

x2 - 9

Multiply by the LCD. Noticethat and 5cannot be solutions.


5


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Solve:2x

x - 4=

8x - 4

+ 1

Solut ion

CLASSROOM EXAMPLE

Solve:

answer: no solution

5x

x - 1=

5x - 1

+ 3

CONCEPT CHECKWhen can we clear fractions by multiplying through by the LCD?

a. When adding or subtracting rational expressionsb. When solving an equation containing rational expressionsc. Both of these d. Neither of these

Multiply both sides by the LCD,

Multiply by the LCD. Notice that 4 cannot be a solution.


Simplify.

Notice that 4 makes the denominator 0 in the original equation.Therefore, 4 is not a so-lution and this equation has no solution.

x = 4

2x = 4 + x

2x = 8 + 1x - 42 1x - 42 # 2x

x - 4= 1x - 42 # 8

x - 4+ 1x - 42 # 1

1x - 42a 2x

x - 4b = 1x - 42a 8

x - 4+ 1b

x - 4.


Solve: x +14

x - 2=

7x

x - 2+ 1

Solut ion Notice the denominators in this equation. We can see that 2 can’t be a solution. TheLCD is so we multiply both sides of the equation by

1x - 221x2 + 1x - 22a 14x - 2

b = 1x - 22a 7x

x - 2b + 1x - 22112

1x - 22ax +14

x - 2b = 1x - 22a 7x

x - 2+ 1b

x - 2.x - 2,CLASSROOM EXAMPLE

Solve:

answer: 4

x -6

x + 3=

2x

x + 3+ 2

H e l p f u l H i n tAs we can see from Example 5, it is important to check the proposed solution(s) inthe original equation.

Check Check by replacing x with in the original equation. The solution is -3.-3

Simplify.

Use the distributive property.Combine like terms.

Divide both sides by 5. x = -3

5x = -15

6x + 10 = x - 5 4x + 2x + 10 = x - 5

4x + 21x + 52 = x - 5

E X A M P L E 5

E X A M P L E 6

TEACHING TIPOne of the most important conceptsthat you can help students with isthe difference between an equationand an expression.This is a greattime to reinforce that difference.


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Simplify.Combine like terms.Write the quadratic

equation in standard form.

Factor.or Set each factor equal to 0.

Solve.

As we have already noted, 2 can’t be a solution of the original equation. So we needonly replace x with 8 in the original equation. We find that 8 is a solution; the only so-lution is 8.

2 The last example in this section is an equation containing several variables.Weare directed to solve for one of them.The steps used in the preceeding examples can beapplied to solve equations for a specified variable as well.

x = 2 x = 8x - 2 = 0 x - 8 = 0

1x - 821x - 22 = 0

x2 - 10x + 16 = 0x2 - 2x + 14 = 8x - 2x2 - 2x + 14 = 7x + x - 2

Solve for x.1a

+1b

=1x

E X A M P L E 7

Solut ion (This type of equation often models a work problem, as we shall see in Section 7.6.)The LCD is abx, so we multiply both sides by abx.

Simplify.Factor out x from each term on the left side.


Simplify.

This equation is now solved for x.

x =ab

b + a

b a . x1b + a2

b + a=

ab

b + a

x1b + a2 = ab

bx + ax = ab

abxa 1ab + abxa 1

bb = abx # 1

x

abxa 1a

+1bb = abxa 1

xbCLASSROOM EXAMPLE

Solve for a.

answer: a =bx

b - x

1a

+1b

=1x

TEACHING TIPAsk students why Example 6 sim-ply states “Solve” whereas Exam-ple 7 states “Solve for x.” SinceExample 7 has more than 1 un-known, the unknown we need tofind must be specified.

Á

A grapher may be used to check solutions of equa-tions containing rational expressions. For example,to check the solution of Example 1,graph and Use TRACE and ZOOM, or use INTERSECT, tofind the point of intersection.The point of intersec-tion has an x-value of so the solution of theequation is

Use a grapher to check the examples of this section.1. Example 2 2. Example 3 3. Example 44. Example 5 5. Example 6

-5.-5,

y2 = 1/6.y1 = x/2 + 8/3

x2 + 8

3 = 16 ,


10

10

10

10


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MENTAL MATHSolve each equation for the variable.

1. 10 2. 32 3. 36 4. 56y

7= 8

z

6= 6

x

8= 4

x

5= 2

STUDY SKILLS REMINDERIs Your Notebook Still Organized?

Is your notebook still organized? If it’s not, it’s not too late tostart organizing it. Start writing your notes and completing yourhomework assignment in a notebook with pockets (spiral or ringbinder). Take class notes in this notebook, and then follow thenotes with your completed homework assignment. When you re-ceive graded papers or handouts, place them in the notebookpocket so that you will not lose them.

Remember to mark (possibly with an exclamation point) anynote(s) that seems extra important to you. Also remember tomark (possibly with a question mark) any notes or homeworkthat you are having trouble with. Don’t forget to see yourinstructor or a math tutor to help you with the concepts orexercises that you are having trouble understanding.

Also—don’t forget to write neatly.



7.6 P R O P O R T I O N A N D P R O B L E M S O LV I N G W I T H R AT I O N A L E Q UAT I O N S

O b j e c t i v e s

1 Use proportions to solve problems.

2 Solve problems about numbers.

3 Solve problems about work.

4 Solve problems about distance.

1 A ratio is the quotient of two numbers or two quantities. For example, theratio of 2 to 5 can be written as the quotient of 2 and 5.

If two ratios are equal, we say the ratios are in proportion to each other. A pro-portion is a mathematical statement that two ratios are equal.

For example, the equation is a proportion, as is because both sides ofthe equations are ratios. When we want to emphasize the equation as a proportion, we

In a proportion, cross products are equal.To understand cross products, let’s startwith the proportion

and multiply both sides by the LCD, bd.

Multiply both sides by the LCD, bd.

Simplify.

Cross product Cross productNotice why ad and bc are called cross products.

∆˚

ad"

= bc"

bda a

bb = bda c

db

a

b=

c

d

read the proportion 12

48

as “one is to two as four is to eight”

x5 = 8

10 ,12 = 4

8

25 ,

ab

cd

bc

ad

TEACHING TIPAsk students to mentally calculatewhether the statements below aretrue or false by using cross products.

3

12=

28

57

=7

10

46

=1015

23

=6

10

10. expression; 11. expression; 12. expression;7p + 55x + 7z



PROPORTION AND PROBLEM SOLVING WITH RATIONAL EQUATIONS SECTION 7.6 447

CALCULATING THE COST OF RECORDABLE COMPACT DISCS

Three boxes of CD-Rs (recordable compact discs) cost $37.47. How much should 5boxes cost?

Solut ion 1. UNDERSTAND. Read and reread the problem.We know that the cost of 5 boxesis more than the cost of 3 boxes, or $37.47, and less than the cost of 6 boxes, which isdouble the cost of 3 boxes, or Let’s suppose that 5 boxes cost$60.00. To check, we see if 3 boxes is to 5 boxes as the price of 3 boxes is to the price

21$37.472 = $74.94.

For example, if

12

=48

, then 1 # 8 = 2 # 48 = 8

or

Solve for x:45x

=57

CLASSROOM EXAMPLE

Solve for x:

answer: 168

38

=63x

Solut ion This is an equation with rational expressions, and also a proportion. Below are twoways to solve.

Since this is a rational equation, Since this is also a proportion,we can use the methods of the we may set cross products equal.previous section.

Multiply both sidesby LCD 7x.

Set cross products equal.

Divide out common factors. Multiply.

Multiply. Divide both sides by 5.

Divide both sides by 5. Simplify.

Simplify. 63 = x

63 = x 3155

=5x

5

3155

=5x

5 315 = 5x

315 = 5x 7 # 45 = x # 5

45 # 7 = x # 5 7x # 45x

= 7x # 57

45x

=57

Check

Cross Products

If then ad = bc .a

b=

c

d,

45x

57

E X A M P L E 1

Both methods give us a solution of 63. To check, substitute 63 for x in the original pro-portion. The solution is 63.

In this section, if the rational equation is a proportion, we will use cross productsto solve.

Proportions can be used to model and solve many real-life problems. When usingproportions in this way, it is important to judge whether the solution is reasonable.Doing so helps us to decide if the proportion has been formed correctly. We use thesame problem-solving steps that were introduced in Section 2.5.

E X A M P L E 2



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CLASSROOM EXAMPLETo estimate the number of people inJackson, population 50,000, who havea flu shot, 250 people were polled. Ofthose polled, 26 had a flu shot. Howmany people in the city might we expectto have a flu shot?answer: 5200 people

of 5 boxes. In other words, we see if

or

Set cross products equal.or

Not a true statement.

Thus, $60 is not correct, but we now have a better understanding of the problem.

180.00 = 187.35

3160.002 = 5137.472 35

=37.4760.00

3 boxes5 boxes

=price of 3 boxes

price of 5 boxes

2. TRANSLATE.

3. SOLVE.

Set cross products equal.

Divide both sides by 3.4. INTERPRET.

Check: Verify that 3 boxes is to 5 boxes as $37.47 is to $62.45. Also, notice that oursolution is a reasonable one as discussed in Step 1.State: Five boxes of CD-Rs cost $62.45.

x = 62.45 3x = 187.35 3x = 5137.472 35

=37.47

x

35

=37.47

x

3 boxes5 boxes

=price of 3 boxes

price of 5 boxes

Let x = price of 5 boxes of CD-Rs.


The proportion could also have been used to solve

the problem above. Notice that the cross products are the same.

5 boxes3 boxes

=price of 5 boxes

price of 3 boxes



FINDING THE LENGTH OF A SIDE OF A TRIANGLE

If the following two triangles are similar, find the missing length x.

Solut ion 1. UNDERSTAND. Read the problem and study the figure.2. TRANSLATE. Since the triangles are similar, their corresponding sides are in

proportion and we have

3. SOLVE. To solve, we multiply both sides by the LCD, 3x, or cross multiply.

Divide both sides by 2.4. INTERPRET.

Check: To check, replace x with 15 in the original proportion and see that a truestatement results.State: The missing length is 15 yards.

x = 15 2x = 30

23

=10x

CLASSROOM EXAMPLEFor the similar triangles, find x.

answer: 15

2 yards 10 yards

9 yards

3 yards x yards

Similar triangles have the same shape but not necessarily the same size. In similar tri-angles, the measures of corresponding angles are equal, and corresponding sides are inproportion.

If triangle ABC and triangle XYZ shown are similar, then we know that themeasure of angle measure of angle X, the measure of angle measureof angle Y, and the measure of angle measure of angle Z. We also know thatcorresponding sides are in proportion: a

x = by = c

z .C = the

B = theA = the

In this section, we will position similar triangles so that they have the same orientation.To show that corresponding sides are in proportion for the triangles above, we

write the ratios of the corresponding sides.

ax

=186

= 3 by

=124

= 3 cz

=155

= 3

(15 in.) c

b (12 in.)

a (18 in.)

A

B

C(5 in.) z

y (4 in.)

x (6 in.)

X

Y

Z

E X A M P L E 3

10

3x

2



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1. UNDERSTAND. Read and reread the problem. Suppose that the unknown num-ber is 2, then we see if the quotient of 2 and 6, or minus is equal to the quotientof 2 and 2, or

Don’t forget that the purpose of a proposed solution is to better understand theproblem.Let unknown number.

2. TRANSLATE.

In words:

Translate:

3. SOLVE. Here, we solve the equation We begin by multiplying both

sides of the equation by the LCD, 6.


Simplify.Subtract x from both sides.


Simplify. -5 = x

- 102

=2x

2

-10 = 2x x - 10 = 3x

6ax

6b - 6a5

3b = 6ax

2b

6 ax

6-

53b = 6ax

2b

x

6-

53

=x

2.

x

2=

53

-x

6

TTTTT

the quotientof x and 2

is53

minusthe quotientof x and 6

x = the

26

-53

=13

-53

= - 43

, not 22

22 .

53

26 ,

2 Let’s continue to solve problems. The remaining problems are all modeled byrational equations.

FINDING AN UNKNOWN NUMBERThe quotient of a number and 6, minus is the quotient of the number and 2. Find the

number.

53

,

CLASSROOM EXAMPLEThe quotient of a number and 2, minus

is the quotient of the number and 6.answer: 1

13 ,

Solut ion

Suppose you coach your company’s softball team. Two new employees are in-terested in playing on the company team, and you have only one open posi-tion. Employee A reports that last season he had 32 hits in 122 times at bat.Employee B reports that last season she had 19 hits in 56 times at bat. Whichwould you try to recruit first? Why? What other factors would you want toconsider?

E X A M P L E 4


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2. TRANSLATE.

In words:

T T T T T

Translate:

3. SOLVE. Here, we solve the equation We begin by multiplying both

sides of the equation by the LCD, 21x.

13

+17

=1x

.

1x

=17

+13

part of jobthey completed

together in1 hour

isequal

to

part of jobFrank

completedin 1 hour

addedto

part ofjob Sam

completedin 1 hour


FINDING WORK RATES

Sam Waterton and Frank Schaffer work in a plant that manufactures automobiles. Samcan complete a quality control tour of the plant in 3 hours while his assistant, Frank,needs 7 hours to complete the same job. The regional manager is coming to inspect theplant facilities, so both Sam and Frank are directed to complete a quality control tourtogether. How long will this take?

1. UNDERSTAND. Read and reread the problem.The key idea here is the relation-ship between the time (hours) it takes to complete the job and the part of the jobcompleted in 1 unit of time (hour). For example, if the time it takes Sam to completethe job is 3 hours, the part of the job he can complete in 1 hour is Similarly, Frankcan complete of the job in 1 hour.

Let time in hours it takes Sam and Frank to complete the job together.

Then part of the job they complete in 1 hour.1x

= the

x = the

17

13 .

Hours to Part of JobComplete Total Job Completed in 1 Hour

Sam 3

Frank 7

Together x1x

17

13

4. INTERPRET.Check: To check, we verify that “the quotient of and 6 minus is the quotientof and 2,” or State: The unknown number is

3 The next example is often called a work problem. Work problems usually in-volve people or machines doing a certain task.

-5.-

56 - 5

3 = - 52 .-5

53-5

TEACHING TIPTake some time to review the con-cept of a “work” problem. Ask stu-dents a few questions like thefollowing: If it takes you 2 hours tocomplete a job, what part of thejob has been completed in 1 hour?After a few of these, see if you cansuccessfully insert the variable x.

Solut ion

E X A M P L E 5



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CLASSROOM EXAMPLEAndrew and Timothy Larson volunteer ata local recycling plant. Andrew can sorta batch of recyclables in 2 hours alonewhile his brother Timothy needs 3 hoursto complete the same job. If they worktogether, how long will it take them tosort one batch?answer: 1

15 hr

Simplify.

4. INTERPRET.Check: Our proposed solution is This proposed solution is reasonablesince is more than half of Sam’s time and less than half of Frank’s time.Check this solution in the originally stated problem.State: Sam and Frank can complete the quality control tour in

4 Next we look at a problem solved by the distance formula.

2 110 hours.

2 110 hours

2 110 hours.

x =2110 or 2

110

hours

10x = 21 7x + 3x = 21

21xa13b + 21xa1

7b = 21xa 1

xb

FINDING SPEEDS OF VEHICLES

A car travels 180 miles in the same time that a truck travels 120 miles. If the car’s speedis 20 miles per hour faster than the truck’s, find the car’s speed and the truck’s speed.

Solut ion 1. UNDERSTAND. Read and reread the problem. Suppose that the truck’s speed is45 miles per hour.Then the car’s speed is 20 miles per hour more, or 65 miles per hour.

