ALGEBRA - CengageIC4a: Solve linear equations IC4b: Solve linear inequalities IC5: Use given formulas to compute results, when geometric measurements are not involved IC6: Find particular

2

ALGEBRA CLAST MATHEMATICS COMPETENCIES

IC1a: Add and subtract real numbers IClb: Multiply and divide real numbers IC2: Apply order-of-operations agreement to computations involving numbers and variables IC3: Use scientific notation in calculations involving very large or very small measurements IC4a: Solve linear equations IC4b: Solve linear inequalities IC5: Use given formulas to compute results, when geometric measurements are not involved IC6: Find particular values of a function IC7: Factor a quadratic expression IC8: Find solutions to quadratic equation IC9: Solve a system of two linear equations in two unknowns IIC1: Use properties of operations correctly IIC2: Determine whether a particular number is among the solutions of a given

equation or inequality IIC3: Recognize statements and conditions of proportionality and variation IIC4: Identify regions of the coordinate plane which correspond to specified conditions and vice versa IIIC2: Use applicable properties to select equivalent equations and inequalities IVC1: Solve real world problems involving the use of variables aside from commonly used geometric formulas IVC2: Solve problems that involve the structure and logic of algebra

63

Copyright © Houghton Mifflin Company. All rights reserved.

2.1 REAL NUMBERS, SCIENTIFIC NOTATION AND ORDER OF OPERATIONS

We are now ready to expand our knowledge of the rational numbers to a new set of numbers called the real numbers. The real numbers include the rational numbers and the irrational numbers. The

irrationals are numbers that cannot be written as the ratio of two integers. For example, 2 , 3

2

and π are irrational numbers. We will start by performing the four fundamental operations using irrational numbers.

A. Operations with Irrational Numbers Objectives IC1a, IC1b CLAST SAMPLE PROBLEMS 1. 4π - 5 + 2π = 2. 5 2 + 18 = 3. 3 × 6 = 4.

363 = 5.

65 =

T TERMINOLOGY--EXPRESSIONS AND LIKE TERMS EXPRESSIONS

An expression is a collection of numbers and letters (variables) connected by operation signs. The parts to be added or subtracted are called terms. The numerical part of a term (the 9 in 9xy2, for example) is the coefficient of the term, and the letters x and y2 are variables.

EXAMPLES 9xy2, x + y, 3 x + 5 and π7 - 3x2 + 5 are expressions. 9xy2 has one term. x + y has two terms: x and y 3 x + 5 has two terms: 3 x and 5 π7 - 3x2 + 5 has three terms: π7 - 3x2 and 5

LIKE TERMS

When two or more terms have exactly the same variable factors, (except possibly for their coefficients or the order in which the factors are multiplied), they are like terms.

EXAMPLES 3x and -5x are like terms. -7x2y and 32x2y are like terms. 18π and 27π are like terms. 3 7 and -8 7 are like terms.

Irrational numbers can be added or subtracted using the idea of like terms. For example, 3π + 4π = 7π , 9 7 - 5 7 = 4 7 , 8π - 15π = - 7π . Note that the expression 2π + 4 cannot be simplified any further because it involves an irrational number (2π ) and a rational number (4). Radicals such as 9 7 and 5 7 can be added or subtracted because the radicands, the numbers under the radical , are equal. However, 3 2 and 8 cannot be added unless we simplify 8 . Here is the rule we need to simplify 8 .

ANSWERS 1. 6π - 5 2. 8 2 3. 3 2 4. 2 3 5.

305

64 CHAPTER 2 Algebra


1 RULES TO SIMPLIFY RADICALS PRODUCTS

baba ×=× Note: The expressions under the radical must be positive.

EXAMPLES 8 = 24× = 24 × = 2 2

45 = 59× = 59 × = 3 5

QUOTIENTS ab =

ab

When the denominator of ab is irrational,

rationalize the denominator b by multiplying numerator and denominator by b . Note that b × b = b2 = b.

354 =

354 =

35 2

6364 =

6364 =

638

=8

79× =8

79 × = 3 7

8

35 = 5

5××

53 =

2553× =

3 55 .

To rationalize the denominator in 35 we

multiplied the numerator and denominator by 5 .

�CLAST EXAMPLES

Example Solution 1. 2π + 11π - 1 = A. 13π 2 - 1 B. 13π - 1 C. π + 12 D. 12π

2π and 11π are like terms and can be added to obtain 13π . Since 1 is a rational number, you cannot combine it with the 13π . The final answer is B, 13π - 1.

Example

2. 75 - 3 = A. 5 B. 66 C. 4 3 D. 72 Note: To combine 75 and 3 , rewrite 75 using 3 as a factor.

Solution

To combine 75 and 3 we have to write 75 and 3 as like terms.

Now, 75 = 325× = 25 × 3 = 5 3 Thus, 75 - 3 = 5 3 - 3 = 4 3 . The correct answer is C.

Example

3. 5 × 10 = A. 5 2 B. 25 2 C. 5 D. 50

Solution

Using the multiplication property, 5 × 10 = 50 = 225×

= 25 × 2 = 5 2 The correct answer is A.

SECTION 2.1 Real Numbers, Scientific Notation and Order of Operations 65


Example Solution

4. 43 =

A. 4 3

9 B. 4 3 C. 4 3

3 D. 4

3

Since the denominator is irrational, we multiply numerator and denominator by 3 obtaining: 43 = 33

3××4 =

4 39 =

4 33 . The correct

answer is C.

Example

5. 402 =

A. 40 B. 2 10 C. 4 5 D. 2 5 Note: the question involves a quotient, so use the quotient property to simplify

Solution

Using the quotient property,

402 =

402 = 20 . Since 20 = 4 × 5 and 4

is a perfect square, by the product property 20 = 5×4 = 4 × 5 = 2 5 . The answer is D.

Example Solution

6. 35 =

A. 3

5 B. 155 C.

35 D.

95

To rationalize the denominator, multiply the 35

under the radical by 55 = 1 obtaining:

35 =

5•35•5 =

155 . The answer is B.

B. Scientific Notation Objective IC3 CLAST SAMPLE PROBLEMS Write in Scientific Notation:

1. 26,200,000 2. (2.3 × 103) × (2.0 × 104) 3. 0.0004 × 23,000 4. 53

102.7106.3

×× 5.

0.000082000

6. Write 6.02 × 10-5 as a decimal Scientific notation and exponents are used to write very large or very small numbers. Here are the rules and terminology CLAST uses.

ANSWERS 1. 2.62 × 107 2. 4.6 × 107 3. 9.2 × 100

4. 5 × 10-3 5. 4.0 × 10-8 6. 0.0000602



T SCIENTIFIC NOTATION SCIENTIFIC NOTATION

A number in scientific notation is written as M × 10n where M is a number between 1 and 10 and n is an integer. Note that when n is a positive integer,

x -n = 1xn .

EXAMPLES 3.48 × 104 is written in scientific notation. In this number M = 3.48 and n = 4. 9.325 × 10-5 is written in scientific notation with M = 9.325 and n = -5.

1 WRITING NUMBERS IN SCIENTIFIC NOTATION

RULE 1. The M is obtained by placing the decimal

point so that there is exactly one nonzero digit to its left.

2. The n is the number of places the decimal

point must be moved to be at its original position (n is positive if the point must be moved right, and negative if moved left.)

3. Write the answer in the form M × 10n

EXAMPLES Write 68,347.09 in scientific notation 1. 68,347.09 = 6.834709 × 10n 2. The decimal point must be moved n = 4 places right to get 68,347.09 3. 68,347.09 = 6.834709 × 104. Write 0.087 in scientific notation 1. 0.087 = 8.7 × 10n 2. The decimal point must be moved 2 places

left to get 0.087, so n = -2 3. 0.087 = 8.7 × 10-2

2 LAWS OF EXPONENTS

RULE (1) ax × ay = ax+y When multiplying expressions with the same bases, we add the exponents.

(2) ax ay = a

x - y

When dividing expressions with the same bases, we subtract the exponents (3) ( )ax y = ax×y= axy

When raising a power to a power, we multiply the exponents.

EXAMPLES 83 × 82 = 82 + 3 = 85 and 46 × 49 = 46 + 9 = 415 Note that: 82 × 49 is not 3211 and 46 × 49 is not 454 3632 = 3

6 - 2 = 34 and 3236 = 3

2 - 6 = 3-4 = 134

( )32 5 = 32×5 = 310 and ( )73 2 = 73×2 = 76 Note that: (83)7 is not 810. (83)7 = 83•7 = 821



�CLAST EXAMPLES

Example Solution

7. (6.1 × 1016) × (1.4 × 10-14) = A. 854 B. 8540 C. 85.4 D. -854 The best way to do this problem is to multiply the decimal parts (6.1 and 1.4) first and then multiply the powers of 10. Remember that when multiplying by 10n you move the decimal n places right, if n is positive and n places left if n is negative.

We first multiply 6.1 by 1.4 and then 1016 by 10-14. Thus, (6.1 × 1016) × (1.4 × 10-14) = (6.1 × 1.4) × ( 1016 × 10-14) = (8.54) × (1016 × 10-14) = (8.54) × (1016 +(-14)) = (8.54) × (102) = 854 Note that when we multiplied by 102, we moved the decimal 2 places right The answer is A.

Example 8. 0.000904 ÷ 2,260,000 = A. 4.00 × 102 B. 4.00 × 1010 C. 4.00 × 109 D. 4.00 × 10-10 Here it is easier to first write 0.000904 and 2,260,000 in scientific notation and then perform the division.

Solution 0.000904 = 9.04 × 10-4 and 2,260,000 = 2.26 × 106. Thus, 0.000904 ÷ 2,260,000

= 64

1026.21004.9×× −

= 9.042.26 ×

10-4106

= 4.00 × 10-4 - 6 = 4.00 × 10-10 The answer is D.