We are given that the car travels 180 miles in the same time that the truck travels120 miles. To find the time it takes the car to travel 180 miles, remember that since

we know that

Car’s Time Truck’s Time

Since the times are not the same, our proposed solution is not correct. But we havea better understanding of the problem.Let speed of the truck.Since the car’s speed is 20 miles per hour faster than the truck’s, then

Use the formula or Prepare a chart to organize theinformation in the problem.

distance = rate # time.d = r # t

x + 20 = the speed of the car

x = the

t =dr

=12045

= 2 3045

= 2 23

hourst =dr

=18065

= 2 5065

= 2 1013

hours

dr = t.d = rt,

E X A M P L E 6

CLASSROOM EXAMPLEA car travels 600 miles in the same timethat a motorcycle travels 450 miles. If thecar’s speed is 15 miles per hour morethan the motorcycle’s, find the speed ofthe car and the speed of the motorcycle.answer: car: 60 mph; motorcycle: 45 mph

Distance Rate Time

Truck 120 x

Car 180180

x + 20; distance; rate

x + 20

120x

; distance; rate

# H e l p f u l H i n tIf

then

or time =distance

rate.

t =d

r

d = r # t ,



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2. TRANSLATE. Since the car and the truck traveled the same amount of time, wehave thatIn words:

Translate:

3. SOLVE. We begin by multiplying both sides of the equation by the LCD,or cross multiplying.

Use the distributive property.Subtract 120x from both sides.Divide both sides by 60.

4. INTERPRET. The speed of the truck is 40 miles per hour. The speed of the carmust then be or 60 miles per hour.Check: Find the time it takes the car to travel 180 miles and the time it takes thetruck to travel 120 miles.

Car’s Time Truck’s Time

Since both travel the same amount of time, the proposed solution is correct.State: The car’s speed is 60 miles per hour and the truck’s speed is 40 miles perhour.

t =dr

=12040

= 3 hourst =dr

=18060

= 3 hours

x + 20

x = 40 60x = 2400

180x = 120x + 2400

180x = 1201x + 202

180x + 20

=120x

x1x + 202 ,

120x

=180

x + 20

TTtruck’s time=car’s time

Suppose you must selecta child care center foryour two-year-old daugh-ter. You have compiledinformation on two pos-sible choices in the tableshown at the right.Which

child care center would you choose? Why?

Center A Center B

Weekly cost $130 $110Number of 2-year-olds 7 13Number of adults for 2-year-olds 2 3Distance from home 6 miles 11 miles


VARIAT ION AND PROBLEM SOLVING SECTION 7.7 457

O b j e c t i v e s

1 Solve problems involving direct variation.

2 Solve problems involving inverse variation.

3 Other types of direct and inverse variation.

4 Variation and problem solving.

In Chapter 3, we studied linear equations in two variables. Recall that such anequation can be written in the form where A and B are not both 0.Ax + By = C,

7.7 VA R I AT I O N A N D P R O B L E M S O LV I N G



Hours Worked 0 1 2 3 4

Money Earned(before deductions) 0 7.25 14.50 21.75 29.00


Also recall that the graph of a linear equation in two variables is a line. In this section,we begin by looking at a particular family of linear equations—those that can be writ-ten in the form

where k is a constant. This family of equations is called direct variation.

1 Let’s suppose that you are earning $7.25 per hour at a part-time job. Theamount of money you earn depends on the number of hours you work.This is illustrat-ed by the following table:

and so on

y = kx,

In general, to calculate your earnings (before deductions) multiply the constant $7.25by the number of hours you work. If we let y represent the amount of money earnedand x represent the number of hours worked, we get the direct variation equation

Notice that in this direct variation equation, as the number of hours increases, the payincreases as well.

earnings = $7.25 # hours worked∆q˚

y = 7.25 # x

Direct Variationy varies directly as x, or y is directly proportional to x, if there is a nonzeroconstant k such that

The number k is called the constant of variation or the constant of proportionality.

y = kx

In our direct variation example: the constant of variation is 7.25.Let’s use the previous table to graph We begin our graph at the ordered-

pair solution (0, 0). Why? We assume that the least amount of hours worked is 0. If 0hours are worked, then the pay is $0.

y = 7.25x .y = 7.25x ,

0

Pay

Hours Worked0 1 2 3 4 5 6 7

0

10

20

30

40

(1, 7.25)

(2, 14.50)

(3, 21.75)

(4, 29.00)

As illustrated in this graph, a direct variation equation is linear. Also noticethat is a function since its graph passes the vertical line test.y = 7.25x

y = kx



Suppose that y varies directly as x. If y is 17 when x is 34, find the constant of variationand the direct variation equation. Then find y when x is 12.

CLASSROOM EXAMPLEIf y varies directly as x and y is 15 whenx is 45, find the constant of variation andthe direct variation equation. Then find ywhen x is 3.

Answer: y =13

x ; y = 1

Solut ion Let’s use the same method as in Example 1 to find x. Since we are told that y varies di-rectly as x, we know the relationship is of the form

Let and and solve for k.

Solve for k.

Thus, the constant of variation is and the equation is

To find y when use and replace x with 12.

Replace x with 12.

Thus, when x is 12, y is 6.

Let’s review a few facts about linear equations of the form y = kx.

y = 6

y =12

# 12

y =12

x

y = 12 xx = 12,

y = 12 x .1

2

12

= k

1734

=k # 34

34

17 = k # 34

x = 34y = 17

y = kx.

Write a direct variation equation of the form that satisfies the ordered pairs inthe table below.

y = kx

CLASSROOM EXAMPLEWrite a direct variation equation that sat-isfies:

x 4 1.5 6

y 1 12

answer: y = 2x

38

12

Solut ion We are given that there is a direct variation relationship between x and y.This means that

By studying the given values, you may be able to mentally calculate k. If not, to find k,we simply substitute one given ordered pair into this equation and solve for k.We’ll usethe given pair (2, 6).

Solve for k.

Since we have the equation

To check, see that each given y is 3 times the given x.

Let’s try another type of direct variation example.

y = 3x .k = 3,

3 = k

62

=k # 2

2

6 = k # 2

y = kx

y = kx

x 2 9 1.5

y 6 27 4.5 -3

-1

E X A M P L E 1

E X A M P L E 2



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Direct Variation: y = kx• There is a direct variation relationship between x and y.

• The graph is a line.• The line will always go through the origin (0, 0). Why?

Let Then or • The slope of the graph of is k, the constant of variation. Why?

Remember that the slope of an equation of the form is m,the coefficient of x.

• The equation describes a function. Each x has a unique y and itsgraph passes the vertical line test.

y = kx

y = mx + by = kx

y = 0.y = k # 0x = 0.

The line is the graph of a direct variation equation. Find the constant of variation andthe direct variation equation.

CLASSROOM EXAMPLEFind the constant of variation and the di-rect variation equation for the line below.

answer: k = 2; y = 2x

6 7

67

1123

12345

2 3 4 5123

(4, 5)

(0, 0)

x

y

E X A M P L E 3

Solut ion Recall that k, the constant of variation is the same as the slope of the line. Thus, to findk, we use the slope formula and find slope.

Using the given points (0, 0), and (4, 5), we have

Thus, and the variation equation is

2 In this section, we will introduce another type of variation, called inversevariation.

Let’s suppose you need to drive a distance of 40 miles. You know that the fasteryou drive the distance, the sooner you arrive at your destination. Recall that there is amathematical relationship between distance, rate, and time. It is In our exam-ple, distance is a constant 40 miles, so we have or

For example, if you drive 10 mph, the time to drive the 40 miles is

If you drive 20 mph, the time is

t =40r

=4020

= 2 hours

t =40r

=4010

= 4 hours

t = 40r .40 = r # td = r # t .

y = 54 x .k = 5

4

slope =5 - 04 - 0

=54

.

(0, 0)

(1, 2) x

y


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Write an inverse variation equation of the form that satisfies the ordered pairs inthe table below.

y = kx

x 2 4

y 6 3 24

12

Again, notice that as speed increases, time decreases. Below are some ordered-pair solutions of and its graph.t = 40

r

Notice that the graph of this variation is not a line, but it passes the vertical linetest so does describe a function. This is an example of inverse variation.t = 40

r

In our inverse variation example, or the constant of variation is 40.We can immediately see differences and similarities in direct variation and in-

verse variation.

y = 40x ,t = 40

r

Remember that is a rational equation and not a linear equation. Also notice thatbecause x is in the denominator, that x can be any value except 0.

We can still derive an inverse variation equation from a table of values.

y = kx

Tim

e (h

ours

)

Rate (miles per hour)

0

10

9

8

7

6

5

4

3

2

1

0 10 20 30 40 50 60 70 80 90 100

rate (mph) r 5 10 20 40 60 80

time (hr) t 8 4 2 1 12

23

Inverse Variationy varies inversely as x, or y is inversely proportional to x, if there is a nonzeroconstant k such that

The number k is called the constant of variation or the constant of proportionality.

y =kx

Direct variation linear equationboth

Inverse variation rational equationfunctions

y =kx

y = kx

E X A M P L E 4


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x 2 4

y 6 3 2412# 24 = 124 # 3 = 122 # 6 = 12

12


CLASSROOM EXAMPLEWrite an inverse variation equation thatsatisfies:

x 4 10 40

y 5 2

answer: y =20x

-1012

-2

Solut ion Since there is an inverse variation relationship between x and y, we know that

To find k, choose one given ordered pair and substitute the values into the equation.We’ll use (2, 6).


Solve.

Since we have the equation y = 12x .k = 12,

12 = k

2 # 6 = 2 # k

2

6 =k

2

y =kx

y = kx .

Suppose that y varies inversely as x. If when find the constant of vari-ation and the inverse variation equation. Then find y when x is 30.

x = 75,y = 0.02

CLASSROOM EXAMPLEIf y varies inversely as x and y is 4 whenx is 0.8, find the constant of variationand the direct variation equation. Thenfind y when x is 20.

answer: y =3.2x

; y = 0.16

Solut ion

H e l p f u l H i n tMultiply both sides of the inverse variation relationship equation by x(as long as x is not 0), and we have This means that if y varies inverse-ly as x, their product is always the constant of variation k. For an example ofthis, check the table below.

xy = k .y = k

x

E X A M P L E 5

Since y varies inversely as x, the constant of variation may be found by simply findingthe product of the given x and y.

To check, we will use the inverse variation equation

Let and and solve for k.


Solve for k.

Thus, the constant of variation is 1.5 and the equation is

To find y when use and replace x with 30.

Replace x with 30.

Thus, when x is 30, y is 0.05.

y = 0.05

y =1.530

y =1.5x

y = 1.5xx = 30

y = 1.5x .

1.5 = k

7510.022 = 75 # k

75

0.02 =k

75

x = 75y = 0.02

y =kx

.

k = xy = 7510.022 = 1.5



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CLASSROOM EXAMPLEThe area of a circle varies directly as thesquare of the radius. A circle with radius7 inches has an area of squareinches. Find the area of a circle whereradius is 4 feet.answer: 16p sq. ft

49p

Solut ion Since the surface area A varies directly as the square of side s, we have

To find k, let and

Let and

Divide by 9.

The formula for surface area of a cube is then

To find the surface area when substitute.

The surface area of a cube whose side measures 4.2 units is 105.84 square units.

4 There are many real-life applications of direct and inverse variation.

A = 105.84 A = 6 # 14.222 A = 6s2

s = 4.2,

A = 6s2 where s is the length of a side.

6 = k32 9 . 54 = 9k

s 3 .A 54 54 = k # 32 A = k # s2

s = 3.A = 54A = ks2 .

The weight of a body w varies inversely with the square of its distance from the centerof Earth d. If a person weighs 160 pounds on the surface of Earth, what is the person’sweight 200 miles above the surface? (Assume that the radius of Earth is 4000 miles.)

s

200 miles

160 poundsEarth

? pounds

The surface area of a cube A varies directly as the square of a length of its side s. If A is54 when s is 3, find A when s = 4.2.

Direct and InverseVariation as nth Powers of x

y varies directly as a power of x if there is a nonzero constant k and a naturalnumber n such that

y varies inversely as a power of x if there is a nonzero constant k and a naturalnumber n such that

y =k

xn

y = kxn

3 It is possible for y to vary directly or inversely as powers of x. In this section,our powers of x will be natural numbers only.

E X A M P L E 6

E X A M P L E 7


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Solut ion 1. UNDERSTAND. Make sure you read and reread the problem.

2. TRANSLATE. Since we are told that weight w varies inversely with the square ofits distance from the center of Earth, d, we have

3. SOLVE. To solve the problem, we first find k. To do so, use the fact that the personweighs 160 pounds on Earth’s surface, which is a distance of 4000 miles from Earth’scenter.

Thus, we have

Since we want to know the person’s weight 200 miles above the Earth’s surface, welet and find .

A person 200 miles abovethe Earth’s surface is 4200miles from the Earth’s center.

Simplify.

4. INTERPRET. Check: Your answer is reasonable since the further a person isfrom Earth, the less the person weighs.

State: Thus, 200 miles above the surface of the Earth, a 160-pound person weighsapproximately 145 pounds.

w L 145

w =2,560,000,000

1420022

w =2,560,000,000

d2

wd = 4200

w =2,560,000,000

d2

2,560,000,000 = k

160 =k

1400022 w =

k

d2

w =k

d2 .

CLASSROOM EXAMPLEThe distance d that an object falls is di-rectly proportional to the square of thetime of the fall, t. If an object falls 144feet in 3 seconds, find how for the objectfalls in 5 seconds.answer: 400 feet

MENTAL MATHState whether each equation represents direct or indirect variation.

1. 2. 3. 4. 5. 6. 7. 8. y =20

x3y = 12x2y = 18xy =11x

y = 6.5x4y =7

x2y =5x

y = 5x

1. direct 2. inverse 3. inverse 4. direct 5. inverse 6. direct 7. direct 8. inverse



7.8 S I M P L I F Y I N G C O M P L E X F R AC T I O N S

O b j e c t i v e s

1 Simplify complex fractions using method 1.

2 Simplify complex fractions using method 2.

1 A rational expression whose numerator or denominator or both numeratorand denominator contain fractions is called a complex rational expression or acomplex fraction. Some examples are

4

2 -12

, 32

47

- x

, 1x + 2

x + 2 -1x

f ; Numerator of complex fraction; Main fraction barf ; Denominator of complex fraction



SIMPLIFYING COMPLEX FRACTIONS SECTION 7.8 467

CLASSROOM EXAMPLE

Simplify:

answer: 221

34

-23

12

+38

Simplify the complex fraction 5823

.

Solut ion Since the numerator and denominator of the complex fraction are already single frac-tions, we proceed to step 2: perform the indicated division by multiplying the numera-tor by the reciprocal of the denominator

The reciprocal of is q32 .2

3

5823

=58

# 32

=1516

23 .5

8CLASSROOM EXAMPLE

Simplify:

answer:2735

37

59

Simplify:

23

+15

23

-29

.

Solut ion Simplify above and below the main fraction bar separately. First, add and to obtaina single fraction in the numerator; then subtract from to obtain asingle fraction in the denominator.

The LCD of the numerator’s fractions is 15.

The LCD of the denominator’s fractions is 9.

Simplify. =

1015

+315

69

-29

23

+15

23

-29

=

21523152 +

11325132

21323132 -

29

23

29

15

23

Method 1: Simplifying a Complex FractionStep 1. Add or subtract fractions in the numerator or denominator so that the

numerator is a single fraction and the denominator is a single fraction.

Step 2. Perform the indicated division by multiplying the numerator of thecomplex fraction by the reciprocal of the denominator of the complexfraction.

Step 3. Write the rational expression in simplest form.

Our goal in this section is to write complex fractions in simplest form. A complexfraction is in simplest form when it is in the form where P and Q are polynomialsthat have no common factors.

In this section, two methods of simplifying complex fractions are represented.The first method presented uses the fact that the main fraction bar indicates division.

PQ,

E X A M P L E 1

E X A M P L E 2


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Add the numerator’s fractions.

Subtract the denominator’s fractions.

Next, perform the indicated division by multiplying the numerator of the complex frac-tion by the reciprocal of the denominator of the complex fraction.

The reciprocal of is

=13 # 3 # 33 # 5 # 4

=3920

9

4.