C. Order of Operations Objective IC2 CLAST SAMPLE PROBLEMS 1. 4 × (2 + 6) - 16 ÷ 8 = 2. 4(x + y) - 8(2x - y)2 = 3. (5 - 8)2 + 42 = What does 1 + 6 × 2 mean? If we add 1 and 6 first, the expression becomes 7 × 2 = 14. If we multiply first the expression is 1 + 12 = 13. To avoid confusion, we agree to do operations in a certain order. Here is the rule: ANSWERS 1. 30 2. - 4x + 8y 3. 25



3 ORDER OF OPERATIONS RULE

1. Do the operations inside Parentheses 2. Do Exponentiations 3. Do Multiplications and Divisions as they

occur from left to right. 4. Do Additions and Subtractions as they occur

from left to right. You can remember the order if you remember

PEMDAS as "Please Excuse My Dear Aunt Sally"

EXAMPLES

To find 20 - 12(13 +

12 ) + 2

3 ÷ 4 we proceed as

follows:

20 - 12(56 ) + 2

3 ÷ 4 Added inside ( )

= 20 - 12(56 ) + 8 ÷ 4 Since 2

3 = 8

= 20 - 10 + 8 ÷ 4 Multiplied = 20 - 10 + 2 Divided = 10 + 2 Subtracted = 12 Added

Note that we subtracted before we added because addition and subtraction are done as they occur from left to right.

�CLAST EXAMPLES

Example Solution

9. 37 - 2(

14 + 3) =

A. 128 B. 6

114 C. - 6

114 D. 2

1314

You can remember the order of operations if you remember PEMDAS!

37 - 2(

14 +

124 ) Note: 3 =

124

= 37 - 2(

134 ) Add inside ( )

= 37 -

264 Multiply

= 37 -

132 Since

264 =

132

= 614 -

9114 14 is the LCD

= - 8514 = - 6

114 The answer is C.

Example Solution

10. 10t + t × 2 + 14t2 ÷ 7 × 5 = A. 10t2 + 22t B. 10t2 + 12t

C. 25 t

2 + 18t D. t2

Note that variables are being used here, but the order of operations must be the same: PEMDAS.

10t + t × 2 + 14t2 ÷ 7 × 5 = 10t + 2t + 2t2 × 5 M t × 2, D 14t2 ÷ 7 = 10t + 2t + 10t2 M 2t2 × 5 = 12t + 10t2 A 10t + 2t An answer equivalent to 12t + 10t2 is 10t2 + 12t. The correct answer is B. Answers involving variables are written in descending order in the CLAST and are sometimes rewritten using the commutative law of addition which we shall study in Section 2.2.



Section 2.1 Exercises

�WARM-UPS A 1. 6π + 8π - 5 = 2. 9π + 5π - 3 = 3. 9π - 7 - 2π = 4. 3π - 2 - 8π = 5. 3 2 + 8 = 6. 4 3 + 27 = 7. 112 - 3 7 = 8. 250 - 3 10 = 9. 7 × 35 = 10. 11 × 33 = 11. 10 × 15 = 12. 35 × 10 =

13. 45 = 14.

76 =

15. 38 = 16.

527 =

17. 318 = 18.

763 =

�CLAST PRACTICE A PRACTICE PROBLEMS: Chapter 2, # 1-5

19. 11π + 15π - 4 = A. 26π - 4 B. 26π 2 - 4 C. 22π D. 7π + 22 20. 10 + 250 =

A. 2 65 B. 6 10 C. 62,600 D. 50 21. 63 - 7 = A. 14 B. 56 C. 3 D. 2 7



22. 11 × 55 = A. 121 5 B. 11 C. 11 5 D. 605

23. 28 × 21 = A. 14 3 B. 294 C. 14 D. 96 3

24. 15

7 =

A. 15 7

49 B. 157 C.

15 77 D. 15 7

25. 211 =

A. 2 11 B. 2 11121 C.

211 D.

2 1111

26. 4 2

7 =

A. 8

7 B. 4 14

7 C. 4 14

49 D. 4 14

�WARM-UPS B 27. (5.5 × 1015) × (3.1 × 10-10) = 28. (9.0 × 1018) × (2.4 × 10-15) = 29. (9.3 × 1017) (4.5 × 10 -20) = 30. (6.6 × 105) × (2.4 × 10-9) = 31. 0.0000276 ÷ 1380 = 32. 0.0258 ÷ 1290 = 33. 0.0882 ÷ 21,000 = 34. 0.00246 ÷ 1,230,000 =

×



�CLAST PRACTICE B PRACTICE PROBLEMS: Chapter 2, # 6-8 35. (2.5 × 109) × (5.7 × 10-3) = A. 14,250,000 B. 142,500,000 C. -14,250,000 D. 1,425,000 36. (6.6 × 1019) × (3.7 × 10-13) = A. 2,442,000 B. -24,420,000 C. 244,200,000 D. 24,420,000 37. 0.000504 ÷ 1,260,000 =

A. 4.00 × 102 B. 4.00 × 10-10 C. 4.00 × 10-2 D. 4.00 × 1010 38. 0.00494 ÷ 2,470,000 A. 2.00 × 10-9 B. 2.00 × 10-3 C. 2.00 × 109 D. 2.00 × 103 �WARM-UPS C

39. 13 - 5(

13 + 7) = 40.

56 - 6(

14 + 4) =

41. 12 - 4(

35 + 2) = 42.

23 - 2(

13 + 5) =

43. 9t - 3t × 4 + 10t2 ÷ 5 × 3 = 44. 8t - t × 3 + 14t2 ÷ 7 × 4 = 45. 7t - 4t × 5 + 10t2 ÷ 5 × 4 = 46. 9t + 4 × t + 21t2 ÷ 3 × 2 =



�CLAST PRACTICE C PRACTICE PROBLEMS: Chapter 2, # 9-10

47. 56 - 3(

14 + 4) =

A. - 111112 B. 4

112 C. - 4

112 D. 11

1112

48. 15 - 4(

35 + 7) =

A. 3015 B.

245 C. -

2425 D. - 30

15

49. 9t - 3t × 3 + 8t2 ÷ 4 × 2 = A. 4t2 B. 4t2 + 18t C. t2 + 18t D. t2 50. 5t - 2t × 4 + 24t2 ÷ 6 × 2 =

A. 2t2 + 12t B. 8t2 + 12t C. 8t2 - 3t D. 2t2 - 3t EXTRA CLAST PRACTICE

51. 3(x + y) - 2

)2(6 yx − =

A. 0 B. - 3x + 6y C. 3x + 6y D. - 3x 52. (10 - 4)2 + 32 = A. 15 B. - 25 C. 45 D. 25

53. πππ

223 2× =

A. 3π 2 B. 2π 2 C. 3π D. 2π 54. 2π + 16 π = A. 2 + 16 π B. 6π C. 18π D. 258π

SECTION 2.2 Real Number Properties 73


2.2 REAL NUMBER PROPERTIES In Section 2.1, Example 10, we used the commutative law of addition to write 12t + 10t2 as 10t2 + 12t. We shall now discuss the properties of the operations of addition and multiplication. In the statement of these laws, a, b and c are real numbers. T PROPERTIES OF ADDITION AND MULTIPLICATION

PROPERTIES OF ADDITION

Commutative a + b = b + a (The order in which two numbers are added makes no difference in the answer.) Associative a + (b + c)=(a + b) + c (The grouping of numbers in addition makes no difference in the answer.) Identity a + 0 = 0 + a = a If 0 is added to any real number a, the answer is still a. Inverse a + (-a) = 0 a and - a are additive inverses. (When a and -a are added, the answer is 0.)

EXAMPLES

2 + 3 = 3 + 2 2x + y2 = y2 + 2x 3xy(y2 + 2x) = 3xy(2x + y2) 2 + (7 + 5) = (2 + 7) + 5 (x + y) + 7 = x + (y + 7) 2 3 + (5 3 - 9) = (2 3 + 5 3 ) - 9 8 + 0 = 0 + 8 = 8 x2 + 0 = 0 + x2 = x2 3xy + 0 = 0 + 3xy = 3xy 8 + (-8) = 0 yx2 + (-xy2) = 0

PROPERTIES OF MULTIPLICATION

Commutative a × b = b × a (The order in which two numbers are multiplied makes no difference in the answer.) Associative a × (b × c)=(a × b) × c (The grouping of numbers in multiplication makes no difference in the answer.) Identity a × 1 = 1 × a = a If 1 is multiplied by any real number a, the answer is still a.

Inverse a × ⎝⎜⎛⎠⎟⎞1

a = 1, a � 0

a and 1a are inverses. (When a and

1a are

multiplied, the answer is 1.)

EXAMPLES

3 × 8 = 8 × 3 z(3x - y) = (3x - y)z (3x - y)(2y + z) = (2y + z)(3x - y) 3 × (8 × 2) = (3 × 8) × 2 3x2(x3y4) = (3x2x3)y4 (6x3y4)z5 = 6x3(y4z5) 5 × 1 = 1 × 5 = 5 (x + y2) × 1 = 1 × ( x + y2) = x + y2 (x2y) × 1 = 1 × (x2y) = x2y

15 × 115 = 1

(x + 2y) × 1

x + 2y = 1



There is one more property that involves both addition and multiplication, it is called the distributive property of multiplication over addition. T DISTRIBUTIVE PROPERTY OF MULTIPLICATION OVER ADDITION

DISTRIBUTIVE PROPERTY

a(b + c) = ab + ac

EXAMPLES 3(4 + y) = 3 × 4 + 3 × y or 12x + 3y 6a + 2b = 2(3a + b)

A. Using Real Number Properties

Objective IIC1 CLAST SAMPLE PROBLEMS Identify the property used

1. 3(x + 5) = 3(5 + x) 2. 3(x + 5) = 3x + 15 3. 3 + (4 + 5) = 3 + (5 + 4)

4. 3 + (4 + 5) = (3 + 4) + 5 5. x2⎝⎜⎛

⎠⎟⎞1

x2 = 1

6. Use the distributive law to write an expression equivalent to 3x + 3y

7. Choose the expression that is NOT true for all real numbers: A. 4x + 4y = 4(x + y) B. (x + y)(x - y) = (x - y)(x + y) C. x + (-x) = 0 D. 4xy(5x + y) = 4xy(5y + x)

The CLAST does not ask to identify properties by name, but concentrates on their proper use as we shall illustrate next.

�CLAST EXAMPLES

Example Solution 1. Choose the expression equivalent to the following: 15(13) + 15(10)

A. 15(13 + 10) B. 15(15) + 13(10)

C. (15+15)(13+10) D. 30(13)(10)

Note that in the original expression 15(13) + 15(10) the 15 occurs in both the first and second terms, that is, it has been distributed to the first and second terms.

Thus, the correct answer is A Example Solution

2. Choose the expression equivalent to the

following: 8 + 19 + 4

A. 8(19) + 4 B. 8(19 + 4)

C. 19 + 8 + 4 D. 4 - 19 - 8

There are no multiplications in 8 + 19 + 4, so the answer does not involve the distributive property. Changing the order of addition in 8 + 19 + 4 yields 19 + 8 + 4. The correct answer is C.

ANSWERS 1. Commutative Property of Addition 2. Distributive Property

3. Commutative Property of Addition 4. Associative Property of Add. 5. Inverse Property of Multiplication 6. 3(x + y) 7. D is not true for all real numbers



� CLAST EXAMPLES

Example Solution

3. Choose the statement which is not true for all real numbers

A. 5(x) + 5(y) = 5(x + y)

B. (x - y)(x + y) = (x + y)(x - y)

C. 7(xy) = (7x)y

D. 8yz(2y + z) = 8yz(2z + y)

Here we have to be careful, because we are seeking the statement which is not true for all real numbers. A is an example of the distributive property, B illustrates the commutative property of multiplication and C uses the associative law of multiplication. D is not true since there is no law that says that 2y + z = 2z + y. Try it for y = 1 and z = 2.

B. Properties for Solving Equations and Inequalities Objective IIIC2 CLAST SAMPLE PROBLEMS

Find an equation equivalent to the given one and with all variables on the left:

1. 2x + 3 = 14 2. 2x - 1 = - 16 x 3. 6x - 7 > 4 - 4x

4. If both sides of 6xy < y + 2 are multiplied by y > 0, the resulting inequality is:

5. If both sides of 6xy < y + 2 are multiplied by y < 0, the resulting inequality is:

6. If -4x < 28 is divided by -4, the resulting inequality is: 7. If -2 < -8x < 16 is divided by -8, the resulting inequality is: 8. If 3 is added to all terms of - 6 < x - 3 < 10, the resulting inequality is: There are some properties of the real numbers that are used to solve equations and inequalities. What do we mean by an equation or an inequality? Here are the definitions.

T TERMINOLOGY--FIRST DEGREE EQUATIONS EQUATION

An equation is a sentence using "=" for its verb. A first-degree or linear equation is an equation which can be written as

ax + b = c, where a, b and c are real numbers and x is the variable.

EXAMPLES 2x + 1 = 7 is an equation. 3x + 7 = 4 - 2x is an equation. 3x + 7 = 4 - 2x is a linear (first-degree) equation since we can add 2x to both sides and write it as 5x + 7 = 4, which is of the form ax + b = c, where a, b and c are real numbers and x is the variable.

ANSWERS 1. 2x = 11 2. 13x = 6 3. 10x > 11 4. 6x < y2 + 2y

5. 6x > y2 + 2y 6. x > - 7 7. 14 > x > -2 or - 2 < x <

14

8. - 3 < x < 13



T TERMINOLOGY--LINEAR INEQUALITIES INEQUALITY

An inequality is a sentence using , < or > for its verb. A first-degree or linear inequality is an inequality which can be written as

ax + b < c or ax + b < c where a, b and c are real numbers and x is the variable.

EXAMPLES 2x + 1 < 7 and 8x + 2 > 9 are inequalities. -3x < 6, -2x > 8 and 5(x + 2) < 3x - 7 are first-degree inequalities.

SOLUTIONS The solutions of an equation or inequality are the replacements of the variable that make the equation or inequality a true statement. When we find the solutions of an equation or inequality, we say that we have solved the equation or inequality.

EXAMPLES The solution of: 2x + 1 = 7 is 3. If x is replaced by 3 in 2x + 1 = 7, we get the true statement 2•3 + 1 = 7 The solution of 5(x + 2) - 3 = 6[10 - 2(x + 3)] is 1, because 5(1 + 2) - 3 = 6[10 - 2(1 + 3)] since 5(3) - 3 = 6[10 - 2(4)] or 12 = 12 The solution of 7x + 2 > 9 consists of all real numbers x such that x > 1 (Try it for x = 1.5, x = 2 or x = 5.7.) The solution of -2x > 8 consists of all real numbers x such that x < -4 (Try it for x = - 4.5 or x = - 5.)

To prove that 5(x + 2) - 3 = 6[10 - 2(x + 3)] is a first-degree equation, we have to find an equivalent equation of the form ax + b = c. Here are the rules used to find equivalent equations.

1 RULES FOR FINDING EQUIVALENT EQUATIONS RULE

The equation a = b is equivalent to:

a + c = b + c

a - c = b - c ac = bc, c ≠ 0

ac =

bc c ≠ 0

This means that you can add or subtract the same number c on both sides of an equation and multiply or divide both sides of an equation by the same nonzero number c.

EXAMPLES 3x - 7 = 13 is equivalent to 3x - 7 + 7 = 13 + 7 8x + 1 = 9 is equivalent to 8x + 1 - 1 = 9 - 1

13 x = 6

is equivalent to 3 • 13 x = 3 •6

4x = 12

is equivalent to 4x4

= 124



We are now ready to prove that 5(x + 2) - 3 = 6[10 - 2(x + 3)] is a linear equation by using properties of the real numbers to write it as ax + b = c. Given: 5(x + 2) - 3 = 6[10 - 2(x + 3)] By the Distributive Property 5x + 10 - 3 = 6[10 - 2x - 6] Combining like terms 5x + 7 = 6[4 - 2x] Using the Distributive Property 5x + 7 = 24 - 12x Adding 12x to both sides 5x + 7 + 12x = 24 - 12x + 12x Combining like terms 17x + 7 = 24 which is of the required form. �CLAST EXAMPLE

Example Solution 4. Choose the equation equivalent to: 4x - 7 = 3x + 6 A. 7x - 7 = 6 B. x - 7 = 6 C. 4x - 6 = 3x + 1 D. 4x - 1 = 3x + 6

Look for answers with all variables on one side, so B is the probable answer. Subtracting 3x on both sides we get, 4x - 7 - 3x = 3x + 6 - 3x or x - 7 = 6 The answer is B.

The rules used for finding equivalent inequalities are similar to those used to find equivalent equations with one important exception: multiplication or division by a negative number reverses the inequality. For example, consider the true inequality 2 < 4 If we multiply both sides by -2, we get -2• 2 < -2•4 or -4 < -8 (FALSE!) which is not true. To obtain a true statement, we must reverse -4 > -8 (TRUE) the inequality after multiplying by -2 as shown. Similarly, 3 < 6

Dividing both sides by -3 3-3 <

6-3

or -1 < -2 (FALSE) which is not true. To obtain a true statement, we must reverse -1 > -2 (TRUE) the inequality after dividing by -3 as shown. Here are the rules for finding equivalent inequalities.



2 RULES FOR FINDING EQUIVALENT INEQUALITIES RULE

The inequality a < b is equivalent to:

a + c < b + c

EXAMPLES

5x - 3 < 6 is equivalent to 5x - 3 + 3 < 6 + 3


a - c < b - c

4 - 3x > 7 is equivalent to 4 - 3x - 4 > 7 - 4


ac < bc, c > 0

13 x < 4

is equivalent to 3 •13 x < 3 •4


ac <

bc c > 0

5 x < 10

is equivalent to 5 x5

< 105


ac > bc, for c < 0 Note that when you multiply by c < 0, you have to reverse the inequality.

- 15 x < 2

is equivalent to -5 •- 15 x > -5 •2


ac >

bc for c < 0

Dividing by a negative number c < 0, reverses the inequality.

-2 x > 4

is equivalent to -2 x-2

< 4-2

Note: The rules are similar for and >.

�CLAST EXAMPLES

Example

Solution

5. Choose the inequality equivalent to the following: 4 - 2x > 8

A. -2x > 4 B. -2x < 4 C. 2x > 4 D. -2x < - 4

CLAST questions usually require answers in which all variables are on one side. If we subtract 4 from both sides of 4 - 2x > 8 we obtain 4 - 2x - 4 > 8 - 4 or -2x > - 4

The answer is A.



Example 6. Choose the inequality equivalent to the

following: -4x < 12 A. x > -3 B. x < -3 C. x > 3 D. x < 3

Solution Since all the answers have x by itself on the left, we must divide both sides of the inequality by -4 (remember, you must then reverse the sign of the inequality). We have -4x < 12

Thus, -4x-4 >

12-4

or x > -3 The answer is A.

Example

7. If x < 0, then xy < x2 + x is equivalent to: A. y < x + 1 B. y > x + 1 C. y > -x - 1 D. y < -x - 1

Solution Remember that if you divide by a negative number (x < 0), you must reverse the inequality. Thus,

if x < 0 then xyx >

x2x +

xx

or y > x + 1 The answer is B.

Example

8. Choose the inequality equivalent to the

following: 5 < x + 4 < 8 A. 9 < x < 12 B. 1 > x > 4 C. 20 < x < 32 D. 1 < x < 4 This time we cannot have x by itself on one side of the equation, so we isolate it in the middle by subtracting 4 from each term.

Solution Since all answers have x by itself as the middle expression, we subtract 4 from each expression. We have 5 < x + 4 < 4 Thus, 5 - 4 < x + 4 - 4 < 8 - 4 or 1 < x < 4 The answer is D.