4

9

131549

=1315

# 94

=

131549

Simplify:

1z

-12

13

-z

6

Solut ion Subtract to get a single fraction in the numerator and a single fraction in the denomi-nator of the complex fraction.

The LCD of the numerator’s fractions is 2z.

The LCD of the denominator’s fractions is 6.

Multiply by the reciprocal of

Factor.

Write in simplest form.

2 Next we study a second method for simplifying complex fractions. In thismethod, we multiply the numerator and the denominator of the complex fraction bythe LCD of all fractions in the complex fraction.

=3z

=2 # 3 # 12 - z22 # z # 12 - z2

2 - z

6. =

2 - z

2z# 62 - z

=

2 - z

2z

2 - z

6

1z

-12

13

-z

6

=

22z

-z

2z

26

-z

6

CLASSROOM EXAMPLE

Simplify:

answer:612x - 52

x13x - 102

25

-1x

x

10-

13

E X A M P L E 3

Method 2: Simplifying a Complex FractionStep 1. Find the LCD of all the fractions in the complex fraction.

Step 2. Multiply both the numerator and the denominator of the complex frac-tion by the LCD from Step 1.

Step 3. Perform the indicated operations and write the result in simplest form.

We use method 2 to rework Example 2.


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SIMPLIFYING COMPLEX FRACTIONS SECTION 7.8 469

TEACHING TIPFor Method 2, remind studentsthat they must multiply the numer-ator and the denominator of thecomplex fraction by the same num-ber or expression.

Simplify:

x + 1y

xy

+ 2

Solut ion The LCD of and is y, so we multiply the numerator and the denominator

of the complex fraction by y.

Apply the distributive property in the denominator.

Simplify. =x + 1

x + 2y

=yax + 1

yb

yaxyb + y # 2

x + 1y

xy

+ 2=

yax + 1yb

yaxy

+ 2b

21

x + 1y

, xy

,

CLASSROOM EXAMPLE

Simplify:

answer:y + x

2x + 1

1 +x

y

2x + 1y

H e l p f u l H i n tThe same complex fraction was simplified using two different methods in Examples 2 and 4. Notice that each time the simplified result is the same.

Simplify:

23

+15

23

-29

Solut ion

CLASSROOM EXAMPLE

Use method 2 to simplify:

answer: 221

34 - 2

3

12 + 3

8

The LCD of and is 45, so we multiply the numerator and the denominator ofthe complex fraction by 45. Then we perform the indicated operations, and write insimplest form.


Simplify. =30 + 930 - 10

=3920

=45a2

3b + 45a1

5b

45a23b - 45a2

9b

23

+15

23

-29

=45a2

3+

15b

45a23

-29b

29

23 , 15 , 23

E X A M P L E 4

E X A M P L E 5



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E X A M P L E 6

Simplify:

xy

+3

2xx

2+ y

Solut ion The LCD of and is 2xy, so we multiply both the numerator and the de-

nominator of the complex fraction by 2xy.


or 2x2 + 3y

xy1x + 2y2

=2x2 + 3y

x2y + 2xy2

=2xyax

yb + 2xya 3

2xb

2xyax

2b + 2xy1y2

xy

+3

2xx

2+ y

=2xyax

y+

32xb

2xyax

2+ yb

y

1xy

, 3

2x,

x

2,

CLASSROOM EXAMPLE

Simplify:

answer:5x + 6y2

2xy1y - 3x2

56y

+y

x

y

3- x

MENTAL MATHComplete the steps by stating the simplified complex fraction.

1. 2. 3. 4.2a

b

a

10b

20

=20a a

10b

20a b

20b

=??

3x

5

3x

5

x2

=x2a 3

xb

x2a 5

x2 b=

??

10z

10xz

x

=xa10

xb

xa z

xb

=??

y

5x

y

25x

2

=2ay

2b

2a5x

2b

=??


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484 CHAPTER 8 ROOTS AND RADICALS

8.1 I N T R O D U C T I O N TO R A D I CA L S

O b j e c t i v e s

Find square roots of perfect squares.

Approximate irrational square roots.

Simplify square roots containing variables.

Find higher roots.

1 In this section, we define finding the root of a number by its reverse operation,raising a number to a power. We begin with squares and square roots.

The reverse operation of squaring a number is finding the square root of a num-ber. For example,

In general, a number b is a square root of a number a if

Notice that both 5 and are square roots of 25. The symbol is used to de-note the positive or principal square root of a number. For example,

The symbol is used to denote the negative square root. For example,

The symbol is called a radical or radical sign. The expression within or undera radical sign is called the radicand. An expression containing a radical is called aradical expression. 2a

radical sign radicand

2

-225 = -5

-2

225 = 5 since 52 = 25 and 5 is positive.

2 -5

b2 a .

A square root of 14

is 12

, because a12b2

=14

.

A square root of 25 is also -5, because 1-522 = 25.

A square root of 25 is 5 , because 52 = 25.

The square of 12

is a12b2

=14

.

The square of -5 is 1-522 = 25.

The square of 5 is 52 = 25.

4

3

2

1

Square RootThe positive or principal square root of a positive number a is written as The negative square root of a is written as

Also, the square root of 0, written as is 0.20,

2a = b only if b2 = a and b 7 0

-2a .2a .



INTRODUCTION TO RADICALS SECTION 8.1 485

E X A M P L E 1Find each square root.

a. b. c. d. e.

a. because and 6 is positive.

b. because and 8 is positive.

c. The negative sign in front of the radical indicates the negative squareroot of 16.

d. because and is positive.

e. because

Is the square root of a negative number a real number? For example, is areal number? To answer this question, we ask ourselves, is there a real number whose square is Since there is no real number whose square is we say that isnot a real number. In general,

2-4-4,-4?

2-4

02 = 0.20 = 0

310

a 310b2

=9

100A 9100

=310

-216 = -4.

82 = 64264 = 8,

62 = 36236 = 6,

20A 9100

-216264236

CLASSROOM EXAMPLE

Find each square root.

a. b.

c. d.

e.answer:

a. 10 b. 3

c. d.

e. 1

59

-6

21A25

81-236

292100

A square root of a negative number is not a real number.

Solut ion

We will discuss numbers such as in Chapter 9.

2 Recall that numbers such as 1, 4, 9, 25, and are called perfect squares, since

and Square roots of perfect squareradicands simplify to rational numbers. What happens when we try to simplify a rootsuch as Since 3 is not a perfect square, is not a rational number. It cannot bewritten as a quotient of integers. It is called an irrational number and we can find adecimal approximation of it. To find decimal approximations, use a calculator or anappendix. (For calculator help, see the box at the end of this section.)

2323?

A25 B2 = 425 .12 = 1, 22 = 4, 32 = 9, 52 = 25,

425

2-4

E X A M P L E 2

Use a calculator or an appendix to approximate to three decimal places.

We may use an appendix or a calculator to approximate To use a calculator, findthe square root key

To three decimal places,

3 Radicals can also contain variables. To simplify radicals containing variables,special care must be taken.To see how we simplify let’s look at a few examples inthis form.

If we have or x.

If x is 5, we have or x.252 = 225 = 5,

232 = 29 = 3,x = 3,

2x2 ,

23 L 1.732.

23 L 1.732050808

1 .23.

23

Solut ion

CLASSROOM EXAMPLE

Approximate to 3 decimal places.answer: 3.162

210




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For any real number a, 2a2 = ƒ a ƒ .

Solut ion

From these two examples, you may think that simplifies to x. Let’s now look at an example where x is a negative number. If we have not our original x. To make sure that simplifies to a nonnegative number, wehave the following.

2x2-3,21-322 = 29 = 3,x = -3,

2x2

E X A M P L E 3

Thus,

To avoid this, for the rest of the chapter we assume that if a variable appears inthe radicand of a radical expression, it represents positive numbers only. Then

Because

Because

Because 13x22 9x2 29x2 = 3x

1x 422 x 8 2x8 = x4

1y22 y2 2y2 = y

2x2 = ƒ x ƒ = x since x is a positive number.

217y22 = ƒ 7y ƒ , and so on.

21-822 = ƒ -8 ƒ = 8

2x2 = ƒ x ƒ ,

Simplify each expression. Assume that all variables represent positive numbers.

a. b. c. d.

a. because x times itself equals

b. because

c. because

d. because

4 We can find roots other than square roots. For example, since we call 2the cube root of 8. In symbols, we write

Also,

Since

Since

Notice that unlike the square root of a negative number, the cube root of a negativenumber is a real number. This is so because while we cannot find a real number whosesquare is negative, we can find a real number whose cube is negative. In fact, the cubeof a negative number is a negative number. Therefore, the cube root of a negativenumber is a negative number.

142 3 64 23 -64 = -4

3 3 27 23 27 = 3

T——————ƒ23 8 = 2 The number 3 is called the index .

23 = 8,

ax2

5b

2

=x4

25.Bx4

25=

x2

5

14x822 = 16x16 .216x16 = 4x8

1x322 = x6 .2x6 = x3

x2 .2x2 = x

Bx4

25216x162x62x2

CLASSROOM EXAMPLE

Simplify each expression. Assume that allvariables represent positive numbers.

a. b.

c. d.

e.

answer:

a. b.

c. d.

e.x5

5

8y62x3

x10x4

Bx10

25

264y1224x6

2x202x8



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Find each cube root.

a. b. c.

a. because

b. because

c. because

Just as we can raise a real number to powers other than 2 or 3, we can find roots otherthan square roots and cube roots. In fact, we can take the nth root of a number where nis any natural number. An nth root of a number a is a number whose nth power is a.The natural number n is called the index.

In symbols, the nth root of a is written as The index 2 is usually omitted forsquare roots.

2n a .

a15b3

=1

125.A3 1

125=

15

1-323 = -27.23 -27 = -3

13 = 1.23 1 = 1

A3 1125

23 -2723 1

Solut ion

E X A M P L E 4

H e l p f u l H i n tIf the index is even, such as and so on, the radicand must benonnegative for the root to be a real number. For example,

26 64 = 2 but 26 -64 is not a real number

24 16 = 2 but 24 -16 is not a real number

2 , 24 , 26 ,

CLASSROOM EXAMPLE

Find each cube root.

a. b. c.answer:

a. 5 b. c.14

-2

A3 164

23 -823 125


CONCEPT CHECKWhich of the following is a real number?

a.

b.

c.

d. 26 -64

25 -64

24 -64

2-64

Find each root.

a. b. c. d.

a. because and 2 is positive.

b. because

c. since

d. is not a real number since the index 4 is even and the radicand is

negative.

-8124 -81

23 8 = 2.-23 8 = -2

1-225 = -32.25 -32 = -2

24 = 1624 16 = 2

24 -81-23 825 -3224 16

CLASSROOM EXAMPLE

Find each root.

a. b.

c. d.answer: a. not a real number

b. c. 3 d. not a real number-1

26 -6424 81

25 -124 -16

E X A M P L E 5

Solut ion




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To simplify or approximate square roots using a calculator, locate the

key marked To simplify using a scientific calculator, press

The display should read To simplify

using a graphing calculator, press

To approximate press (or

). The display should read This is an approxi-

mation for A three-decimal-place approximation is

Is this answer reasonable? Since 30 is between perfect squares 25 and36, is between and The calculator result isthen reasonable since 5.4772256 is between 5 and 6.

Use a calculator to approximate each expression to three decimal places.Decide whether each result is reasonable.

1. 2. 3.4. 5. 6.

Many scientific calculators have a key, such as that can beused to approximate roots other than square roots. To approximatethese roots using a graphing calculator, look under the menu or consult your manual.

Use a calculator to approximate each expression to three decimal places.Decide whether each result is reasonable.

7. 8. 9.10. 11. 12.G

26 225 1824 1524 2023 7123 40

MATH

1 x

y ,

246282220021121427

236 = 6.225 = 5230

230 L 5.477

230.

5.4772256 . ENTER 30 1 1 30 230,

ENTER . 25 1 225 5 . 1 . 25

225 1 .


Suppose you are a highway maintenance supervisor. You and your crew areheading out to post signs at a new cloverleaf exit ramp on the highway. Oneof the signs needed is a suggested ramp speed limit. Once you are at theramp, you realize that you forgot to check with the highway engineers aboutwhich sign to post. You know that the formula can be used to es-timate the maximum safe speed S, in miles per hour, at which a car can travelon a curved road with radius of curvature, r, in feet. Using a tape measure,

your crew measures the radius of curvature as 400 feet. Which sign should you post? Explain your reasoning.

S = 22.5r



MENTAL MATHAnswer each exercise true or false.

1. simplifies to a real number. false 2. while true

3. The number 9 has two square roots. true 4. and true

5. If x is a positive number, true 6. If x is a positive number, false2x16 = x4 .2x10 = x5 .

21 = 1.20 = 0

23 64 = 4.264 = 82-16


SIMPLIFYING RADICALS SECTION 8.2 491

O b j e c t i v e s

Use the product rule to simplify square roots.

Use the quotient rule to simplify square roots.

Simplify radicals containing variables.

Simplify higher roots.4

3

2

1

8.2 S I M P L I F Y I N G R A D I CA L S

1 A square root is simplified when the radicand contains no perfect square factors (other than 1). For example, is not simplified because and 4 is aperfect square.

To begin simplifying square roots, we notice the following pattern.

Since both expressions simplify to 12, we can write

This suggests the following product rule for square roots.

29 # 16 = 29 # 216

29 # 216 = 3 # 4 = 12

29 # 16 = 2144 = 12

220 = 24 # 5220

Product Rule for Square RootsIf and are real numbers, then2a # b = 2a # 2b

2b2a

In other words, the square root of a product is equal to the product of the square roots.

To simplify for example, we factor 20 so that one of its factors is a perfectsquare factor.

Factor 20.

Use the product rule.

Write as 2.

The notation means Since the radicand 5 has no perfect square fac-tor other than 1 then 2 is in simplest form.25

2 # 25.225

24 = 225

= 24 # 25

220 = 24 # 5

220,

H e l p f u l H i n tA radical expression in simplest form does not mean a decimal approximation.The simplest form of a radical expression is an exact form and may still containa radical.

exact decimal approximation

220 = 2253 220 L 4.473




E X A M P L E 1

Simplify.

a. b. c. d.

a. Try to factor 54 so that at least one of the factors is a perfect square. Since 9 is a per-fect square and

Factor 54.

Apply the product rule.

Write as 3.

b. Factor 12.


Write as 2.

c. The largest perfect square factor of 200 is 100.

Factor 200.


Write as 10.

d. The radicand 35 contains no perfect square factors other than 1.Thus is in sim-plest form.

In Example 1, part (c), what happens if we don’t use the largest perfect square factor of200? Although using the largest perfect square factor saves time, the result is the sameno matter what perfect square factor is used. For example, it is also true that

Then

Since is not in simplest form, we continue.

2 Next, let’s examine the square root of a quotient.

Also,

Since both expressions equal 2, we can writeA164

=21624

21624=

42

= 2

A164

= 24 = 2

= 1022

= 2 # 5 # 22

= 2 # 225 # 22

2200 = 2 # 250

250

= 2 # 250

2200 = 24 # 250

200 = 4 # 50.

235

2100 = 1022

= 2100 # 22

2200 = 2100 # 2

24 = 223

= 24 # 23

212 = 24 # 3

29 = 326

= 29 # 26

254 = 29 # 6

54 = 9 # 6,

2352200212254

Solut ion

CLASSROOM EXAMPLE

Simplify.

a. b.c. d.answer:

a. b.c. d. 2151027

3222210

2152700218240



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This suggests the following quotient rule.

Quotient Rule for Square RootsIf and are real numbers and thenAa

b=2a2b

b Z 0,2b2a

In other words, the square root of a quotient is equal to the quotient of the squareroots.

E X A M P L E 2

Solut ion

CLASSROOM EXAMPLE

Simplify.

a. b. c.

answer:

a. b. c.325

7225

,49

,

A4549A 2

25A1681

Simplify.

a. b. c.


a. b.

c. Use the quotient rule.

Apply the product rule and write as 9.