�WARM-UPS A FILL IN THE BLANK WITH AN EQUIVALENT EXPRESSION 1. 20(5) + 20(7) = 2. a(-4) + a(-6) = 3. a(5) + a(-3) = 4. -c(7) + -c(9) = 5. 5 + (3 + 2) = 6. 9 + (3 + x) = 7. 3(a + b ) = 8. 7(x + z) = 9. 3a2(a5b4) = ______ 10. -5x3(x7y2) = �CLAST PRACTICE A PRACTICE PROBLEMS: Chapter 2, # 11-13

11. Choose the expression equivalent to the following: 7(x + y)

A. 7xy B. 7x + 7y C. 7x + 7 D. 7 + (x + y)

12. Choose the expression equivalent to the following: 18 + 16 + 11

A. 18(16 + 11) B. 11 - 16 - 18 C. 18(16) + 11 D. 16 + 18 + 11

13. Choose the expression equivalent to the following: 9y + 3x

A. 12xy B. 3(3y + x) C. 3y + x D. y(9 + 3x)

14. Choose the expression equivalent to the following: (3x + 6y)(3x - 6y)

A. (3x - 6y)(3x + 6y) B. 3(x + 2y)(x - 2y) C. (3x + 6y)(6y - 3x) D. 9xy(3x - 6y)

15. Choose the expression equivalent to the following: 6xy(4x + y2)

A. 24x2y + 6xy B. 10xy + 6xy2 C. 4x(6xy + y2) D. 6xy(y2 + 4x)

16. Choose the statement which is not true for all real numbers

A. (a + b)(a - b) = (a - b)(a + b) B. 5(xy) = (5x)y

C. 5xy(3x + y) = 5xy(3y + x) D. 9(a) + 9(b) = 9(a + b)

17. Choose the expression equivalent to the following: 7a3(a3b4)

A. (7a3a3)b4 B. 7a3(b3a4) C. 7a3b4 D. 7a2(ab)7



�WARM-UPS B WRITE AN EQUIVALENT EQUATION OF THE FORM ax + b = c, (a > 0) 18. 5x - 7 = 4x + 2 19. 6x - 9 = 5x + 4 20. 6x - (17 - 7x) = 20x - 8x + 20 21. 9x - (6 - 6x) = 13 - 8x + 14 22. 6a + 7 = 9[a - (10 - a)] 23. 3(a + 1) = 8[a - (7 - a)] WRITE AN EQUIVALENT INEQUALITY OF THE FORM ax < c, ax > c, ax < c or ax > c and with a > 0 24. -3x + 4 < 7 25. -5x - 2 < - 3 26. 7 - 2x > 5x + 2 27. 4 - 3x > 6x + 1 28. 8 + 4x < 7x + 2 29. 9 + 5x < 9x + 3 30. 15x + 16 > 14x - 19 31. 17x + 15 > 16x - 9 �CLAST PRACTICE B PRACTICE PROBLEMS: Chapter 2, #14 32. Choose the equation equivalent to the following: 4x - 8 = 3x + 5 A 7x - 8 = 5 B. 4x - 5 = 3x + 3 C. 4x - 3 = 3x + 5 D. x - 8 = 5 33. Choose the inequality equivalent to the following: -5x < 15 A. x > -3 B. x < -3 C. x > 3 D. x < 3 34. Choose the inequality equivalent to the following: 2 - 3x > 6 A. - 3x > 4 B. - 3x < 4 C. 3x > -4 D. -3x < -4 35. If x < 0, then x2 > xy + 2x is equivalent to which of the following:

A. x > y + 2 B. x < -y - 2 C. x > -y - 2 D. x < y + 2



EXTRA CLAST PRACTICE 36. Choose the expression equivalent to: 4y + 8x A. y + 2x B. 4(y + 2x) C. 12xy D. 4(x + 2y) 37. Identify the property of addition illustrated by the following: x + (y + z) = (x + y) + z A. Associative B. Commutative C. Distributive D. Inverse 38. Identify the property of addition illustrated by the following: x + (y + z) = x + (z + y) A. Commutative B. Associative C. Distributive D. Inverse

39. Identify the property of multiplication illustrated by the following: n2⎝⎜⎛

⎠⎟⎞1

n2 = 1

A. Commutative B. Associative C. Inverse D. Identity 40. Choose the expression equivalent to: 3a4(a2b3) A. (3a4b2) + b3 B. 3a4a2 + 3a4b3 C. 3a4b3 D. 3(a4a2)b3 41. Choose the statement that is not true for all real numbers:

A. a × 1a = 1 for a ≠ 0 B. a × 0 = a for a ≠ 0 C. a(1 + b) = a + ab D. a + (-a) = 0

42. Choose the equation equivalent to the following: x2 - 3 > x A. x2 - x - 3 > 0 B. x2 - x - 3 < 0 C. x2 + x - 3 > 0 D. x2 + x - 3 < 0

43. Given that y < 0, choose the inequality equivalent to the following: 1 < xy < 3

A. 3y > x > y B. y < x < 3y C. y > x > 3y D. 1 > x > 3

SECTION 2.3 Solving Equations and Inequalities 83


2.3 SOLVING EQUATIONS AND INEQUALITIES In the preceding section we learned some properties of the real numbers used to solve linear equations and inequalities. We now discuss a general procedure to solve equations. The object of this procedure is to have all variables (unknowns) on one side of the equation and all numbers on the other side so that the solution can be written as x = or = x.

A. Solving Linear Equations

Objective IC4a CLAST SAMPLE PROBLEMS SOLVE:

1. -5 = 5 - x 2. 3(x - 1) + 2x = 7 - (2 - x) 3. 6(2m - 1) - (m + 4) = 3 Here is the procedure we need to solve any linear equation:

1 PROCEDURE FOR SOLVING EQUATIONS PROCEDURE

1. If there are fractions, multiply both sides of the equation by the LCD. (The LCD of 4, 6 and 12 is 12.)

2. Remove parentheses and simplify, if

necessary. 3. Add or subtract the same number on both

sides of the equation so that one side has only variables.

4. Add or subtract the same expression on

both sides so that the other side has only numbers.

5. If the coefficient of the variable is not 1,

divide both sides of the equation by this coefficient.

6. Check your answer by substituting it in the

original equation.

EXAMPLES

Given: x4 -

16 =

712(x - 2)

M 12 12•⎝⎜⎛

⎠⎟⎞x

4 - 16 = 12•

712(x - 2)

12•x4 - 12•

16 = 7(x - 2)

3x - 2 = 7x - 14 A 2 + 2 = + 2 3x = 7x - 12 S 7x -7x= -7x - 4x = -12

D -4 -4x-4 =

-12-4

x = 3

You should verify that:

34 -

16 =

712(3 - 2)

is a true statement. ANSWERS 1. x = 10 2. x = 2 3. m = 1311



�CLAST EXAMPLES

Example Solution 1. If 7x - 6 = 3x + 20, then

A. x = 54 B. x =

52

C. x = 134 D. x =

132

Note that there are no fractions in this equation and the equation is simplified, so we may skip steps 1 and 2 and go to step 3 of the procedure.

Given 7x – 6 = 3x + 20 A 6 + 6 = + 6 7x = 3x + 26 S 3x -3x = -3x 4x = 26

D 4 4x4 = 4

26

x = 132

The answer is D.

Example 2. If 6(a + 7) = 9[a - (10 - a)], then a =

A. 312 B. 11 C. -11 D. - 25

Remember: The object of this procedure is to have all the variables on one side and all the numbers on the other side. Start by simplifying both sides and then work toward the objective of having a = or = a.

Solution Given 6(a + 7) = 9[a - (10 - a)] Simplify 6a + 42 = 9[a - 10 + a] 6a + 42 = 9[2a - 10] 6a + 42 = 18a - 90 S 42 -42 = -42 6a = 18a - 132 S 18a -18a = -18a - 12a = - 132

D -12 -12a-12 =

-132-12

a = 11 The answer is B.

B. Solving Linear Inequalities

Objective IC4b CLAST SAMPLE PROBLEMS 1. Solve: 5(1 - a) - 4 < 6 2. Solve: 5 - (x + 1) > 4(1 - x) + 4

Just as we solved equations, we can solve inequalities but remember that when you multiply or divide an inequality by a negative number, the inequality must be reversed. For example, to solve -3x < 9

Divide both sides by -3 (to get the x by itself) -3x-3 >

9-3

The answer is x > - 3 Note that the solution of an inequality is written with the variables on the left, that is,

x < , x < or x > , x > .

ANSWERS 1. a > - 1 2. x > 4/3



�CLAST EXAMPLES

Example Solution 3. If 14b - 19 < 15b + 16 then, A. b = -35 B. b < -35 C. b > -35 D. b < - 25 Note that answer A is not possible. The answer to an inequality is an inequality. To avoid dividing or multiplying by negative numbers, try to have the variables on the side of the inequality in which the variable has the largest coefficient (there are 14 b's on the left, and 15 b's on the right, so we want the variables on the right). However, if you insist on having all variables on the left all the time, you still get the same answer!

Remember, our objective is to have all variables on one side of the inequality, and all numbers on the other side. We can do this by subtracting 16 and 14b from both sides. Here are the steps:

Given: 14b - 19 < 15b + 16 S 16 -16 - 16 14b - 35 < 15b S 14b -14b -14b -35 < b But there is no answer like this! Why? Because the solution of an inequality is written with the variable first. Simply rewrite - 35 < b as b > -35 and you can see that the answer is C. Note that in either case, b is greater than or equal to -35.

Example

4. If 20x - 8x + 20 > 6x - (17 - 7x), then A. x < 37 B. x > 2

C. x < - 3725 D. x > 37

Simplify before deciding on which side the variables must go. You will then see that there are more x's on the right, so we may want the x's on the right, the numbers on the left. Remember that - (17 - 7x) = -1(17 - 7x) = - 17 + 7x

Solution Given: 20x - 8x + 20 > 6x - (17 - 7x) Simplify 12x + 20 > 6x - 17 + 7x 12x + 20 > 13x - 17 A 17 17 17 12x + 37 > 13x S 12x -12x -12x 37 > x or equivalently x < 37 The answer is A. If you have enough time, you should check the answer. Since x < 37, let x = 0 in the original inequality. You get 20•0 - 8•0 + 20 > 6•0 - (17 - 7•0) or 20 > - 17 (true!)

Checking solutions of equations and inequalities is so important that it is a separate CLAST competency. This competency asks to determine if a particular number is among the solutions of an equation or an inequality. Some of these equations and inequalities involve the concept of absolute value. We shall discuss absolute values and checking solutions of equations and inequalities next.