Write as 2.

3 Recall that because If an odd exponent occurs, we writethe exponential expression so that one factor is the greatest even power contained inthe expression. Then we use the product rule to simplify.

1x322 = x6 .2x6 = x3

24 =2210

9

281 =24 # 210

9

A4081

=240281

A 364

=23264

=238A25

36=225236

=56

A4081A 3

64A2536

E X A M P L E 3

Simplify. Assume that all variables represent positive numbers.

a. b. c.

a.

b.

c. A45

x6 =2452x6

=29 # 5

x3 =29 # 25

x3 =325

x3

28y2 = 24 # 2 # y2 = 24y2 # 2 = 24y2 # 22 = 2y22

2x5 = 2x4 # x = 2x4 # 2x = x22x

A45

x628y22x5

Solut ion

CLASSROOM EXAMPLE

Simplify. Assume that all variables represent positive numbers.

a. b. c.

answer:

a. b. c.323

x43x222x52x

A27

x8218x42x11




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Product Rule for RadicalsIf and are real numbers, then

Quotient Rule for RadicalsIf and are real numbers and thenAn a

b=2n a2n b

b Z 0,2n b2n a

2n a # b = 2n a # 2n b

2n b2n a

Simplify.

a. b. c. d.

a.

b. The number 18 contains no perfect cube factors, so cannot be simplified further.

c.

d.

To simplify fourth roots, look for perfect fourth powers of the radicand. For ex-ample, 16 is a perfect fourth power since

To simplify factor 32 as

Factor 32.


Write as 2.24 16 = 224 2

= 24 16 # 24 2

24 32 = 24 16 # 2

16 # 2.24 32,24 = 16.

A3 4027

=23 4023 27

=23 8 # 5

3=23 8 # 23 5

3=

223 53

A3 78

=23 723 8

=23 72

23 18

23 54 = 23 27 # 2 = 23 27 # 23 2 = 323 2

A3 4027A3 7

823 1823 54

Solut ion

CLASSROOM EXAMPLE

Simplify.

a. b.

c. d.

answer:

a. b.c. d.

323 32

23 103

23 50223 5

A3 818A3 10

27

23 5023 40

E X A M P L E 4

4 The product and quotient rules also apply to roots other than square roots. Ingeneral, we have the following product and quotient rules for radicals.

To simplify cube roots, look for perfect cube factors of the radicand. For example,8 is a perfect cube, since

To simplify factor 48 as

Factor 48.


Write as 2.is in simplest form since the radicand 6 contains no perfect cube factors other

than 1.223 6

23 8 = 223 6

= 23 8 # 23 6

23 48 = 23 8 # 6

8 # 6.23 48,

23 = 8.




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Simplify.

a. b. c.

a.

b.

c. 25 64 = 25 32 # 2 = 25 32 # 25 2 = 225 2

A4 316

=24 324 16

=24 32

24 243 = 24 81 # 3 = 24 81 # 24 3 = 324 3

25 64A4 316

24 243

Solut ion

E X A M P L E 5

MENTAL MATH

CLASSROOM EXAMPLE

Simplify.

a. b. c.

answer: a. b. c. 225 324 5

3224 5

25 96A4 581

24 80

Simplify each expression. Assume that all variables represent nonnegative real numbers.

1. 6 2. 18 3. x 4.

5. 0 6. 1 7. 8. 7x249x25x2225x42120

y22y42x229 # 3624 # 9


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ADDING AND SUBTRACTING RADICALS SECTION 8.3 497

8.3 A D D I N G A N D S U B T R AC T I N G R A D I CA L S

O b j e c t i v e s

Add or subtract like radicals.

Simplify radical expressions, and then add or subtract any like radicals.

1 To combine like terms, we use the distributive property.

The distributive property can also be applied to expressions containing radicals.For example,

Also,

Radical terms and are like radicals, as are and 625.925322522

925 - 625 = 19 - 6225 = 325

522 + 322 = 15 + 3222 = 822

5x + 3x = 15 + 32x = 8x

2

1




Like RadicalsLike radicals are radical expressions that have the same index and the sameradicand.

From the examples above, we can see that only like radicals can be combined inthis way. For example, the expression cannot be further simplified sincethe radicals are not like radicals. Also, the expression cannot be furthersimplified because the radicals are not like radicals since the indices are different.

427 + 423 7223 + 322

Solut ion

CLASSROOM EXAMPLE

Simplify by combining like radical terms.

a. b.

c. d.answer:a. b.

c. d. 323 - 322222

-22715211

323 - 32222 + 22

27 - 3276211 + 9211

Concept Check Answer:d

E X A M P L E 1Simplify by combining like radical terms.

a. b. c. d.

a.

b.

c.

d. cannot be simplified further since the indices are not the same.226 + 223 6

223 7 - 523 7 - 323 7 = 12 - 5 - 3223 7 = -623 7

210 - 6210 = 1210 - 6210 = 11 - 62210 = -5210

425 + 325 = 14 + 3225 = 725

226 + 223 6223 7 - 523 7 - 323 7210 - 6210425 + 325

CONCEPT CHECKWhich is true?

a.

b.

c.

d. None of the above is true. In each case, the left-hand side cannot be simplified further.

23 + 25 = 28

223 + 227 = 2210

2 + 325 = 525

2 At first glance, it appears that the expression cannot be simplifiedfurther because the radicands are different. However, the product rule can be used tosimplify each radical, and then further simplification might be possible.

250 + 28

E X A M P L E 2Add or subtract by first simplifying each radical.

a. b. c.

a. First simplify each radical.

Factor radicands.


Simplify and

Add like radicals.

b. Factor radicands.


Simplify and

Multiply.

Subtract like radicals. = 923

= 1423 - 523

225 .24 = 7 # 223 - 523

= 724 # 23 - 225 # 23

7212 - 275 = 724 # 3 - 225 # 3

= 722

24 .225 = 522 + 222

= 225 # 22 + 24 # 22

250 + 28 = 225 # 2 + 24 # 2

225 - 227 - 2218 - 2167212 - 275250 + 28

Solut ion

CLASSROOM EXAMPLE

Add or subtract by first simplifying eachradical.a. b.

c.answer:a. b.

c. 3 - 823

-1525823

236 - 248 - 423 - 29

3220 - 7245227 + 275




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ADDING AND SUBTRACTING RADICALS SECTION 8.3 499

c.

Factor radicands.


Simplify.

Write as 1 and as 6.

If radical expressions contain variables, we proceed in a similar way. Simplify rad-icals using the product and quotient rules. Then add or subtract any like radicals.

2 # 35 4 = 1 - 323 - 622

= 5 - 323 - 2 # 322 - 4

= 5 - 29 # 23 - 229 # 22 - 4

= 5 - 29 # 3 - 229 # 2 - 4

225 - 227 - 2218 - 216

E X A M P L E 3

Simplify Assume variables represent positive numbers.

Write as x and apply the product rule.

Simplify.

Add like radicals. = 2x - 42x

= 2x - 52x + 12x

2x 2 = 2x - 225 # 2x + 2x

22x2 - 225x + 2x

22x2 - 225x + 2x .

CLASSROOM EXAMPLE

Simplify .answer: 3x2 - 5x2x

29x4 - 236x3 + 2x3

Solution

Add or subtract by first simplifying each radical.

Simplify and factor 54.


Simplify 23 27. = 6 - 323 2()*

= 6 - 23 27 # 23 2

23 27 223 27 - 23 54 = 2 # 3 - 23 27 # 2

223 27 - 23 54CLASSROOM EXAMPLE

Simplify .answer: 323 4

523 4 - 23 32

Solution

H e l p f u l H i n tThese two terms may notbe combined. They areunlike terms.

E X A M P L E 4

Simplify each expression by combining like radicals.

1. 2. 3.

4. 5. 6. 326826 - 526327527 - 227112x82x + 32x

72x52x + 22x923223 + 723822322 + 522

MENTAL MATH


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MULTIPLYING AND DIV IDING RADICALS SECTION 8.4 501

8.4 M U LT I P LY I N G A N D D I V I D I N G R A D I CA L S

Product Rule for RadicalsIf and are real numbers, then2n a # 2n b = 2n a # b

2n b2n a

O b j e c t i v e s

Multiply radicals.

Divide radicals.

Rationalize denominators.

Rationalize using conjugates.

1 In Section 8.2 we used the product and quotient rules for radicals to help ussimplify radicals. In this section, we use these rules to simplify products and quotientsof radicals.

4

3

2

1

This property says that the product of the nth roots of two numbers is the nth root ofthe product of the two numbers. For example,

Also, 23 5 # 23 7 = 23 5 # 7 = 23 35

23 # 22 = 23 # 2 = 26




Multiply. Then simplify if possible.

a. b. c. d.

a.

b. Next, simplify

c. Next, simplify

d. A322 B2 = 32 # A22 B2 = 9 # 2 = 18

10212 = 1024 # 3 = 1024 # 23 = 10 # 2 # 23 = 2023212. 226 # 522 = 2 # 526 # 2 = 10212.

245 = 29 # 5 = 29 # 25 = 325245. 23 # 215 = 245.

27 # 23 = 27 # 3 = 221

A322 B2226 # 52223 # 21527 # 23

CLASSROOM EXAMPLE


a. b.

c. d.answer:

a. b. c. d. 758425322210

A523 B2723 # 4215

26 # 2325 # 22

Solut ion

Multiply Then simplify if possible.

When multiplying radical expressions containing more than one term, use the sametechniques we use to multiply other algebraic expressions with more than one term.

23 4 # 23 18 = 23 4 # 18 = 23 4 # 2 # 9 = 23 8 # 9 = 23 8 # 23 9 = 223 9

23 4 # 23 18.

CLASSROOM EXAMPLE

Multiply. answer: 323 4

23 6 # 23 18

Solut ion


a.

b.

a. Using the distributive property, we have

b. Use the FOIL method of multiplication.

F O I L


Simplify.

Special products can be used to multiply expressions containing radicals.

Within Example 3, we found that

This is true in general.

25 # 25 = 5 and 22 # 22 = 2

= 23x - 22x + 26 - 2

= 23x - 22x + 26 - 24

(2x + 22)(23 - 22) = 2x # 23 - 2x # 22 + 22 # 23 - 22 # 22

= 5 - 210

25 A25 - 22 B = 25 # 25 - 25 # 22

A2x + 22 B A23 - 22 B25 A25 - 22 B

Solut ion

CLASSROOM EXAMPLE

Multiply.

a.

b.

answer: a.

b. x - 23x + 25x - 215

7 - 221

A2x + 25 B A2x - 23 B27 A27 - 23 B

E X A M P L E 1

E X A M P L E 3

E X A M P L E 2




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Concept Check Answer:a, c, d

CLASSROOM EXAMPLE

Multiply.

a.

b.answer: a. b. 5x + 825x + 16-33A25x + 4 B2A23 + 6 B A23 - 6 B

Solut ion

If a is a positive number, 2a # 2a = a

CONCEPT CHECKIdentify the true statement(s).

a. b. c.

d. (Here x is a positive number.)25x # 25x = 5x

2131 # 2131 = 13122 # 23 = 627 # 27 = 7

E X A M P L E 4Multiply. Then simplify if possible.

a. b.

a. Recall from Chapter 5 that Then

b. Recall that Then

2 To simplify quotients of radical expressions, we use the quotient rule.

= 7x + 427x + 4

A27x + 2 B2 = A27x B2 + 2 A27x B122 + 12221a + b22 = a2 + 2ab + b2 .

= -44 = 5 - 49

A25 - 7 B A25 + 7 B = A25 B2 - 72

1a - b21a + b2 = a2 - b2 .

A27x + 2 B2A25 - 7 B A25 + 7 B

Quotient Rule for RadicalsIf and are real numbers and then2n a2n b

= An a

b, providing b Z 0

b Z 0,2n b2n a

Divide. Then simplify if possible.

a. b. c.

Use the quotient rule and then simplify the resulting radicand.

a.

b.

c.212x323x

= B12x3

3x= 24x2 = 2x

210025= A100

5= 220 = 24 # 5 = 24 # 25 = 225

21422= A14

2= 27

212x323x

210025

21422

CLASSROOM EXAMPLEDivide.

a. b. c.

answer: a. b. c. x21532525

275x325x

29022

21523

Solut ion

E X A M P L E 5



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CLASSROOM EXAMPLE

Divide.

answer: 5

23 25023 2

Solut ion

Rationalize each denominator.

a. b. c.

a. To eliminate the radical in the denominator of multiply the numerator and the

denominator by

b. We can multiply the numerator and denominator by but see what happens ifwe simplify first.

To rationalize the denominator now, multiply the numerator and the denominatorby

c.

To rationalize the denominator, multiply the numerator and denominator by

1

322x=

1 # 22x

322x # 22x=22x

3 # 2x=22x

6x

22x .

A 118x

=21218x

=129 # 22x

=1

322x

25

223=25 # 23

223 # 23=2152 # 3

=215

6

23.

25212=2524 # 3

=25

223

212,

227=

2 # 2727 # 27=

2277

27.

227,

A 118x

25212

227

CLASSROOM EXAMPLE


a. b. c.

answer:

a. b. c.210x

15x

23510

5233

A 245x

27220

523

Solut ion

E X A M P L E 7

E X A M P L E 6

Divide Then simplify if possible.

3 It is sometimes easier to work with radical expressions if the denominator doesnot contain a radical. To eliminate the radical in the denominator of a radicalexpression, we use the fact that we can multiply the numerator and the denominator ofa fraction by the same nonzero number. This is equivalent to multiplying the fraction

by 1. To eliminate the radical in the denominator of multiply the numerator and

the denominator by Then

This process is called rationalizing the denominator.

2522=25 # 2222 # 22

=210

2

22.

2522,

23 3223 4= A3 32

4= 23 8 = 2

23 3223 4.



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CLASSROOM EXAMPLE


a. b.

answer:

a. b.23 275

5723 21

3

23 1123 5

723 9

Solut ion

E X A M P L E 8Rationalize each denominator.

a. b.

a. Since the denominator contains a cube root, we multiply the numerator and the denominator by a factor that gives the cube root of a perfect cube in the de-nominator. Recall that and that the denominator multiplied by is

or

b. Recall that Multiply the denominator by and the result is or

4 To rationalize a denominator that is a sum, such as the denominator in

we multiply the numerator and the denominator by The expressions and are called conjugates of each other. When a radical expression such as

is multiplied by its conjugate the product simplifies to an expressionthat contains no radicals.

Then

2

4 + 23=

2 A4 - 23 BA4 + 23 B A4 - 23 B =

2 A4 - 23 B13

A4 + 23 B A4 - 23 B = 42 - A23 B2 = 16 - 3 = 13

1a + b21a - b2 = a2 - b2

4 - 23,4 + 234 - 23

4 + 234 - 23.

2

4 + 23

23 723 3=23 7 # 23 923 3 # 23 9

=23 6323 27

=23 63

3

23 27.23 3 # 923 923 323 27 = 3.

523 4=

5 # 23 223 4 # 23 2=

523 223 8=

523 22

23 8.23 4 # 223 223 423 8 = 2

23 723 3

523 4

Rationalize each denominator and simplify.

a. b.25 + 425 - 1

2

1 + 23

E X A M P L E 9

As a general rule, simplify a radical expression first and then rationalize the denominator.


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Simplify

First simplify

Next, factor out a common factor of 3 from the terms in the numerator and the de-nominator and simplify.

12 - 3229

=3 A4 - 22 B

3 # 3=

4 - 223

12 - 2189

=12 - 29 # 2

9=

12 - 3229

218.

12 - 2189

.

Solut ion

CLASSROOM EXAMPLE

Simplify .

answer:3 - 23

5

15 - 27525

Find each product. Assume that variables represent nonnegative real numbers.

1. 2. 3.

4. 5. 6. 2xy2x # 2y210y210 # 2y27x27 # 2x

2621 # 2623525 # 272622 # 23

E X A M P L E 1 0

Solut ion

CLASSROOM EXAMPLE

Rationalize each denominator and simplify.

a. b.

answer:

a. b. 7 + 622-1 + 27

2

22 + 522 - 1

3

1 + 27

a. Multiply the numerator and the denominator of this fraction by the conjugate of

that is, by

Simplify.

b.Multiply the numerator and denominator by

the conjugate of

Multiply. =5 + 25 + 425 + 4

5 - 1

25 1 .25 1 , 25 + 425 - 1

=A25 + 4 B A25 + 1 BA25 - 1 B A25 + 1 B

= -1 + 23

= -1 A1 - 23 Ba

-b= -

a

b = -

2 A1 - 23 B2

=2 A1 - 23 B

-2

=2 A1 - 23 B

1 - 3

=2 A1 - 23 B12 - A23 B2

2

1 + 23=

2 A1 - 23 BA1 + 23 B A1 - 23 B

1 - 23.1 + 23,

MENTAL MATH


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SOLVING EQUATIONS CONTAINING RADICALS SECTION 8.5 509

O b j e c t i v e s

Solve radical equations by using the squaring property of equality once.

Solve radical equations by using the squaring property of equality twice.

1 In this section, we solve radical equations such as2x + 3 = 5 and 22x + 1 = 23x

2

1

8.5 S O LV I N G E Q UAT I O N S C O N TA I N I N G R A D I CA L S




The Squaring Property of EqualityIf then a2 = b2a = b ,

Unfortunately, this squaring property does not guarantee that all solutions of thenew equation are solutions of the original equation. For example, if we square bothsides of the equation

we have

This new equation has two solutions, 2 and while the original equation hasonly one solution. Thus, raising both sides of the original equation to the second powerresulted in an equation that has an extraneous solution that isn’t a solution of the orig-inal equation. For this reason, we must always check proposed solutions of radicalequations in the original equation. If a proposed solution does not work, we call thatvalue an extraneous solution.

x = 2-2,

x2 = 4

x = 2

Solve.

To solve this radical equation, we use the squaring property of equality and squareboth sides of the equation.

Square both sides.

Simplify.


We replace x with 22 in the original equation.

Original equation

Let

True

Since a true statement results, 22 is the solution.

5 = 5 225 5

x 22 . 222 + 3 5

2x + 3 = 5

x = 22

x + 3 = 25

A2x + 3 B2 = 52

2x + 3 = 5

2x + 3 = 5

Solut ion

CLASSROOM EXAMPLE

Solve. answer: 51

2x - 2 = 7

Check

H e l p f u l H i n tDon’t forget to check theproposed solutions ofradical equations in theoriginal equation.

Solve.

First we set the radical by itself on one side of the equation.Then we square both sides.

Subtract 6 from both sides to get the radical by itself. 2x = -2

2x + 6 = 4

2x + 6 = 4

Solut ion

CLASSROOM EXAMPLE

Solve. answer: no solution

2x + 9 = 2

Radical equations contain variables in the radicand. To solve these equations, we relyon the following squaring property.

E X A M P L E 1

E X A M P L E 2


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Recall that is the principal or nonnegative square root of x so that cannotequal and thus this equation has no solution.We arrive at the same conclusion if wecontinue by applying the squaring property.

Square both sides.

Simplify.

We replace x with 4 in the original equation.

Original equation

Let False

Since 4 does not satisfy the original equation, this equation has no solution.

Example 2 makes it very clear that we must check proposed solutions in the orig-inal equation to determine if they are truly solutions. Remember, if a proposed solu-tion is not an actual solution, we say that the value is an extraneous solution.

The following steps can be used to solve radical equations containing square roots.

2 + 6 = 4x 4 . 24 + 6 4

2x + 6 = 4

x = 4

A2x B2 = 1-222 2x = -2

-22x2x

Solving a Radical Equation Containing Square Roots

Step 1: Arrange terms so that one radical is by itself on one side of the equa-tion. That is, isolate a radical.

Step 2: Square both sides of the equation.

Step 3: Simplify both sides of the equation.

Step 4: If the equation still contains a radical term, repeat steps 1 through 3.

Step 5: Solve the equation.

Step 6: Check all solutions in the original equation for extraneous solutions.

Solve .

Each of the radicals is already isolated, since each is by itself on one side of the equa-tion. So we begin solving by squaring both sides.

Original equation

Square both sides.

Simplify.Subtract 5x from both sides.

Divide both sides by and simplify.4 x =-2-4

=12

-4x = -2 x = 5x - 2

A2x B2 = A25x - 2 B2 2x = 25x - 2

2x = 25x - 2

Solut ion

CLASSROOM EXAMPLE

Solve .

answer:15

26x - 1 = 2x

E X A M P L E 3

Check


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We replace x with in the original equation.

Original equation

Let

Multiply.

Write 2 as

True

This statement is true, so the solution is 12

.

A12

= A12

42

. A12

A52

-42

A12

A52

- 2

x 1

2. A1

2 A5 # 1

2- 2

2x = 25x - 2

12

Check

Solve.

The radical is already isolated, so we start by squaring both sides.

Square both sides.

Simplify.Subtract from both sides.

Add 15 to both sides.Divide both sides by 5.

We replace y with 3 in the original equation.

Original equation

Let Simplify.

True

This statement is true, so the solution is 3.

6 = 6 236 6

24 # 9 + 15 - 15 6y 3 . 24 # 32 + 5 # 3 - 15 2 # 3

24y2 + 5y - 15 = 2y

y = 3 5y = 15

4y2 5y - 15 = 0 4y2 + 5y - 15 = 4y2

A24y2 + 5y - 15 B2 = 12y22 24y2 + 5y - 15 = 2y

24y2 + 5y - 15 = 2y

CLASSROOM EXAMPLE

Solve. answer: 5

29y2 + 2y - 10 = 3y

Check

E X A M P L E 4

TEACHING TIPBefore solving Example 5, ask stu-dents, “Will squaring both sides ofan equation always eliminate theradical sign?” Demonstrate thatthe radical is not eliminated whenboth sides of are squared directly.

2x + 3 - x = -3

Solut ion

H e l p f u l H i n tDon’t forget that 1x - 322 = 1x - 321x - 32 = x2 - 6x + 9

Solve.

First we isolate the radical by adding x to both sides. Then we square both sides.

Add x to both sides.Square both sides.

x + 3 =

x2 - 6x + 9

5

A2x + 3 B2 = 1x - 322 2x + 3 = x - 3

2x + 3 - x = -3

2x + 3 - x = -3

Solut ion

E X A M P L E 5




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Check

Solve.

Square both sides.


Square both sides again. 4 = x

4 . 2 = 2x

-8 = -42x

x - 4 = x - 42x + 45

A2x - 4 B2 = A2x - 2 B2 2x - 4 = 2x - 2

2x - 4 = 2x - 2

Solut ion

CLASSROOM EXAMPLE

Solve. answer: 36

2x - 5 = 2x - 35

E X A M P L E 6

To solve the resulting quadratic equation, we write the equation in standard form bysubtracting x and 3 from both sides.

Subtract x from both sides.


Factor.

Set each factor equal to zero.

Solve for x.

We replace x with 6 and then x with 1 in the original equation.

True False

Since replacing x with 1 resulted in a false statement, 1 is an extraneous solution. The only solution is 6.

2 If a radical equation contains two radicals, we may need to use the squaringproperty twice.

1 = -3 -3 = -3

2 - 1 -3 3 - 6 -3

24 - 1 -3 29 - 6 -3

21 + 3 - 1 -3 26 + 3 - 6 -3

2x + 3 - x = -3 2x + 3 - x = -3

Let x = 1. Let x = 6.

6 = x 1 = x

0 = x - 6 or 0 = x - 1

0 = 1x - 621x - 12 0 = x2 - 7x + 6

3 = x2 - 7x + 9


= x - 42x + 4

= 2x # 2x - 22x - 22x + 4

A2x - 2 B2 = A2x - 2 B A2x - 2 B

Check the proposed solution in the original equation. The solution is 4.

CLASSROOM EXAMPLE

Solve. answer: 8

2x + 1 - x = -5


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8.6 R A D I CA L E Q UAT I O N S A N D P R O B L E M S O LV I N G

O b j e c t i v e s

Use the Pythagorean formula to solve problems.

Use the distance formula.

Solve problems using formulas containing radicals.

1 Applications of radicals can be found in geometry, finance, science, and otherareas of technology. Our first application involves the Pythagorean theorem, giving aformula that relates the lengths of the three sides of a right triangle. We first studiedthe Pythagorean theorem in Chapter 6 and we review it here.

3

2

1

The Pythagorean TheoremIf a and b are lengths of the legs of a right triangleand c is the length of the hypotenuse, thena2 + b2 = c2 .

cb

a Legs

Hypotenuse

Find the length of the hypotenuse of a right triangle whose legs are 6 inches and 8 inch-es long.

Because this is a right triangle, we use the Pythagorean theorem. We let inchesand inches. Length c must be the length of the hypotenuse.

Use the Pythagorean theorem.

Substitute the lengths of the legs.

Simplify.

100 = c2

36 + 64 = c2

62 + 82 = c2

a2 + b2 = c2

b = 8a = 6

CLASSROOM EXAMPLE

Find the length of the hypotenuse of theright triangle shown.

answer: 5 cm

6 inches

8 inches

Hypotenuse

E X A M P L E 1

Solut ion

Since c represents a length, we know that c is positive and is the principal square rootof 100.

Use the definition of principal square root.Simplify.

The hypotenuse has a length of 10 inches.

10 = c 2100 = c

100 = c2

36 in.2

100 in.2

64 in.2

8

610

TEACHING TIPAfter Example 1, demonstrate thePythagorean theorem by drawingsquares on each side of the trian-gle, finding the area of each square,and noting that the sum of the areaof the squares of each leg equalsthe area of the square of the hy-potenuse.

Hypotenuse4 centimeters

3 centimeters

That is, the square of the length of the hypotenuse is equal to the sum of the squares ofthe lengths of the legs.



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RADICAL EQUATIONS AND PROBLEM SOLVING SECTION 8.6 517

Find the length of the leg of the right triangle shown. Give the exact length and a two-decimal-place approximation.

We let meters and b be the unknown length of the other leg. The hypotenuse ismeters.


Let and

b = 221 L 4.58 meters

b2 = 21

4 + b2 = 25

c 5 .a 2 22 + b2 = 52

a2 + b2 = c2

c = 5a = 2Solut ion

CLASSROOM EXAMPLE

Find the length of the leg of the right trian-gle shown. Give the exact length and atwo-decimal-place approximation.

answer: 323 L 5.20 mi

Leg5 meters

2 meters

E X A M P L E 2

3 miles 6 miles

Leg

E X A M P L E 3

FINDING A DISTANCE

A surveyor must determine the distance across a lake at points P and Q as shown inthe figure.To do this, she finds a third point R perpendicular to line PQ. If the length of

is 320 feet and the length of is 240 feet, what is the distance across the lake?Approximate this distance to the nearest whole foot.

QRPR

1. UNDERSTAND. Read and reread the problem. We will set up the problem usingthe Pythagorean theorem. By creating a line perpendicular to line PQ, the surveyordeliberately constructed a right triangle. The hypotenuse, has a length of 320feet, so we let in the Pythagorean theorem.The side is one of the legs, sowe let and unknown length.

2. TRANSLATE.


Let and c 320 .a 240 2402 + b2 = 3202

a2 + b2 = c2

b = thea = 240QRc = 320

PR ,

CLASSROOM EXAMPLE

Given the distances in the figure, find thelength of the pond to the nearest tenth ofa foot.

320 feet

240 feet

P

QR

The length of the leg is exactly meters or approximately 4.58 meters.221

a 240

c 320b

QR

P

Solut ion



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Distance FormulaThe distance d between two points with coordinates and is givenby

d = 21x2 - x122 + 1y2 - y1221x2 , y221x1 , y12

Find the distance between and

Use the distance formula with and 1x2 , y22 = 1-3, -52 .1x1 , y12 = 1-1, 921-3, -52 .1-1, 92

The distance formula.

Substitute known values.

Simplify.

Simplify the radical.

The distance is exactly units or approximately 14.1 units.1022

= 2200 = 1022

= 24 + 196

= 21-222 + 1-1422 = 2[-3 - 1-12]2 + 1-5 - 922

d = 21x2 - x122 + 1y2 - y122

Solut ion

CLASSROOM EXAMPLE

Find the distance between and

answer: 2341-11, 52 .

1-6, 22

P2 (x2, y2)

P1 (x1, y1)

x

y

d

6789

112345

12345

2 3 4 512345

(1, 9)

(3, 5)

x

y

d

3. SOLVE.

Subtract 57,600 from both sides.

Use the definition of principal square root.

4. INTERPRET.

Check: See that

State: The distance across the lake is exactly The surveyor can now use a calculator to find that is approximately 211.6601 feet, so the dis-tance across the lake is roughly 212 feet.

2 A second important application of radicals is in finding the distance betweentwo points in the plane. By using the Pythagorean theorem, the following formula canbe derived.

244,800 feet244,800 feet .

2402 + A244,800 B2 = 3202.

b = 244,800

b2 = 44,800

57,600 + b2 = 102,400

E X A M P L E 4



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RADICAL EQUATIONS AND PROBLEM SOLVING SECTION 8.6 519

DETERMINING VELOCITY

A formula used to determine the velocity v, in feet per second, of an object (neglectingair resistance) after it has fallen a certain height is where g is the accelera-tion due to gravity, and h is the height the object has fallen. On Earth, the accelerationg due to gravity is approximately 32 feet per second per second. Find the velocity of aperson after falling 5 feet.

We are told that per second per second. To find the velocity v whenwe use the velocity formula.

Use the velocity formula.

Substitute known values.

Simplify the radicand.

The velocity of the person after falling 5 feet is exactly feet per second, or app-roximately 17.9 feet per second.

825

= 825

= 2320

= 22 # 32 # 5

v = 22gh

h = 5 feet ,g = 32 feet

v = 22gh ,CLASSROOM EXAMPLE

Use the formula in Example 5 to find the ve-locity of an object after falling 10 feet.answer: 8210 L 25.3 ft/sec

E X A M P L E 5

3 The Pythagorean theorem is an extremely important result in mathematics andshould be memorized. But there are other applications involving formulas containingradicals that are not quite as well known, such as the velocity formula used in the nextexample.

Solut ion

10 feet14 feet

Suppose you are a carpenter. You are in-stalling a 10-foot by 14-foot wooden deck at-tached to a client’s house. Before sinking thedeck posts into the ground, you lay out thedimensions and deck placement with stakesand string. It is very important that the stringlayout is “square”—that is, that the edges of

the deck layout meet at right angles. If not, then the deck posts could be sunk in thewrong positions and portions of the deck will not line up properly.

From the Pythagorean theorem, you know that in a right triangle,It’s also true that in a triangle, if you know that the

triangle is a right triangle. This can be used to check that the two edges meet atright angles.

What should the diagonal of the string layout measure if everything is square?

You measure one diagonal of the string layout as 17 feet, Should anyadjustments be made to the string layout? Explain.

2 1532

inches.

a2 + b2 = c2 ,a2 + b2 = c2 .



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RATIONAL EXPONENTS SECTION 8.7 523

8.7 R AT I O N A L E X P O N E N T S

O b j e c t i v e s

Evaluate exponential expressions of the form



Use rules for exponents to simplify expressions containing fractional exponents.

1 Radical notation is widely used, as we’ve seen. In this section, we study analternate notation, one that proves to be more efficient and compact. This alternatenotation makes use of expressions containing an exponent that is a rational numberbut not necessarily an integer, for example,

In giving meaning to rational exponents, keep in mind that we want the rules for oper-ating with them to be the same as the rules for operating with integer exponents. Forthis to be true,

Also, we know that

Since the square of both and is 3, it would be reasonable to say that

In general, we have the following.