C. Checking Solutions

Objective IIC2 CLAST SAMPLE PROBLEMS 1. Determine if x = 3 is a solution of x2 - 2 = 2x + 1 2. Determine if x = -2 is a solution of |3x + 4 | = x 3. Determine if x = 3 is a solution of (x - 3)(x + 4) > 0 4. Determine is x = 1/2 is a solution of - x2 + 5x > 3x + 1 T TERMINOLOGY--ABSOLUTE VALUE

ABSOLUTE VALUE

The absolute value of a number n is its distance from 0 on the number line and is denoted by |n|. Read "the absolute value of n". You do not need a number line to find the absolute value of a number, here is the rule: If the number is negative, its absolute value is positive, so make it positive. If the number is positive, leave it alone.

Note: The absolute value represents a distance, so it is always positive.

EXAMPLES • •

-3 -2 -1 0 1 2 3 | 2 | = 2 because 2 is two units from 0. | -3 | = 3 because -3 is three units from 0. Let us do it now using the rule: -3 is negative so | - 3| = 3. Remember, if the number is negative, make it positive 2 is positive, so | 2 | = 2. Similarly,

|12 | = 12 , |- 0.5| = 0.5 and | 2 | = 2

�CLAST EXAMPLES

Example

Solution

5. For each of the statements below, determine whether -1 is a solution

i. |x - 1| = 0 ii. (t - 3)(t - 6) < 6 iii. y2 + 3y + 17 = 15 A. i only B. ii and iii only C. iii only D. ii only

To determine if -1 is a solution, substitute -1 for the variable and check whether we get a true statement. For x = -1 in i, |-1 - 1| = |-2| = 2, not 0. Statement i is not true. For t = -1 in ii, (-1 - 3)(-1 - 6) = (-4)(-7) = 28, which is not less than or equal to 6. Statement ii is not true. For y = -1 in iii, (-1)2 + 3(-1) + 17 = 1 - 3 + 17 = 15 Statement iii is true. The answer is C.

ANSWERS 1. Yes 2. No 3. Yes 4. No



Example Solution 6. For each of the statements below,

determine whether 12 is a solution

i. 2x - 1 < 0 ii. (2y - 1)(y - 2) = 0 iii. 2t - 2 = -2t A. ii only B. i only C. ii and iii only D. i, ii and iii

Substituting 12 for the variable in i, ii and iii we

have:

i. 2x - 1 = 2(12 ) - 1 = 1 - 1 = 0 < 0, true

Note: 0 = 0, so 0 < 0 is also true.

ii. (2y - 1)(y - 2) = ⎝⎜⎛

⎠⎟⎞2(

12) - 1 ⎝⎜

⎛⎠⎟⎞1

2 - 2

= (1 - 1)⎝⎜⎛

⎠⎟⎞1

2 - 2

= 0(-32 ) = 0, true

iii. 2t - 2 = - 2t becomes

2(12 ) - 2 = -2(

12 )

1 - 2 = -1 - 1 = -1, a true statement Thus, i, ii and iii are true. The answer is D.


�WARM-UPS A In Problems 1- 20 solve for the unknown 1. 10x = 8x + 28 2. 8x - 7 = 2x - 37 3. 8x + 33 = 4x - 3 4. 6x - 3 = 9x + 27 5. 10 + 7x = 10x + 1 6. 7x - 20 = 12x - 5 7. 3y + 2 = 10 8. 4y - 3 = 7 9. 8y - 6 = 4y + 20 10. 9y - 7 = 5y + 6 11. 6y + 2 = 4(2y - 1) 12. 6y + 4 = 3(4y - 1) 13. 6(z + 4) + 4 - 3z = 4z 14. 6(z - 1) - 2 + z = 8z + 4



15. 32 b + 3 - b = (b - 2) + 2 16. b + 2 -

b3 = (b - 2) + 3

17. 3(a + 1) = 8[a - (7 - a)] 18. 2(a + 1) = 10[a - (5 - a)] 19. 2(a + 7) = 10[a - (3 - a)] 20. 4(5a - 4) + 8 = 24 - 2(6a - 6)

�CLAST PRACTICE A PRACTICE PROBLEMS: Chapter 2, # 15-16

21. If 2x + 1 = 9, then

A. x = 5 B. x = 4 C. x = 52 D. x = 2

22. If 4y + 2 = y + 9, then

A . y = 3 B. y = 83 C. y =

73 D. y = 2

23. If 6x + 2 = 4(3x - 1), then

A. x = 32 B. x =

23 C. x = 1 D. x =

13

24. If 4(z - 2) - 3(2z + 7) = z - 19, then

A. z = - 103 B. z =

23 C. z =

53 D. z =

103

25. If 3(a + 7) = 15[a - (3 - a)] then,

A. a = 249 B. a = -32 C. a =

79 D. a = - 2

49

26. If 8(a + 3) = 2[a - (5 - a)], then

A. a = 812 B. a = -6 C. a = - 8

12 D. a = - 4

34



�WARM-UPS B In Problems 27-34 solve for the unknown 27. 5x < 20 28. 7x > 14 29. -3x < 9 30. -4x > 8 31. 2b - 13 > 3b + 12 32. 16b - 9 < 17b + 15 33. 13x - 8x + 14 > 9x - (6 - 6x) 34. 5x - 4x + 20 < 3x - (5 - 3x) �CLAST PRACTICE B PRACTICE PROBLEMS: Chapter # 17-18 35. If 19y - 18 < 20y + 16, then A. y > 34 B. y > - 34 C. y = 34 D. y < - 34 36. If 3(x - 2) - 4(x + 1) > 0, then A. 10 < x B. -10 > x C. -10 < x D. 10 > x 37. If 4x - 7x + 12 > 8x - (5 - 6x), then

A. x < - 1711 B. x > -

1711 C. x < 1 D. x > 1

38. If 3x - 6x + 9 > 4x - (16 - 5x), then

A. x < - 256 B. x >

2512 C. x > -

256 D. x <

2512

39. If 4(x - 1) + 5x < 7x - 4, then

A. x < 0 B. x > 0 C. x = 0 D. 1x > 0

40. If 4x - 3 < 3(x - 1) + 3x, then

A. x < 0 B. x > 0 C. x = 0 D. 1x < 0



�WARM-UPS C In Problems 41-46, determine whether -2 is a solution. 41. | x - 2 | = 0 42. (t + 5)(t - 1) < 9 43. y2 + 5y + 21 = 15 44. | x + 2| = 0 45. (t - 8)(t - 3) < 2 46. y3 + 8 = 0

In Problems 47-50, determine whether 12 is a solution.

47. 2x - 1 < 0 48. (2y - 1)(y - 5) = 0 49. 2t - 2 = - 2t 50. y3 - 2 > 0 �CLAST PRACTICE C PRACTICE PROBLEMS: Chapter 2, # 19-21 51. For each of the statements below, determine whether -1 is a solution.

i. | x - 1 | = 0 ii. (t - 3)(t - 6) < 6 iii. y2 + 4y + 16 = 13 A. iii only B. i only C. ii only D. ii and iii only

52. For each of statements below, determine whether 13 is a solution.

i. 2x - 23 < 0 ii (3y - 1)(y - 4) = 0 iii. 3t - 2 = - 3t

A. i only B. i and iii only C. ii and iii only D. i, ii, and iii 53. For each of the statements below, determine whether 6 is a solution. i. x3 > 0 ii. (x - 6)(x + 5) = 0 iii. 6(x - 5) > 0 A. i only B. i and ii only C. ii and iii only D. i, ii, and iii 54. For each of the statements below, determine whether -8 is a solution. i. x3 < 0 ii. (x - 8)(x + 4) = 0 iii. 8(x - 4) < 0 A. i only B. i and ii only C. i and iii only D. ii and iii only

SECTION 2.4 Evaluating Formulas and Functions 91


2.4 EVALUATING FORMULAS AND FUNCTIONS

Do you know what your systolic blood pressure S should be? Since blood pressure increases with age, some people claim that your blood pressure should be given by the formula S = 100 + A, where A is your age. If you are 23 years old, what should your blood pressure be? To find out, substitute 23 for A in S = 100 + A obtaining S = 100 + 23 or 123 Since we have found the value of S, we have evaluated the formula.

A. Using Formulas to Compute Results

Objective IC5 CLAST SAMPLE PROBLEMS Given I = Prt, I the interest, P the principal, r the rate and t the time (in years) 1. Find the interest paid on a $1000 principal at 10% for 3 years. 2. Find the time required to earn $200 on a $5000 investment at an 8% rate.

The formula S = 100 + A has two variables, A and S. If the value of S or A is given, the value of the other variable can be found. The CLAST uses two, three or four variables in a given formula but the objective is the same: The values for all but one of the variables are given and you have to find the value of the missing variable. Here is the procedure you need.

1 EVALUATING FORMULAS PROCEDURE

1. Replace the given values for the corresponding variables in the formula.

2. Simplify the expression. (Remember that arithmetic operations have to be performed using PEMDAS, Parentheses, Exponents, Multiplication and Division and, Addition and Subtraction as they occur from left to right.)

EXAMPLES If a = (b + 3)2 and b = 5, find a. The given value is b = 5. We replace b by 5 in the formula. Thus, a = (b + 3)2 becomes a = (5 + 3)2 a = (8)2 a = 64 Note that we did the addition inside the parentheses first.

�CLAST EXAMPLES

Example Solution 1. If b = (3 - a)2 and a = -2, find b. A. 1 B. 25 C. -1 D. -25

Replacing a by -2 in b = (3 - a)2 gives b = (3 - (-2))2 = (3 + 2)2 Recall that - (-2) = 2 = 52 = 25 The answer is B.

ANSWERS 1. $300 2. 1/2 year



� CLAST EXAMPLES

Example Solution

2. The formula for converting a Celsius temperature to Fahrenheit is

F = 95 C + 32

o

What is the temperature on the Fahrenheit scale when the Celsius temperature is 20o?

A. 26o B. 68o C. 40.4o D. 212o

Since the given temperature is 20o Celsius, substitute 20o for C in the equation

F = 95 C + 32

o, obtaining

F = 95 (20

o) + 32o

F = 36o + 32o = 68o It is easier to divide 20 by 5 first and then multiply by 9, instead of multiplying 9 by 20 and then dividing by 5. In either case, the answer is B.

Example Solution

3. The formula for finding the simple interest

(I) on a loan of P dollars at a rate r, after t years is I = Prt. How much interest will be paid on a $10,000, 4 year loan if the rate is 8%?