31/2 means 23

2331/2

A23 B2 = 3

131/222 = 31/2 #2 = 31 = 3

31/2 , 2-3/4 , and y5/6

4

a-m/n .3

am/n .2

a1/n .1

Definition of If n is a positive integer and is a real number, then

a1/n = 2n a

2n a

a1/n

Notice that the denominator of the rational exponent is the same as the index of thecorresponding radical.

Write in radical notation. Then simplify.

a. b. c. d. e. a19b1/21-2721/3-161/481/3251/2

E X A M P L E 1




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CLASSROOM EXAMPLE

Write in radical notation. Then simplify.

a. b. c.

d. e.

answer:

a. 10 b. 4 c.d. e. 2/5-2

-3

a 425b1/21-821/3

-811/4641/31001/2

Solut ion

Definition of If m and n are integers with and if a is a positive number, then

Also,

am/n = 1am21/n = 2n am

am/n = 1a1/n2m = A2n a Bmn 7 0

a m/n


a. b. c.

a. b.

c. The negative sign is not affected by the exponent since the base of the exponent is

16. Thus, -163/4 = -1161/423 = - A24 16 B3 = -23 = -8.

272/3 = 1271/322 = A23 27 B2 = 32 = 943/2 = 141/223 = A24 B3 = 23 = 8

-163/4272/343/2

CLASSROOM EXAMPLE

Simplify.

a. b. c.answer:

a. 64 b. 8 c. -9

-272/3163/4163/2

E X A M P L E 2

Solut ion

H e l p f u l H i n tRecall that

and

In other words, without parentheses the exponent 2 applies to the base 3, notThe same is true of rational exponents. For example,

and

1-2721/3 = 23 -27 = -3

-161/2 = -216 = -4

-3.

1-322 = 1-321-32 = 9

-32 = -13 # 32 = -9

a.

b.

c. In the base of the exponent is 16. Thus the negative sign is not affected by

the exponent; so

d. The parentheses show that is the base.

e.

2 In Example 1, each rational exponent has a numerator of 1.What happens if thenumerator is some other positive integer? Consider Since is the same wereason that

The denominator 3 of the rational exponent is the same as the index of the rad-ical. The numerator 2 of the fractional exponent indicates that the radical base is tobe squared.

82/3 = 8A1/322 = 181/322 = 123 8)2 = 22 = 4

13# 2,2

382/3 .

a19b1/2

= A19

=13

1-2721/3 = 23 -27 = -3.-27

-161/4 = -24 16 = -2.

-161/4 ,

81/3 = 23 8 = 2

251/2 = 225 = 5



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RATIONAL EXPONENTS SECTION 8.7 525

3 If the exponent is a negative rational number, use the following definition.

Write each expression with a positive exponent and then simplify.

a. b. c. d.

a.

b.

c.

d.

4 It can be shown that the properties of integer exponents hold for rationalexponents. By using these properties and definitions, we can now simplify products andquotients of expressions containing rational exponents.

32-4/5 =1

324/5 =1

A25 32 B4 =1

24 =1

16

-91/2 = -29 = -3

16-3/4 =1

163/4 =1

A24 16 B3 =1

23 =18

36-1/2 =1

361/2 =1236

=16

32-4/5-91/216-3/436-1/2

E X A M P L E 3

Definition of If is a nonzero real number, then

a-m/n =1

am/n

a-m/n

a-m/n

Solut ion

CLASSROOM EXAMPLE

Write with a positive exponent, then simplify.

a. b.

c. d.answer:

a. b.

c. d.1

81-4

127

17

243-4/5-161/2

81-3/449-1/2

Simplify each expression. Write results with positive exponents only. Assume that allvariables represent positive numbers.

a. b. c. d. e.

a.

b.

c.

d.

e. ay3/5

z1/4 b2

=y13/522z11/422 =

y6/5

z1/2

x1/5

x-4/5 = x11/52-1-4/52 = x5/5 = x1 or x

1x1/4212 = x11/4212 = x3

51/3

52/3 = 511/32-12/32 = 5-1/3 =1

51/3

31/2 # 33/2 = 311/22+13/22 = 34/2 = 32 = 9

ay3/5

z1/4 b2x1/5

x-4/51x1/421251/3

52/331/2 # 33/2

Solut ion

E X A M P L E 4

CLASSROOM EXAMPLE

Simplify.

a. b. c.

d. e.

answer:

a. 25 b. c. d. e.a2

b1/2x2y31

131/2

aa2/3

b1/6b

3x2/3

x-4/3

1y1/5215131/4

133/453/4 # 55/4




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CLASSROOM EXAMPLESolveanswer: -7 and 7

x2 - 49 = 0.

TEACHING TIPConsider pointing out the similari-ties and differences in structure of

andx2 - 4 = 0.x - 4 = 0, 2x - 4 = 0,

538 CHAPTER 9 SOLVING QUADRATIC EQUATIONS

E X A M P L E 1

Use the square root property to solve

First we solve for by adding 9 to both sides.


Next we use the square root property.

x = 3 x = -3

x = 29 or x = -29

x2 = 9

x2 - 9 = 0

x2

x2 - 9 = 0.

Square Root PropertyIf for then

x = 2a or x = -2a

a Ú 0,x2 = a

O b j e c t i v e s

Use the square root property to solve quadratic equations.

Solve problems modeled by quadratic equations.

1 Recall that a quadratic equation is an equation that can be written in the form

where a, b, and c are real numbers and To solve quadratic equations by factoring, use the zero factor theorem: If the

product of two numbers is zero, then at least one of the two numbers is zero. For ex-ample, to solve we first factor the left side of the equation and then seteach factor equal to 0.

Factor.

Apply the zero factor theorem.

Solve each equation.

The solutions are and 2.Now let’s solve another way. First, add 4 to both sides of the equation.

Add 4 both sides.

Now we see that the value for x must be a number whose square is 4. Thereforeor This reasoning is an example of the square root

property.x = -24 = -2.x = 24 = 2

x2 = 4

x2 - 4 = 0

x2 - 4 = 0-2

x = -2 or x = 2

x + 2 = 0 or x - 2 = 0

1x + 221x - 22 = 0

x2 - 4 = 0

x2 - 4 = 0,

a Z 0.

ax2 + bx + c = 0

2

1

Solut ion

S O LV I N G Q UA D R AT I C E Q UAT I O N S B Y T H E S Q UA R E R O OT M E T H O D9.1




CLASSROOM EXAMPLESolve answer: -3 and 11

1x - 422 = 49.


answer:233

3 and -

2333

3x2 = 11.

SOLVING QUADRATIC EQUATIONS BY THE SQUARE ROOT METHOD SECTION 9.1 539

Check: Original equation Original equation

Let Let

True True

The solutions are 3 and -3.

0 = 0 0 = 0

x 3 . 1-322 - 9 0x 3 . 32 - 9 0

x2 - 9 = 0 x2 - 9 = 0

E X A M P L E 2


First we solve for by dividing both sides by 2. Then we use the square root property.


Use the square root property.

If the denominators are rationalized, we have

Rationalize the denominator.

Simplify.

Remember to check both solutions in the original equation. The solutions are

and - 214

2.

2142

x =214

2 x = -

2142

x =27 # 2222 # 22

or x = - 27 # 2222 # 22

x = A72 or x = -A7

2

x2 =72

2x2 = 7

x2

2x2 = 7.

E X A M P L E 3


Instead of here we have But the square root property can still be used.


Write as 4 and as

Solve.

Check

Original equation Original equation

Let Let

Simplify. Simplify.

True True

Both 7 and are solutions.-1

16 = 16 16 = 16

1-422 16 42 16

x 1 . 1-1 - 322 16x 7 . 17 - 322 16

1x - 322 = 16 1x - 322 = 16

x = 7 x = -1

4 .216216 x - 3 = 4 x - 3 = -4

x - 3 = 116 or x - 3 = -116

1x - 322 = 16

1x - 322 .x2 ,

1x - 322 = 16.

Solut ion

Solut ion


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answer:-1 ; 215

4

14x + 122 = 15.

Solut ion

CLASSROOM EXAMPLESolveanswer: no real solution

1x + 322 = -5.

Solut ion

CLASSROOM EXAMPLESolveanswer: 5 ; 322

1x - 522 = 18.

Solut ion


E X A M P L E 4




Solve for x.

Check both solutions in the original equation. The solutions are andThis can be written compactly as The notation is read

as “plus or minus.”;-1 ; 222.-1 - 222.

-1 + 222

x = -1 + 222 x = -1 - 222

x + 1 = 222 x + 1 = -222

x + 1 = 28 or x + 1 = -28

1x + 122 = 8

1x + 122 = 8.

H e l p f u l H i n tread “plus or minus”

The notation for example, is just

a shorthand notation for both

and -1 - 25.

-1 + 25

-1 ; 25,

E X A M P L E 5


This equation has no real solution because the square root of is not a realnumber.

-2

1x - 122 = -2.

E X A M P L E 6





Check both solutions in the original equation. The solutions are and

which can be written as 2 ; 210

5.

2 - 2105

,

2 + 2105

x =2 + 210

5 x =

2 - 2105

5x = 2 + 210 5x = 2 - 210

5x - 2 = 210 or 5x - 2 = -210

15x - 222 = 10

15x - 222 = 10.

ø



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SOLVING QUADRATIC EQUATIONS BY THE SQUARE ROOT METHOD SECTION 9.1 541

H e l p f u l H i n tFor some applications and graphing purposes, decimal approximations ofexact solutions to quadratic equations may be desired.

Exact Solutions from Example 6 Decimal Approximations

1.032

-0.232L2 - 210

5

L2 + 210

5

2 Many real-world applications are modeled by quadratic equations.

CLASSROOM EXAMPLEUse the formula to find how longit takes a free-falling body to fall 650 feet.answer: 6.4 sec

h = 16t2

E X A M P L E 7

FINDING THE LENGTH OF TIME OF A DIVE

The record for the highest dive into a lake was made by Harry Froboess of Switzerland. In1936 he dove 394 feet from the airship Hindenburg into Lake Constance. To the nearesttenth of a second, how long did his dive take? (Source:The Guiness Book of Records)

1. UNDERSTAND. To approximate the time of the dive, we use the formula* where t is time in seconds and h is the distance in feet, traveled by a free-

falling body or object. For example, to find the distance traveled in 1 second, or 3seconds, we let and then

Since a body travels 144 feet in 3 seconds, we now know the dive of 394 feet lastedlonger than 3 seconds.

2. TRANSLATE. Use the formula let the distance and wehave the equation

3. SOLVE. To solve for t, we will use the square root property.


Simplify.


Approximate.

4. INTERPRET.

Check: We reject the solution since the length of the dive is not a negativenumber.

State: The dive lasted approximately 5 seconds.

-5.0

5.0 L t or -5.0 L t

224.625 = t or -224.625 = t

24.625 = t2

39416

= t2

394 = 16t2

394 = 16t2

394 = 16t2 .h = 394,h = 16t2 ,

If t = 3, h = 161322 = 16 # 9 = 144 feet If t = 1, h = 161122 = 16 # 1 = 16 feet

t = 3.t = 1

h = 16t2

*The formula does not take into account air resistance.h = 16t2

Solut ion


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S O LV I N G Q UA D R AT I C E Q UAT I O N S B Y C O M P L E T I N G T H E S Q UA R E9.2

O b j e c t i v e s

Find perfect square trinomials.

Solve quadratic equations by completing the square.

1 In the last section, we used the square root property to solve equations such as

1x + 122 = 8 and 15x - 222 = 3

2

1

SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE SECTION 9.2 543




Completing The Square

To complete the square on add To find find half the coef-

ficient of x, then square the result.

ab

2b2

,ab

2b2

.x2 + bx ,

Notice that one side of each equation is a quantity squared and that the other side is aconstant. To solve

notice that if we add 1 to both sides of the equation, the left side is a perfect square tri-nomial that can be factored.


Factor.

Now we can solve this equation as we did in the previous section by using the squareroot property.


Solve.

The solutions are Adding a number to to form a perfect square trinomial is called

completing the square on In general, we have the following.

x2 + 2x .x2 + 2x

-1 ; 25.

x = -1 + 25 x = -1 - 25

x + 1 = 25 or x + 1 = -25

1x + 122 = 5

x2 + 2x + 1 = 4 + 1

x2 + 2x = 4

E X A M P L E 1

Solut ion

Complete the square for each expression and then factor the resulting perfect squaretrinomial.

a. b. c.

a. The coefficient of the x-term is 10. Half of 10 is 5, and Add 25.

b. Half the coefficient of m is and is 9. Add 9.

c. Half the coefficient of x is and Add

2 By completing the square, a quadratic equation can be solved using the squareroot property.

x2 + x +14

= ax +12b2

14

.a12b2

=14

.12

m2 - 6m + 9 = 1m - 3221-322-3,

x2 + 10x + 25 = 1x + 52252 = 25.

x2 + xm2 - 6mx2 + 10x

CLASSROOM EXAMPLEComplete the square and then factor.

a. b. c.answer:

a.

b.

c. z2 + 3z + 94 = Az + 3

2 B2y2 - 14y + 49 = 1y - 722x2 + 16x + 64 = 1x + 822

z2 + 3zy2 - 14yx2 + 16x



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Solut ion

CLASSROOM EXAMPLESolve answer: -4 ; 215

x2 + 8x + 1 = 0.

H e l p f u l H i n tRemember, when completing the square, add the number that completes thesquare to both sides of the equation. In Example 2, we added 9 to both sides tocomplete the square.

E X A M P L E 2

Solve by completing the square.

First we get the variable terms alone by subtracting 3 from both sides of the equation.


Next we find half the coefficient of the x-term, then square it. Add this result to bothsides of the equation.This will make the left side a perfect square trinomial.The coeffi-cient of x is 6, and half of 6 is 3. So we add or 9 to both sides.

Complete the square.

Factor the trinomial



Check by substituting and in the original equation. The solutionsare -3 ; 26.

-3 - 26-3 + 26

x = -3 + 26 x = -3 - 26

x + 3 = 26 or x + 3 = -26

x2 6x 9 . 1x + 322 = 6

x2 + 6x + 9 = -3 + 9

32

x2 + 6x = -3

x2 + 6x + 3 = 0

x2 + 6x + 3 = 0

E X A M P L E 3


The variable terms are already alone on one side of the equation.The coefficient of x isHalf of is and So we add 25 to both sides.

Factor the trinomial and simplify



The solutions are

The method of completing the square can be used to solve any quadratic equa-tion whether the coefficient of the squared variable is 1 or not. When the coefficientof the squared variable is not 1, we first divide both sides of the equation by the co-efficient of the squared variable so that the coefficient is 1. Then we complete thesquare.

5 ; 211.

x = 5 + 211 x = 5 - 211

x - 5 = 211 or x - 5 = -211

14 25 . 1x - 522 = 11

x2 - 10x + 25 = -14 + 25

x2 - 10x = -14

1-522 = 25.-5,-10-10.

x2 - 10x = -14

Solut ion

H e l p f u l H i n tAdd 25 to both sides ofthe equation.

CLASSROOM EXAMPLESolve answer: 7 ; 217

x2 + 14x = -32.



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Solut ion

CLASSROOM EXAMPLESolveanswer: -

12 , and 92

4x2 - 16x - 9 = 0.

Solut ion

Solving a Quadratic Equation in x by Completing the Square

Step 1: If the coefficient of is 1, go to Step 2. If not, divide both sides of theequation by the coefficient of

Step 2: Get all terms with variables on one side of the equation and constantson the other side.

Step 3: Find half the coefficient of x and then square the result. Add this num-ber to both sides of the equation.

Step 4: Factor the resulting perfect square trinomial.

Step 5: Use the square root property to solve the equation.

x2 .x2

E X A M P L E 4



Get the variable terms alone on one side of the equation.

The coefficient of x is Half of is and So we add 1 to both sides.

Factor and simplify


Add 1 to both sides and simplify the radical.

Simplify.

Both and are solutions.

The following steps may be used to solve a quadratic equation in x by completingthe square.