A. $32,000 B. $2000 C. $200 D. $3200

The formula I = Prt has four variables and we are asked to find the interest I. We substitute the value of P = $10,000, r = 8% or 0.08, and t = 4 in the formula I = Prt. We then have:

I = 10,000 × 0.08 × 4 = 10,000 × 0.32 = $3200

The answer is D.

B. Finding the Value of a Function Objective IC6 CLAST SAMPLE PROBLEMS 1. Given f(x) = 3x + 1, find f(4) 2. Given f(x) = - x2 - 2x - 5, find f(-1) The blood pressure formula S = 100 + A suggests that your blood pressure is a function of your age A. Thus, we can write S = f(A) = 100 + A. The notation f(A) is called function notation and is read as "f of A." To find the blood pressure for a 23 year old, that is, to find f(23), we proceed as before:

S = f(23) = 100 + 23 = 123.

The CLAST asks us to find particular values of a given function. For example, if we know that f(x) = 100 + x and are asked to find f(23), we substitute 23 for x obtaining:

f(23) = 100 + 23 = 123

ANSWERS 1. 13 2. -4



�CLAST EXAMPLE

Example Solution

4. Find f(-3) given f(x) = x2 - 4x + 3 A. 9 B. 6 C. 24 D. 6 Caution: After you substitute -3 for x, you have to perform the operations in the correct order.

Since f(x) = x2 - 4x + 3, we substitute -3 for x in the function to get

f(-3) = (-3)2 - 4(-3) + 3 Now use the order of operations and proceed from left to right by taking care of exponents, multiplications, and then additions . Thus, f(-3) = (-3)2 - 4(-3) + 3 = 9 + 12 + 3 = 24 The answer is C.


�WARM-UPS A 1. If a = (b + 4)2 and b = 10, find a. 2. If b = (a + 5)2 and a = 8, find b. 3. If (b - 4)2 = c and b = -5, find c. 4. If d = (5 - e) and e = - 1, find d. 5. The formula for finding a man's weight W (in pounds) is W = 5H - 190, where H is the height

of the man in inches. If H is 70, what is W? 6. The formula for finding a woman's weight W (in pounds) is W = 4H - 140, where H is the

height of the woman in inches. If H = 60, what is W?

7. The number of hours a growing child should sleep is H = 17 - A2 , where A is the age of the

child in years. If A = 6, what is H? 8. The time t in hours it takes an object moving at a rate of r miles per hour to travel a distance d

is t = d/r. Find t when d = 140 and r = 35.



�CLAST PRACTICE A PRACTICE PROBLEMS: Chapter 2, #22-24 9. If a = (b + 3) and b = 10, find a. A. 13 B. 100 C. 109 D. 169 10. If c = (b - 4)2 and b = -1, find c. A. 25 B. -5 C. 5 D. -25 11. If x = (4 - y)3 and y = -1, then x equals: A. 27. B. 125 C. -125 D. -27 12. The formula for finding the simple interest (I) on a loan is I = PRT. How much interest will

Bill pay on his car loan if he finances $19,000 (P) at a 14% simple interest rate (R) for 4 years (T)?

A. $1064 B. $2660 C. $665 D. $10,640 13. The formula for finding the distance (d) in miles traveled in t hours at a rate of r miles per hour

is given by d = rt. How far did Sue travel if she drove her car at 56 miles per hour for 3 hours? A. 37 B. 336 miles C. 168 miles D. 19 miles

14. The acid-test (AT) ratio for a business is given by AT = C + R

CL , where C is the cash, R is the

amount of receivables and CL is the current liabilities. If C = $5000, R = $2800 and CL = $1000, then AT is: A. 7.8 B. $7800 C. 78 D. 780 15. The cost of a long-distance phone call from New York to Rome is given by

C(t) = 0.80(t - 1) + 1.30, where the cost is $1.30 for the first minute and $0.80 for each additional minute. Find the cost of a 5-minute phone call from New York to Rome.

A. $4.50 B. $4.00 C. $3.20 D. $5.30



�WARM-UPS B 16. Find f(2) given f(x) = 2x2 - x + 5 17. Find f(-2) given f(x) = 2x2 - x + 5. 18. Find f(3) given f(x)= - x2 - 3x + 14 19. Find f(-3) given f(x) = - x2 - 3x + 14 �CLAST PRACTICE B PRACTICE PROBLEMS: Chapter 2, # 25-26 20. Find f(2) given f(x) = 2x2 + 2x - 29 A. -21 B. 41 C. 37 D. -17 21. Find f(-2) given f(x) = - 2x2 - x - 14

A. -20 B. -6 C. -8 D. -26 22. Find f(-1) given f(x) = 2x3 - 2x2 - 22 A. -27 B. - 26 C. 26 D. - 22 23. Find f(-2) given f(x) = - 2x3 - x2 + 9 A. 21 B. 17 C. 33 D. - 15 24. Find f(-1) given f(x) = x10 - 9 A. 1 B. -10 C. - 8 D. 1 25. Find f(-1) given f(x) = - x10 + 1 A. 0 B. 2 C. 1 D. - 9



2.5 SOLVING QUADRATIC EQUATIONS

We have already learned how to solve linear equations which can be written in the form ax + b = c. In this section we shall learn how to solve quadratic equations using two methods: factoring and the quadratic formula. A. Factoring Quadratic Expressions

Objective IC7 CLAST SAMPLE PROBLEMS 1. Find the linear factors of 2x2 + 5x + 2 2. Find a linear factor of 3x2 + 14x - 5 3. Which is a linear factor of 3x2 - 7x + 2? A. x = 2 B. 3x + 1 C. 3x - 1 D. 2x - 2

T TERMINOLOGY--QUADRATIC EQUATIONS QUADRATIC EQUATIONS

A quadratic equation is an equation which can be written in standard form as ax2 + bx + c = 0, where a, b, and c are constants and a ≠ 0

To factor ax2 + bx + c means to write ax2 + bx + c as a product of two linear factors, that is, to write: ax2 + bx + c = (dx + e)(fx + g).

EXAMPLES 3x2 - x - 2 = 0 is a quadratic equation in standard form. -x2 + 3x = 2 is a quadratic equation. To write it in standard form, we subtract 2 from both sides obtaining -x2 + 3x - 2 = 0. 3x2 - x - 2 can be factored by writing 3x2 - x - 2 = (3x + 2)(x - 1) x2 + 5x + 6 can be factored by writing x2 + 5x + 6 = (x + 2)(x + 3)

How do we factor 15? We simply write 15 = 3 × 5, but this process is simple since you know how to multiply 3 × 5. Thus, to learn to factor, we discuss how to multiply expressions. To multiply (x + 5)(x + 2), multiply the terms in the following order:

First terms are multiplied first. Outer terms are multiplied second Inner terms are multiplied third Last terms are multiplied last.

This method is called the FOIL method and we illustrate it next. 1 USING THE FOIL METHOD TO MULTIPLY (x + 5)(x + 2)

RULE 1. Multiply the first terms. 2. Multiply the outer terms. 3. Multiply the inner terms. 4. Multiply the last terms.

Add 2x + 5x = 7x and multiply 5 × 2 = 10.

EXAMPLES (x + 5)(x + 2) → x × x = x2 (x + 5)(x + 2) →x2 + 2x (x + 5)(x + 2) →x2 + 2x + 5x (x + 5)(x + 2) → x2 + 2x + 5x + 5 × 2 (x + 5)(x + 2) = x2 + 7x + 10

ANSWERS 1. (2x + 1)(x + 2) 2. 3x - 1 or x + 5 3. C

SECTION 2.5 Solving Quadratic Equations 97


1 USING THE FOIL METHOD TO MULTIPLY (x - 5)(x - 2) RULE

1. Multiply the first terms. 2. Multiply the outer terms. 3. Multiply the inner terms. 4. Multiply the last terms. Simplify -2x - 5x = - 7x and 5 × 2 = 10.

EXAMPLES (x - 5)(x - 2) → x × x = x2 (x - 5)(x - 2) → x2 - 2x (x - 5)(x – 2) →x2 - 2x - 5x (x - 5)(x - 2) → x2 - 2x - 5x + 5 × 2 (x - 5)(x - 2) = x2 - 7x + 10

If we are asked to factor x2 + 7x + 10 we write x2 + 7x + 10 = (x + 5)(x + 2), where the 10 is the product of 5 and 2 and the 7 is the sum of 5 and 2. How do we factor x2 + 6x + 8? If we follow the same pattern, 8 must be the product of two numbers whose sum is the coefficient of the middle term, that is, 6. What two numbers? 4 and 2, since the product of 4 and 2 is 8 and the sum of 4 and 2 is 6. Thus, we have: x2 + 6x + 8 = (x + 4)(x + 2). (You can check this by using the FOIL method.) In general, we have the following rule:

2 FACTORING x2 + (a + b)x + c RULE

x2 + (a + b)x + c = (x + a)(x + b) That is, x2 + (a + b)x + c is factored by finding two numbers whose product is c and whose sum is a + b. Note: Start by finding numbers whose product is c and see if the sum is a + b. We say that (x + a) and (x + b) are the linear factors of x2 + (a + b)x + c.

EXAMPLES Factor x2 + 8x + 12. We need two numbers whose product is 12 and whose sum is 8. Try 12 and 1 (12 + 1 = 13, not 8). Try 3 and 4 (3 + 4 = 7, not 8) Try 6 and 2 (6 + 2 = 8) Since the sum is 8, 6 and 2 are the correct numbers. Thus, x2 + 8x + 2 = (x + 6)(x + 2)

When c is negative, a and b must have different signs with the larger one having the sign of the middle term. When c is positive, a and b must have the same sign.

Factor x2 - 2x - 8. This time the product must be -8 and the sum -2, which means that one number is positive and the other negative. Since the sum must be -2, the larger number must be the negative one. Try -8 and 1 (-8 + 1 = -7 not -2) Try -4 and 2 (-4 + 2 = -2)

Thus, x2 - 2x - 8 = (x - 4)(x + 2)

The CLAST asks us to factor quadratic expressions of the form ax2 + bx + c. To determine if such expressions are factorable we use the ac test.

3 THE ac TEST RULE

ax2 + bx + c is factorable if there are two integers with product ac and sum b. If the integers do not exist, the expression is prime.

EXAMPLES Is 6x2 + 7x + 2 factorable? Here a = 6, c = 2 and ac = 12. Thus, we need two numbers whose product is 12 and whose sum is 7. A little searching will produce 4 and 3. Thus, 6x2 + 7x + 2 is factorable.



Let us get some more practice in determining if an expression is factorable. Look at the three expressions. Which one is not factorable? EXPRESSION a b c ac 2x2 - 7x - 4 2 -7 -4 -8 6x2 + 8x + 5 6 8 5 30 -3x2 + 2x + 5 -3 2 5 -15 To show that 2x2 - 7x - 4 is factorable, we need two numbers whose product is -8 and whose sum is -7 (the coefficient of the middle term). A little searching will produce -8 and 1. Since -8 � 1 = -8 and -8 + 1 = -7, the expression is factorable. On the other hand, no matter how hard you try there are no factors of 30 whose sum is 8. (Try it!). Thus, 6x2 + 8x + 5 is prime. To show that -3x2 + 2x + 5 is factorable we need two numbers whose product is -15 and whose sum is 2. -3 and 5 will do. Thus, -3x2 + 2x + 5 is factorable. The number ac plays an important part in factoring expressions of the form ax2 + bx + c. Because of that, ac is called the key number. Here is the procedure to factor ax2 + bx + c.

4 TO FACTOR EXPRESSIONS OF THE FORM ax2 + bx + c RULE

1. Find ac, the key number. 2. Find the factors of ac = -8 that add up to b = -7 (-8 and 1) and rewrite the middle

term -7x as a sum involving -8 and 1. 3. Group the terms into pairs. 4. Factor each pair. 5. Factor out the greatest common factor

(GCF), (x - 4). Note: If the first pair has (x - 4) as a

factor, the second pair will have the same factor.

EXAMPLES For 2x2 - 7x - 4, ac = 2 × (-4) = -8 2x2 - 7x - 4 = 2x2 - 8x + 1x - 4 = (2x2 - 8x) + (1x - 4) = 2x(x - 4) + 1(x - 4) = (x - 4)(2x + 1) Thus, 2x2 - 7x - 4 = (x - 4)(2x + 1). Note that (2x + 1)(x - 4) is also correct, since by the commutative law of multiplication, (x - 4)(2x + 1) = (2x + 1)(x - 4)

Some students prefer to use trial-and-error to factor ax2 + bx + c, especially when a or c is a prime number. Start the procedure by writing: _________ ↓ Product c ↓ ax2 + bx + c = (___x + ___)(___x + ___) ↑Product a ↑ Note that: 1. The product of the numbers in the first blanks must be a. 2. The coefficients of the outside product and the inside product must add up to b. 3. The products of the numbers in the last blanks must be c.



For example, to factor 2x2 + 5x + 3, write: __________ ↓ Product 3 ↓ 2x2 + 5x + 3 = (___x + ___)(___x + ___) ↑ Product 2 ↑ We now look for two numbers whose product is 2. The numbers are 2 and 1. Thus, 2x2 + 5x + 3 = (2x + ___)(x + ___) Now, we look for two numbers whose product is 3. The numbers are 3 and 1, which we substitute into the blanks. Here are the possibilities. 2x2 + 5x + 3 = (2x + 3)(x + 1) or 2x2 + 5x + 3 = (2x + 1)(x + 3) Since the middle term must be 5x, the desired factorization is: 2x2 + 5x + 3 = (2x + 3)(x + 1). Note that (2x + 1)(x + 3) = 2x2 + 7x + 3 and not 2x2 + 5x + 3. So far we have factored the trinomial (three terms) ax2 + bx + c. There is a special binomial (two terms) which is easier to factor. Here is the rule.

5 TO FACTOR THE DIFFERENCE OF TWO SQUARES RULE

x2 - y2 = (x + y)(x - y)

This rule says that the difference of the squares of x and y can be factored as the product of the sum of x and y, times the difference of x and y.

EXAMPLES Factor x2 - 9, First write x2 - 32 Now, factor x2 - 32 = (x + 3)(x - 3)

Factor 8x2 - 50 Factor 2 out 2(4x2 - 25) First write 2 [(2x)2 - 52] Now factor, 2(2x + 5)(2x - 5)

�CLAST EXAMPLES

Example Solution 1. Which is a linear factor of 4x2 - 9? A. 2x + 9 B. 2x - 9 C. 2x - 3 D. 3x - 2

The expression 4x2- 9 is the difference of two squares. Write 4x2 - 9 = (2x)2 - 32 Factor = (2x + 3)(2x - 3) The two linear factors of 4x2 - 9 are 2x + 3 and 2x - 3. The answer is C.

2. Which is a linear factor of the following expression?

3x2 - 11x - 4

A. x + 4 B. 3x - 4 C. 3x + 1 D. 3x + 2

1. The key number is 3 � (-4) = -12 2. 1 and -12 are numbers whose product is -12

and whose sum is -11, the coefficient of the middle term -11x., write using 1x and -12x.

3x2 - 11x - 4 = 3x2 + 1x - 12x - 4 3. Group = (3x2 + 1x) + (-12x - 4) 4. Factor = x(3x + 1) -4(3x + 1) 5. Factor GCF = (3x + 1)(x - 4) (3x + 1) is a factor, so the answer is C.



B. Solving Quadratic Equations

Objective IC8 CLAST SAMPLE PROBLEMS 1. Find the real roots of 3x2 - x = 2. 2. Find the real roots of 2x2 + 3x = 1

A quadratic equation in the standard form ax2 + bx + c = 0 can be solved by factoring using the following three steps:

1. Factor 2. Set each of the factors equal to 0 3. Solve the resulting equations.

Thus, to solve 3x2 - x - 4 by factoring, we proceed as follows: 1. Factor 3x2 - x - 4 = 0 has key number 3 × (-4) = -12 3 and -4 are numbers whose product is -12 and whose sum is -1, the coefficient of the middle term -x. Rewrite the middle term using 3 and -4 3x2 - x - 4 = 3x2 + 3x - 4x - 4 = 0 = (3x2 + 3x) + (-4x - 4) = 0 = 3x(x + 1) -4(x + 1) = 0 = (x + 1)(3x - 4) = 0 Now, if the product of (x + 1)(3x - 4) is 0, at least one of the factors must be 0. 2. Set factors equal to 0 : x + 1 = 0 or 3x - 4 = 0

3. Solve for x x = -1 or x = 43

�CLAST EXAMPLE

Example Solution 3. Find the correct solutions to this

equation: 3x2 - 1 = 2x + 4

A. 1 and 35

B. -1 and 35

C. -1 and 53

D. 2 + 62

6 and 2 - 62

6

To write the equation in standard form add -2x - 4 to both sides 3x2 - 1 = 2x + 4 -2x - 4 = -2x - 4 3x2 - 2x - 5 = 0 The key number is -15 and there are two numbers whose product is -15 and whose sum is -2. The numbers are -5 and 3. We now factor 3x2 - 2x - 5 = 0 Write 3x2 - 2x - 5 = 3x2 - 5x + 3x - 5 Group = (3x2- 5x) + (3x - 5) Factor = x(3x - 5) +1(3x - 5) Factor GCF = (3x - 5)(x + 1) Thus, 3x2 - 2x - 5 = (3x - 5)(x + 1) = 0 Solving 3x - 5 = 0 and x + 1 = 0

We have x = 53 and x = -1

The answer is C.

ANSWERS 1. x = -

23 ; x = 1 2. x =

- 3 ± 174



If the given equations are not factorable, we need the quadratic formula.

6 THE QUADRATIC FORMULA RULE

The equation ax2 + bx + c = 0 has two solutions given by

x = -b ± b2 - 4ac

2a

The symbol ± (read "plus or minus") sign means that we have two solutions,

-b + b2 - 4ac

2a and

-b - b2 - 4ac

2a

EXAMPLES

Find the solutions of x2 - 7x + 2 = 0. In this case a = 1, b = -7, c = 2

Thus, x = -(-7) ± (-7)2 - 4(1)(2)

2(1)

= 7 ± 49 - 8

2

= 7 ± 41

2

The two solutions are:

7 + 41

2 and 7 - 41

2

Each of the solutions is called a root of the equation.

One last word. There are many ways in which the CLAST asks you to solve a quadratic equation. Do not be fooled by the terminology. All of these directions mean the same: 1. Find the solutions of the equation. 2. Find the correct solution to the equation. 3. Find the real roots of the equation. 4. Solve.



�CLAST EXAMPLE

Example Solution

4. Find the correct solutions to this equation: 3x2 + 1 = 6x

A. - 3 + 2 3

3 and - 3 - 2 3

3

B. 3 + 2 3

3 and 6 - 30

6

C. -6 + 6

6 and - 3 - 6

3

D. 3 + 6

3 and 3 - 6

3

Note: Make sure you understand the operations with radicals (Section 2.1) before you attempt the problems in this section.

This equation is not in standard form (= 0). We write it in standard form by subtracting 6x from both sides. 3x2 + 1 = 6x -6x = - 6x 3x2 - 6x + 1 = 0 Now, 3x2 - 6x + 1 is not factorable, since there are no numbers whose product is 3 and whose sum -6. Use the quadratic formula. Now, 3x2 - 6x + 1 = 0, so

a = 3, b = -6, c = 1 and x = - b ± b2 - 4ac

2a

Thus, x = -(-6) ± (-6)2 - 4(3)(1)

2(3)

= 6 ± 36 - 12

6

= 6 ± 24

6

= 6 ± 4 ¥ 6

6

= 6 ± 2 6

6

= 3 ± 6

3

Note: To simplify 6 ± 2 6

6 we divided 6, 2 6

and 6 by 2 to obtain 3 ± 6

3 . The answer is D.