- 12

52

x =52 x = -

12

x = 1 +32 x = 1 -

32

x - 1 = A94 or x - 1 = -A9

4

54

1 .x2 2x 1 1x - 122 =94

x2 - 2x + 1 =54

+ 1

1-122 = 1.-1,-2-2.

x2 - 2x =54

x2 - 2x -54

= 0

4x2 - 8x - 5 = 0

4x2 - 8x - 5 = 0

E X A M P L E 5


The coefficient of is not 1. We divide both sides by 2, the coefficient of


Add or to both sides.

Factor the left side and simplify the right.

There is no real solution to this equation since the square root of a negative number isnot a real number.

ax +32b2

= - 54

94

a32b2

x2 + 3x +94

= - 72

+94

x2 + 3x = - 72

2x2 + 6x = -7

x2 .x2

2x2 + 6x = -7


2x2 + 10x = -13.


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Solut ion

E X A M P L E 6


answer: - 3 ; 2174

2x2 = -3x + 1.


First we divide both sides of the equation by 2, the coefficient of


Next we get the variable terms alone by subtracting 5x from both sides.

Add or to both sides.

Factor the left side and simplify theright side.


Simplify.

The solutions are 5 ; 323

2.

x =52

+323

2 x =

52

-323

2

x -52

=323

2 x -

52

= - 323

2

x -52

= A274 or x -

52

= -A274

ax -52b2

=274

254

a 52b2

x2 - 5x +254

=12

+254

x2 - 5x =12

x2 = 5x +12

2x2 = 10x + 1

x2 .

2x2 = 10x + 1

Determine the number to add to make each expression a perfect square trinomial. See Example 1.

1. 16 2. 9 3. 1004. 81 5. 49 6. 1y2 + 2yy2 + 14yx2 + 18x

x2 + 20xp2 + 6pp2 + 8p

MENTAL MATH

13. 14. 15. 19. 21. -4, -11 ; 2-3 ; 34-3 ; 3 2-1 ; 6


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O b j e c t i v e s

Use the quadratic formula to solve quadratic equations.

Determine the number of solutions of a quadratic equation by using the discriminant.

1 We can use the technique of completing the square to develop a formula tofind solutions of any quadratic equation. We develop and use the quadratic formula inthis section.

Recall that a quadratic equation in standard form is

To develop the quadratic formula, let’s complete the square for this quadratic equationin standard form.

First we divide both sides of the equation by the coefficient of and then get thevariable terms alone on one side of the equation.

x2

ax2 + bx + c = 0, a Z 0

2

1

SOLVING QUADRAT IC EQUAT IONS BY THE QUADRAT IC FORMULA9.3



SOLVING QUADRATIC EQUATIONS BY THE QUADRATIC FORMULA SECTION 9.3 549

Quadratic FormulaIf a, b, and c are real numbers and a quadratic equation written in theform has solutions

x =-b ; 2b2 - 4ac

2a

ax2 + bx + c = 0a Z 0,

H e l p f u l H i n tDon’t forget that to correctly identify a, b, and c in the quadratic formula, youshould write the equation in standard form.

Quadratic Equations in Standard Form

22x2 + 25x + 23 = 0 a = 22, b = 25, c = 23

x2 + x = 0 a = 1, b = 1, c = 0

4y2 - 9 = 0 a = 4, b = 0, c = -9

5x2 - 6x + 2 = 0 a = 5, b = -6, c = 2

Divide by a; recall that a cannot be 0.

Get the variable terms alone on one side of the equation.

The coefficient of x is Half of is and So we add to both sides

of the equation.

Add to both sides.

Factor the left side.

Multiply by so that both terms on

the have a right side common denominator.

Simplify the right side.

Now we use the square root property.

Use the square rootproperty.


Subtract from both

sides.

Simplify.

The solutions are This final equation is called the quadratic formula

and gives the solutions of any quadratic equation.

-b ; 2b2 - 4ac

2a.

x =-b + 2b2 - 4ac

2a x =

-b - 2b2 - 4ac

2a

b2a x = -

b

2a+2b2 - 4ac

2a x = -

b

2a-2b2 - 4ac

2a

x +b

2a=2b2 - 4ac

2a x +

b

2a= -

2b2 - 4ac

2a

x +b

2a= Bb2 - 4ac

4a2 or x +b

2a= -Bb2 - 4ac

4a2

ax +b

2ab2

=b2 - 4ac

4a2

4a

4a

c

a ax +b

2ab2

= - 4ac

4a2 +b2

4a2

ax +b

2ab2

= - ca

+b2

4a2

b2

4a2 x2 +ba

x +b2

4a2 = - ca

+b2

4a2

b2

4a2a b

2ab2

=b2

4a2 .b

2a

ba

ba

.

x2 +ba

x = - ca

x2 +ba

x +ca

= 0



Solut ion

Solut ion

H e l p f u l H i n tNotice that the fractionbar is under the entirenumerator of

-b ; 2b2 - 4ac .

Solve using the quadratic formula.

This equation is in standard form with and By the quadraticformula, we have

Let and

Simplify.

Check both solutions in the original equation. The solutions are and

-1 - 2376

.

-1 + 2376

=-1 ; 237

6

=-1 ; 21 + 36

6

c 3 .a 3 , b 1 , x =-1 ; 212 - 4 # 3 # 1-32

2 # 3

x =-b ; 2b2 - 4ac

2a

c = -3.a = 3, b = 1,

3x2 + x - 3 = 0

E X A M P L E 2


First we write the equation in standard form by subtracting 5 from both sides.

Next we note that and We substitute these values intothe quadratic formula.

Substitute in the formula.

Simplify.

Then,

Check and 5 in the original equation. Both and 5 are solutions.

The following steps may be useful when solving a quadratic equation by the qua-dratic formula.

- 12

- 12

x =9 - 11

4= -

12 or x =

9 + 114

= 5

=9 ; 2121

4=

9 ; 114

=9 ; 281 + 40

4

x =-1-92 ; 21-922 - 4 # 2 # 1-52

2 # 2

x =-b ; 2b2 - 4ac

2a

c = -5.a = 2, b = -9,

2x2 - 9x - 5 = 0

2x2 - 9x = 5

2x2 - 9x = 5

E X A M P L E 1


answer:1 ; 241

4

2x2 - x - 5 = 0.


answer: -3 and 13

3x2 + 8x = 3.



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Concept Check Answer:d

E X A M P L E 3


First we write the equation in standard form by subtracting 1 from both sides.

Next we replace a, b, and c with the identified values:


Simplify.

The solutions are and

Notice that the equation in Example 3, could have been easily solved bydividing both sides by 7 and then using the square root property. We solved the equa-tion by the quadratic formula to show that this formula can be used to solve any qua-dratic equation.

7x2 = 1,

- 277

.277

= ;277

=;227

14

=;228

14

x =0 ; 202 - 4 # 7 # 1-12

2 # 7

a = 7, b = 0 and c = -1.

7x2 - 1 = 0

7x2 = 1

7x2 = 1

Solut ion

Solving a Quadratic Equation by the Quadratic Formula

Step 1: Write the quadratic equation in standard form:

Step 2: If necessary, clear the equation of fractions to simplify calculations.

Step 3: Identify a, b, and c.

Step 4: Replace a, b, and c in the quadratic formula with the identified values,and simplify.

ax2 + bx + c = 0.

CONCEPT CHECKFor the quadratic equation if and in the quadratic formula, the value of bis which of the following?

a.

b. 7

c.d. -7

-5

72

c = -5a = 22x2 - 5 = 7x ,


answer: ;210

5

5x2 = 2.

E X A M P L E 4


First we write the equation in standard form.

x2 + x + 1 = 0

x2 = -x - 1

Solut ion


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x2 = -2x - 3.

CLASSROOM EXAMPLE

Solve

answer:3 ; 221

2

13

x2 - x = 1.

Solve by using the quadratic formula.

We write the equation in standard form and then clear the equation of fractions bymultiplying both sides by the LCD, 2.

Write in standard form.


Here, and so we substitute these values into the quadraticformula.

Simplify.

Factor and simplify.

The solutions are and 1 + 25.1 - 25

=211 ; 252

2= 1 ; 25

=2 ; 220

2=

2 ; 2252

x =-1-22 ; 21-222 - 4 # 1 # 1-42

2 # 1

c = -4,a = 1, b = -2,

x2 - 2x - 4 = 0

12

x2 - x - 2 = 0

12

x2 - x = 2

12

x2 - x = 2

Next we replace a, b, and c in the quadratic formula with and


Simplify.

There is no real number solution because is not a real number.2-3

=-1 ; 2-3

2

x =-1 ; 212 - 4 # 1 # 1

2 # 1

c = 1.a = 1, b = 1,

E X A M P L E 5

2 In the quadratic formula, the radicand is

called the discriminant because, by knowing its value, we can discriminate among thepossible number and type of solutions of a quadratic equation. Possible values of thediscriminant and their meanings are summarized next.

b2 - 4acx =-b ; 2b2 - 4ac

2a,

Solut ion

H e l p f u l H i n tWhen simplifying expressions such as

first factor out a common factor from the terms of the numerator and thensimplify.

3 ; 6226

=311 ; 2222

2 # 3=

1 ; 2222

3 ; 6226


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E X A M P L E 6

Use the discriminant to determine the number of solutions of

In and Then

Since the discriminant is 37, a positive number, this equation has two distinct real solutions.

We solved this equation in Example 1 of this section, and the solutions are

and two distinct real solutions.-1 - 237

6,

-1 + 2376

b2 - 4ac = 1122 - 41321-32 = 1 + 36 = 37

c = -3.3x2 + x - 3 = 0, a = 3, b = 1,

3x2 + x - 3 = 0.

DiscriminantThe following table corresponds the discriminant of a quadratic equa-tion of the form with the number of solutions of the equation.ax2 + bx + c = 0

b2 - 4ac

*In this case, the quadratic equation will have two complex (but not real)solutions. See Section 9.4 for a discussion of complex numbers.

E X A M P L E 7

Use the discriminant to determine the number of solutions of each quadratic equation.

a. b.

a. In and

Since the discriminant is 0, this equation has one real solution.

b. In and

Since the discriminant is a negative number, this equation has no real solution.-80,

b2 - 4ac = 02 - 4152142 = 0 - 80 = -80

c = 4.5x2 + 4 = 0, a = 5, b = 0,

b2 - 4ac = 1-622 - 4112192 = 36 - 36 = 0

c = 9.x2 - 6x + 9 = 0, a = 1, b = -6,

5x2 + 4 = 0x2 - 6x + 9 = 0

Spotlight on

DECISIONMAKING

Suppose you are an engineering technician in a manufacturing plant. The engi-neering department has been asked to upgrade any production lines that produceless than 3000 items per hour. Production records show the following:

Production Lines A and B

Lines A and B together: produce 3000 items in 32 minutesLine A alone: takes 11.2 minutes longer than Line B to pro-

duce 3000 items

Decide whether either Line A or Line B should be upgraded. Explain your reasoning.

CLASSROOM EXAMPLEUse the discriminant to determine the num-ber of solutions.

a. b.

answer:

a. no real solution b. one real solution

x2 + 2x + 1 = 0x2 + 2x + 2 = 0

NUMBER OF SOLUTIONS

Positive Two distinct real solutionsZero One real solutionNegative No real solution*

b2 - 4ac

Solut ion

CLASSROOM EXAMPLEUse the discriminant to determine the num-ber of solutions of

answer: two distinct real solutions

5x2 + 2x - 3 = 0

Solut ion



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MENTAL MATHIdentify the value of a, b, and c in each quadratic equation.

1. 2.

3. 4.

5. 6. a = 9, b = 0, c = -49x2 - 4 = 0a = 1, b = 0, c = -6x2 - 6 = 0

a = 1, b = 3, c = -7x2 + 3x - 7 = 0a = 10, b = -13, c = -210x2 - 13x - 2 = 0

a = 5, b = -7, c = 15x2 - 7x + 1 = 0a = 2, b = 5, c = 32x2 + 5x + 3 = 0

3. 4. 9. 10. 13. no real solution 14. no real solution-3 ; 237

14-7 ; 237

6-7 ; 241

4-5 ; 217

2

30. 33. 34. 35. 36. 41. no real solution 42. no real solution 43.3 ; 213

44 ; 26

53 ; 23

25 ; 263 ; 27

9 ; 21776

3 ; 137 ; 1293 1

25.5 ; 233

2

29.-9 ; 2129

12


COMPLEX SOLUTIONS OF QUADRATIC EQUATIONS SECTION 9.4 559

Imaginary Unit iThe imaginary unit, written i, is the number whose square is That is,

i2 = -1 and i = 2-1

-1.

9.4 C O M P L E X S O L U T I O N S O F Q UA D R AT I C E Q UAT I O N S

O b j e c t i v e s

Write complex numbers using i notation.

Add and subtract complex numbers.

Multiply complex numbers.

Divide complex numbers.

Solve quadratic equations that have complex solutions.

In Chapter 8, we learned that for example, is not a real number because there is noreal number whose square is However, our real number system can be extended to in-clude numbers like This extended number system is called the complex number sys-tem.The complex number system includes the imaginary unit i, which is defined next.

2-4.-4.2-4,

5

4

3

2

1

1 We use i to write numbers like as the product of a real number and i.Since we have 2-6 = 2-1 # 6 = 2-1 # 26 = i26

i = 2-1,2-6

E X A M P L E 1

Solut ion

Write each radical as the product of a real number and i.

a. b. c.

Write each negative radicand as a product of a positive number and Then writeas i.

a.

b.

c.

The numbers and are called pure imaginary numbers. Both realnumbers and pure imaginary numbers are complex numbers.

2i252i , i211,

2-20 = 2-1 # 20 = 2-1 # 220 = i # 225 = 2i25

2-11 = 2-1 # 11 = 2-1 # 211 = i211

2-4 = 2-1 # 4 = 2-1 # 24 = i # 2 = 2i

2-1-1.

2-202-112-4

˜CLASSROOM EXAMPLEWrite each radical as the product of a realnumber and i.

a. b. c.

answer:

a. 9i b. c. 4i25i213

2-802-132-81

Complex Numbers and Pure Imaginary NumbersA complex number is a number that can be written in the form

where a and b are real numbers.A complex number that can be written in the form

is also called a pure imaginary number.b Z 0,

0 + bi

a + bi




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A complex number written in the form is in standard form. We call a thereal part and bi the imaginary part of the complex number a + bi .

a + bi

E X A M P L E 2

Solut ion

Identify each number as a complex number by writing it in standard form

a. 7 b. 0 c. d. e.

a. 7 is a complex number since

b. 0 is a complex number since

c. is a complex number since

d. is a complex number since

e. is a complex number since

2 We now present arithmetic operations—addition, subtraction, multiplication,and division—for the complex number system. Complex numbers are added andsubtracted in the same way as we add and subtract polynomials.

2 + 2-4 = 2 + 2i .2 + 2-4

2-27 = i # 323 = 0 + 3i23.2-27

220 = 225 = 225 + 0i .220

0 = 0 + 0i .

7 = 7 + 0i .

2 + 2-42-27220

a + bi .

CLASSROOM EXAMPLE

Write each in the form

a. b. c. 0 d. e. i

answer:

a. b. c.

d. e. ΩΩ0 + 1i-1 + i27

0 + 0i25 + 0i-3 + 0i

-1 + 2-725-3

a + bi .

E X A M P L E 3

Solut ion

Simplify the sum or difference. Write the result in standard form.

a. b. c.

Add the real parts and then add the imaginary parts.

a.

b.

c. 15 - i2 - 4 = 15 - 42 - i = 1 - i

- i + 13 + 7i2 = 3 + 1- i + 7i2 = 3 + 6i

12 + 3i2 + 1-6 - i2 = [2 + 1-62] + 13i - i2 = -4 + 2i

15 - i2 - 4- i + 13 + 7i212 + 3i2 + 1-6 - i2

CLASSROOM EXAMPLEAdd or subtract.

a.

b.

answer:

a. b. 5 - 2i3 - 2i

17 + 6i2 - 12 + 8i21-1 + i2 + 14 - 3i2

E X A M P L E 4

Solut ion

Subtract from

3 Use the distributive property and the FOIL method to multiply complexnumbers.

11 + i2 - 111 - i2 = 1 + i - 11 + i = 11 - 112 + 1i + i2 = -10 + 2i

11 + i2 .111 - i2

CLASSROOM EXAMPLE

Subtract from 9.

answer: 17 - 2i

-8 + 2i

E X A M P L E 5

Find the following products and write in standard form.

a. b. c. 12 + 3i212 - 3i217 - 3i214 + 2i25i12 - i2






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Solut ion

CLASSROOM EXAMPLEMultiply. Write the product in standard form.

a. b.

c.

answer:

a. b. c. 13017 + 28i14 + 2i

19 - 7i219 + 7i212 + 5i216 - i22i11 - 7i2

E X A M P L E 6


To write this quotient as a complex number in the standard form we need tofind an equivalent fraction whose denominator is a real number. By multiplying bothnumerator and denominator by the denominator’s conjugate, we obtain a new fractionthat is an equivalent fraction with a real number denominator.

a + bi ,

4 + i

3 - 4i

Solut ion

a. By the distributive property, we have


Write as


b.

Write as

c.

Write as

The product in part (c) is the real number 13. Notice that one factor is the sum of2 and 3i, and the other factor is the difference of 2 and 3i. When complex number fac-tors are related as these two are, their product is a real number. In general,

The complex numbers and are called complex conjugates of eachother. For example, is the conjugate of and is the conjugate of

Also,

The conjugate of is The conjugate of 5 is 5. (Note that and its conjugate is )The conjugate of 4i is ( is the conjugate of )

4 The fact that the product of a complex number and its conjugate is a realnumber provides a method for dividing by a complex number and for simplifyingfractions whose denominators are complex numbers.

0 + 4i .0 - 4i-4i .5 - 0i = 5.5 = 5 + 0i

3 + 10i .3 - 10i

2 - 3i .2 + 3i2 + 3i ,2 - 3i

a - bia + bi

1a + bi21a - bi2 = a2 + b2

sum difference real number

= 13

1 .i2 = 4 - 91-12 12 + 3i212 - 3i2 = 4 - 6i + 6i - 9i2

= 34 + 2i

= 28 + 2i + 6

1 .i2 = 28 + 2i - 61-12

17 - 3i214 + 2i2 =

F O I L28 + 14i - 12i - 6i2

= 5 + 10i

= 10i + 5

1 .i2 = 10i - 51-12 = 10i - 5i2

5i12 - i2 = 5i # 2 - 5i # i



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Multiply numerator and denominator by

Recall that


Note that our last step was to write in standard form where a and b arereal numbers.

5 Some quadratic equations have complex solutions.

a + bi ,4 + i

3 - 4i

=8

25+

1925

i

=12 + 19i - 4

9 + 16=

8 + 19i

25

i2 1 . =12 + 19i + 41-12

9 - 161-12

=12 + 16i + 3i + 4i2

9 - 16i2

3 4i . 4 + i

3 - 4i=14 + i213 - 4i2 # 13 + 4i2

13 + 4i2CLASSROOM EXAMPLE


answer: - 313

+1113

i

-3 + i

2 + 3i

E X A M P L E 7

Solut ion

Solve

Begin by applying the square root property.

Apply the square root property.

Write as 5i.

The solutions are and -2 - 5i .-2 + 5i

x = -2 ; 5i

225 x + 2 = ;5i

x + 2 = ;2-25

1x + 222 = -25

1x + 222 = -25.

CLASSROOM EXAMPLE

Solve

answer: x = 1 ; 3i

1x - 122 = -9.

E X A M P L E 8

Solut ion

Solve

Write the equation in standard form and use the quadratic formula to solve.

Write the equation in standard form.

Apply the quadratic formula with and

Write as 2i.

The solutions are and 2 + i .2 - i

=212 ; i2

2= 2 ; i

24 =4 ; 2i

2

=4 ; 2-4

2

m =4 ; 216 - 4 # 1 # 5

2 # 1

c = 5.a = 1, b = -4,

m2 - 4m + 5 = 0

m2 = 4m - 5

m2 = 4m - 5.

CLASSROOM EXAMPLE

Solve N

answer: 1 ; i22

x2 = 2x - 3.



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E X A M P L E 9

Solut ion

9. 10.11. 12.13. 14. 15.16. 19.20. 29.

30. 31.3 ; 2i23

22 ; 5i

-1 ; 3i26 + 2i26 - 2i-7 - 2i

9 - 4i-91 - 3i7 + 9i-12 + 6i

-2 - i-3 + 9i

Solve


Apply the quadratic formula with and

The solutions are and -1 + i23

2.

-1 - i232

=-1 ; i23

2

=-1 ; 2-3

2

c 1 .

a 1 , b 1 , x =

-1 ; 21 - 4 # 1 # 12 # 1

x2 + x + 1 = 0

x2 + x = -1

x2 + x = -1.

CLASSROOM EXAMPLE

Solve

answer:-1 ; i27

2

x2 + x = -2.

32. 33. 34. 35.-7 ; i215

81 ; 2i-3 ; 2i

-5 ; 3i223

38. 39. 40.

41. 43. -1 + 3i45 + 63i

-14 + 8i15 - 7i3 ; i226

5



9.5 G R A P H I N G Q UA D R AT I C E Q UAT I O N S

O b j e c t i v e s

Graph quadratic equations of the form

Find the intercepts of a parabola.

Determine the vertex of a parabola.

1 Recall from Section 3.2 that the graph of a linear equation in two variablesis a straight line. Also recall from Section 6.5 that the graph of aAx + By = C

3

2

y = ax2 + bx + c .1



GRAPHING QUADRATIC EQUATIONS SECTION 9.5 565

y x2

(3, 9)

(2, 4)

(1, 1)

89

76

11

12345

2 3 4 512345

(2, 4)

(1, 1)

(3, 9)

Vertex (0, 0)x

y

quadratic equation in two variables is a parabola. In this section, wefurther investigate the graph of a quadratic equation.

To graph the quadratic equation select a few values for x and find thecorresponding y-values. Make a table of values to keep track. Then plot the points cor-responding to these solutions.

y = x2 ,

y = ax2 + bx + c

Clearly, these points are not on one straight line. As we saw in Chapter 6, thegraph of is a smooth curve through the plotted points. This curve is called aparabola. The lowest point on a parabola opening upward is called the vertex. The ver-tex is (0, 0) for the parabola If we fold the graph paper along the y-axis, the twopieces of the parabola match perfectly. For this reason, we say the graph is symmetricabout the y-axis, and we call the y-axis the axis of symmetry.

Notice that the parabola that corresponds to the equation opens upward.Thishappens when the coefficient of is positive. In the equation the coefficient of is 1. Example 1 shows the graph of a quadratic equation whose coefficient of is negative.x2

x2y = x2 ,x2y = x2

y = x2 .

y = x2

x y

9

4

1

0 0

1 1

2 4

3 9

-1

-2

-3

E X A M P L E 1

Solut ion

Graph

Select x-values and calculate the corresponding y-values. Plot the ordered pairs found.Then draw a smooth curve through those points.When the coefficient of is negative,the corresponding parabola opens downward. When a parabola opens downward, thevertex is the highest point of the parabola. The vertex of this parabola is (0, 0) and theaxis of symmetry is again the y-axis.

y = -2x2

x2

y = -2x2 .

CLASSROOM EXAMPLE

Graph

answer: y

x(0, 0)

(1, 3)(1, 3)

y = -3x2 .

6789

112345

1

2 3 4 512345 x

y

y 2x2

x y

0 0

1

2

3

-18-3

-8-2

-2-1

-18

-8

-2

y = x2

If then or 9.

If then or 4.

If then or 1.

If then or 0.

If then or 1.

If then or 4.

If then or 9.y = 32,x = 3,

y = 22,x = 2,

y = 12,x = 1,

y = 02,x = 0,

y = 1-122 ,x = -1,

y = 1-222 ,x = -2,

y = 1-322 ,x = -3,



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H e l p f u l H i n tRecall that:

To find x-intercepts, let and solve for x.

To find y-intercepts, let and solve for y.x = 0

y = 0

112345

12345

2 3 4 512345

(3, 5)

(2, 0)

(0, 4)

(2, 0)

(1, 3)

(3, 5)

(1, 3)

x

y

y x2 4

2 Just as for linear equations, we can use x- and y-intercepts to help graph quadraticequations. Recall from Chapter 3 that an x-intercept is the point where the graphintersects the x-axis.A y-intercept is the point where the graph intersects the y-axis.

E X A M P L E 2

Solut ion

Graph

First, find intercepts. To find the y-intercept, let Then

To find x-intercepts, we let

Thus far, we have the y-intercept and the x-intercepts (2, 0) and Nowwe can select additional x-values, find the corresponding y-values, plot the points, anddraw a smooth curve through the points.

Notice that the vertex of this parabola is 10, -42 .

y = x2 - 4

1-2, 02 .10, -42 x = 2 x = -2

x - 2 = 0 or x + 2 = 0

0 = 1x - 221x + 22 0 = x2 - 4

y = 0.

y = 02 - 4 = -4

x = 0.

y = x2 - 4.

CLASSROOM EXAMPLE

Graph

answer: y

x(3, 0)

(0, 9)

(3, 0)8

8

y = x2 - 9.

H e l p f u l H i n tFor the graph of

If a is positive, the parabola opens upward.If a is negative, the parabola opens downward.

y = ax2 + bx + c ,

x y

0

1

2 0

3 5

0

5-3

-2

-3-1

-3

-4



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GRAPHING QUADRATIC EQUATIONS SECTION 9.5 567

3 Thus far, we have accidentally stumbled upon the vertex of each parabola thatwe have graphed. It would be helpful if we could first find the vertex of a parabola,next determine whether the parabola opens upward or downward, and finally calculateadditional points such as x- and y-intercepts as needed. In fact, there is a formula thatmay be used to find the vertex of a parabola.

One way to develop this formula is to notice that the x-value of the vertex of theparabolas that we are considering lies halfway between its x-intercepts.We can use thisfact to find a formula for the vertex.

Recall that the x-intercepts of a parabola may be found by solvingThese solutions, by the quadratic formula, are

The x-coordinate of the vertex of a parabola is halfway between its x-intercepts,so the x-value of the vertex may be found by computing the average, or of the sum ofthe intercepts.

=-b

2a

=12

a -2b

2ab

=12

a -b - 2b2 - 4ac - b + 2b2 - 4ac

2ab

x =12

a -b - 2b2 - 4ac

2a+

-b + 2b2 - 4ac

2ab

12

x =-b - 2b2 - 4ac

2a, x =

-b + 2b2 - 4ac

2a

0 = ax2 + bx + c .

CONCEPT CHECKFor which of the following graphs of would the value of a be negative?

a. b.

x

y

x

y

y = ax2 + bx + c

y

x

Vertex

b b2 4ac2a

b b2 4ac2a

Samedistance

Vertex FormulaThe vertex of the parabola has x-coordinate

The corresponding y-coordinate of the vertex is found by substituting thex–coordinate into the equation and evaluating y.

-b

2a

y = ax2 + bx + c



E X A M P L E 3

Graph

In the equation and The x-coordinate of the vertex is

Use the vertex formula,

To find the corresponding y-coordinate, we let in the original equation.

The vertex is and the parabola opens upward since a is positive. We now findand plot the intercepts.

To find the x-intercepts, we let

We factor the expression to find The x-interceptsare (4, 0) and (2, 0).

If we let in the original equation, then and the y-intercept is (0, 8).Now we plot the vertex and the intercepts (4, 0), (2, 0), and (0, 8). Then we cansketch the parabola. These and two additional points are shown in the table.

13, -12 y = 8x = 0

1x - 421x - 22 = 0.x2 - 6x + 8

0 = x2 - 6x + 8

y = 0.

13, -12y = x2 - 6x + 8 = 32 - 6 # 3 + 8 = -1

x = 3

x b

2a.x =

-b

2a=

-1-622 # 1

= 3

b = -6.y = x2 - 6x + 8, a = 1

y = x2 - 6x + 8.

Solut ion

NCLASSROOM EXAMPLE

Graph

answer:

(3, 0)(1, 0)

(1, 4)

x

y

y = x2 - 2x - 3.

E X A M P L E 4

6 7

6789

11

12345

2 3 4 5123

(4, 0)(2, 0)

(5, 3)

(6, 8)

(1, 3)

(0, 8)

Axisof

symmetry

x

y

y x2 6x 8

(3, 1)

x y

3

4 0

2 0

0 8

1 3

5 3

6 8

-1

Graph

In the equation and Using the vertex formula, we findthat the x-coordinate of the vertex is

The y-coordinate of the vertex is

y = 1-122 + 21-12 - 5 = -6

x =-b

2a=

-22 # 1

= -1

b = 2.y = x2 + 2x - 5, a = 1

y = x2 + 2x - 5.

Solut ion

y = x2 - 6x + 8



bounce09.html?3_2_l

Thus the vertex is

To find the x-intercepts, we let

This cannot be solved by factoring, so we use the quadratic formula.

Let and


The x-intercepts are and We use a calculator to approx-imate these so that we can easily graph these intercepts.

To find the y-intercept, we let in the original equation and find that Thus the y-intercept is 10, -52 . y = -5.x = 0

-1 + 26 L 1.4 and -1 - 26 L -3.4

1-1 - 26, 02.1-1 + 26, 02 x =

21-1 ; 2622

= -1 ; 26

x =-2 ; 226

2

x =-2 ; 224

2

c 5 .a 1 , b 2 , x =-2 ; 222 - 41121-52

2 # 1

0 = x2 + 2x - 5

y = 0.

1-1, -62 .GRAPHING QUADRATIC EQUATIONS SECTION 9.5 569

CLASSROOM EXAMPLE

Graph

answer:

(0, 1)

(1.5, 1.25)x

y

( (352

( (352 ,0

,0

y = x2 - 3x + 1.

6

67

112345

123

2 3 412345

(0, 5)(2, 5)

(1, 6)

x

y

(1 6, 0)(1 6, 0)

y x2 2x 5

x y

0

0

0

-5-2

-5

-1 - 26

-1 + 26

-6-1

H e l p f u l H i n tNotice that the number of x-intercepts of the graph of the parabola

is the same as the number of real solutions of0 = ax2 + bx + c .y = ax2 + bx + c

Two x-interceptsTwo real solutions of

0 ax2 bx c

y

x

y ax2 bx c

One x-interceptOne real solution of

0 ax2 bx c

y

x

y ax2 bx c

No x-interceptsNo real solutions of

0 ax2 bx c

x

y

y ax2 bx c

y = x2 + 2x - 5



Recall that a graphing calculator may be used to solve quadratic equa-tions. The x-intercepts of the graph of are solutionsof To solve for example, graph

The x-intercepts of the graph are the solutions ofthe equation.

Use a graphing calculator to solve each quadratic equation. Round solu-tions to two decimal places.

1.

2.

3.

4.

5.

6. 7.5x2 - 3.7x - 1.1 = 05.8x2 - 2.6x - 1.9 = 0-5.8x2 + 2.3x - 3.9 = 0-1.7x2 + 5.6x - 3.7 = 02x2 - 11x - 1 = 0x2 - 7x - 3 = 0

y1 = x2 - 7x - 3.x2 - 7x - 3 = 0,0 = ax2 + bx + c .

y = ax2 + bx + c


Spotlight on

DECISIONMAKING

Suppose you are a landscape designer and you have some daffodil bulbs thatyou would like to plant in a homeowner’s yard. It is recommended that daffodilbulbs be planted after the ground temperature has fallen below 50°F. Based onthe following graph of normal ground temperatures in the area, when shouldyou plant the bulbs? Explain.

90

80

70

60

50

40

30

20

10

0

Tem

pera

ture

(deg

rees

Fah

renh

eit)

1

Ground Temperature

122 3 4 5 6 7 8 9 10 11Month (1 = August)

0


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Beginning Algebra