�WARM-UPS A IN PROBLEMS 1-14 FIND THE LINEAR FACTORS. 1. 2x2 + 5x + 3 2. 2x2 + 7x + 3 3. 4y2 - 11y + 6 4. 3y2 - 17y + 10 5. 3y2 - 5y - 2 6. 12y2 - y - 6 7. 5x2 + 2 + 11x 8. 5x2 + 3 + 8x 9. 3x2 - 2 - 5x 10. 5x2 - 8 - 6x 11. x2 - 1 12. x2 - 36 13. 9x2 - 4 14. 16x2 - 25

�CLAST PRACTICE A PRACTICE PROBLEMS: Chapter 2, # 27-29 15. Which is a linear factor of: 4x2 - 7x - 15? A. 4x + 5 B. 4x - 6 C. x - 4 D. x + 4

16. Which is a linear factor of: 5x2 - 27x + 10? A. x + 6 B. 5x + 1 C. x - 6 D. 5x - 2 17. Which is a linear factor of: x2 - 16? A. x - 16 B. x - 4 C. x + 4 D. x - 4

18. Which is a linear factor of: x2 - 81? A. x + 9 B. x - 9 C. x - 9 D. x - 81



�WARM-UPS B IN PROBLEMS 19-30 FIND THE CORRECT SOLUTIONS TO THE GIVEN EQUATION. 19. 2x2 + 7x + 3 = 0 20. 2x2 + 5x + 3 = 0 21. 2x2 + 3 = 3 22. 6x2 - x = 12 23. 3x2 - 2 = 5x 24. 12x2 - 6 = x 25. 3y2 = 17y - 10 26. 3y2 = 2y + 1 27. 2x2 + 7x = - 6 28. 2x2 + 7x = 6 29. 7x2 = 12x - 5 30. 5x2 = -16x - 8 �CLAST PRACTICE B PRACTICE PROBLEMS: Chapter 2, #30-32 31. Find the real roots of the equation: 2x2 - 1 = 7x.

A. 7 - 57

4 and 7 + 57

4 B. -7 - 57

4 and -7 + 57

4

C. -7 - 41

4 and -7 + 41

4 D. 7 - 41

4 and 7 + 41

4

32. Find the real roots of the equation: 3x2 + 1 = 5x.

A. 5 - 13

6 and 5 + 13

6 B. 5 - 3 37

6 and 5 + 37

6

C. - 5 - 3

6 and -5 - 3

6 D. - 5 - 37

6 and - 5 + 37

6



33. Solve: 4x2 = 8x + 2.

A. 2 - 6

2 and 2 + 6

2 B. - 2 - 6

2 and - 2 + 6

2

C. 2 - 2

2 and 2 + 2

2 D. - 2 - 2

2 and - 2 + 2

2

34. Find the correct solutions to the equation: 2x2 - 32 = 0. A. 16 and - 16 B. 4 and - 4 C. 2 and -2 D. 32 and - 32 35. Find the correct solutions to the equation: 2x2 - 162 = 0. A. 9 and - 9 B. 81 and - 81 C. 9 and -9 D. 162 and - 162 EXTRA CLAST PRACTICE 36. Find the real roots of the equation: 2x2 + x = 15

A. - 52 , 3 B. 10, -12 C. 5, -6 D.

52 , -3

37. Find the correct solutions to the equation: x2 = - 3x + 2

A. -3 + 17

2 , -3 - 17

2 B. 3 + 17

2 , 3 - 17

2

C. - 3 + 17 D. -3 - 17



2.6 SYSTEMS OF EQUATIONS AND INEQUALITIES In this section we shall solve systems of equations and inequalities in two variables using two methods: elimination and graphing. Before we proceed to do this, we need to know some of the terminology. T TERMINOLOGY -- EQUATIONS IN TWO VARIABLES

SOLUTIONS OF AN EQUATION The solution of an equation in two variables x and y is an ordered pair (a, b) so that if x is replaced by a and y by b in the equation, the result is a true statement. We say that (a, b) satisfies the equation. Note that the equation ax + by = c has infinitely many solutions.

EXAMPLES The equation 2x + 3y = 12 is an equation in two variables and the ordered pair (3, 2) is a solution (satisfies) the equation. If we replace x by 3 and y by 2, we have 2•3 + 3•2 = 12, a true statement. Note that (0, 4) and (6, 0) are also solutions of 2x + 3y = 12.

SYSTEMS OF LINEAR EQUATIONS A system of two linear equations ax + by = c dx + ey = f is called a system of simultaneous linear equations.

EXAMPLES x + y = 5 x – y = 1 is a system of simultaneous linear equations in the variables x and y.

SOLUTIONS OF A SYSTEM

The solution set of a system of linear equations consists of the ordered pairs satisfying both equations and is written using set notation by listing the ordered pairs separated by commas inside the set symbols { } (read "braces").

EXAMPLES

The solution set of the system x + y = 5 x - y = 1 is {(3, 2)}. Note that the ordered pair (3, 2) satisfies both equations. Thus, 3 + 2 = 5 and 3 - 2 = 1

If a system of equations has no solution, its solution set is the empty set φ

The system x + y = 5 and -x - y = 1 has no solution. (If you add the left sides of both equations, you get 0 while the addition of the right sides of both equations gives 6. This means that 0 = 6, which is impossible. Thus, the solution set is φ

How did we solve the systems in the Examples? The procedure will be explained next.

SECTION 2.6 Systems of Equations and Inequalities 107


A. Solving Systems of Equations

Objective IC9 CLAST SAMPLE PROBLEMS SOLVE THE SYSTEM

1. 2x - y = 2 2. 2x + y = 2 3. 2x - y = 2 3x - 2y = 1 4x + 2y = 3 4x - 2y = 4

1 SOLVING SYSTEMS BY ELIMINATION RULE

You can multiply (or divide) one or both of the equations by any nonzero number you wish, but the idea is to obtain an equivalent system in which the coefficients of the x's (or of the y's) are opposites, thus eliminating x or y when the equations are added.

EXAMPLES Solve the system x + y = 3 x - y = -1 Note that the coefficients of y are opposites already. Thus, we simply add the equations, as shown next: x + y = 3 x - y = -1 Add the equations. 2x = 2 Divide by 2. x = 1 Replace x by 1 in x + y = 3 1 + y = 3 Subtract 1 to find y. y = 2 The solution is (1, 2) or the set is (1,2)}.

�CLAST EXAMPLE

Example Solution 1. Choose the correct solution set for the

system of linear equations:

x + 4y = -1 4x + y = 11

A. {(3,-1)} B. {(3,1)} C. The empty set

D. {(x, y) | y = -4x + 11}

If we want to eliminate the x's, we can multiply the first equation by -4 and then add. Here is the work: x + 4y = -1 Mult. -4 -4x - 16y = 4 4x + y = 11 Leave as is 4x + y = 11 Add the equations. -15y = 15 Divide by -15. y = -1 Replace y by -1 in 4x + y = 11. 4x + (-1) = 11 Add 1 to both sides. 4x = 12 Divide by 4, x = 3 The solution set is {(3, -1)} and the answer is A.

So far, we have multiplied only one of the equations by a number to eliminate the variable. Sometimes we must multiply both equations by numbers that will cause the coefficients of one of the variables to be opposites of each other as illustrated in the next example.

ANSWERS 1. (3, 4) 2. No solution 3. Infinitely many solutions



�CLAST EXAMPLES

Example Solution 2. Choose the correct solution set for the

system of linear equations: 2x + 2y = -4 -3x - 3y = 9 A. {(-1,-1)} B. {(-2,-1)} C. The empty set D. {(x,y) | y = -x - 3 }

Since the coefficient of x in the first equation is 2 and in the second equation -3, we multiply the first equation by 3 and the second one by 2 so that the resulting coefficients of x are 6 and -6, and the x is eliminated by addition. Thus, 2x + 2y = -4 Mult. 3 6x + 6y = -12 -3x - 3y = 9 Mult. 2 -6x - 6y = 18 Add the equations 0 = 6 But this is impossible, so there is no solution and the solution set is empty. The answer is C. You may have also noticed that dividing the first equation by 2 and the second by -3 yields x + y = -2 and x + y = -3, which is also impossible.

Example Solution 3. Find the solution set for the system of

linear equations: 2y - 3x = -10 6x - 4y = 20 A. The empty set B. {(3,-1)}

C. {(x,y) | y = 32 x - 5}

D. {(3,1)} Note: If you multiply the first equation by -2, you get the second equation. The answer has to be C. Look at the solution to see why.

The first equation lists the y's first, then the x's. Rewrite this equation with the x's first obtaining the system: -3x + 2y = -10 6x - 4y = 20 Now, multiply the first equation by 2 to try to eliminate the x's -3x + 2y = -10 Mult. 2 -6x + 4y = -20 6x - 4y = 20 6x - 4y = 20 Add the equations 0 = 0 Since we obtain a true statement regardless of the values of x and y we have infinitely many solutions. In such cases, solve either equation for y. We have: 2y - 3x = -10

Add 3x and divide by 2 y = 32 x - 5

The answer is C.

SECTION 2.6 Systems of Equations and Inequalities 109


B. Graphing Equations and Inequalities Objective IIC4 CLAST SAMPLE PROBLEMS

SHADE THE REGION OF THE COORDINATE PLANE SATISFYING: 1. 2x - y < 4 2. y > 3 3. x + 2y > 0 and x < 2 4. x > 3 or y < 0

The systems of linear equations we have discussed can be solved graphically. Since the solutions of a linear equation in two variables are ordered pair of numbers, we now learn how to make a picture (graph) of these ordered pairs. T GRAPHING ORDERED PAIRS

THE COORDINATE PLANE To graph the ordered pair (a, b) draw a number line and label the points as shown in the Figure. Draw another number line perpendicular to the first one and crossing at 0, the origin. Every point in the plane determined by these lines can be associated with an ordered pair of numbers. The horizontal number line is the x-axis and the vertical number line is the y-axis. The whole system is a Cartesian coordinate system, a coordinate plane, or simply a plane.

EXAMPLES

5 –5

5

–5

x

yx-axis

y-axis

To graph the ordered pair (a, b), start at the origin and go a units to the right if a is positive; to the left if a is negative. Then go b units up if b is positive, down if b is negative.

The point P in the figure is associated with the ordered pair (2, 3) and it is 2 units right and 3 units up. The point Q is associated with the ordered pair (-1, 2), one unit left and 2 units up. 5–5

5

–5

P(2, 3)Q(-1, 2)

ANSWERS

5–5

5

–5

2- 4

1.

5–5

5

–5

2.

5–5

5

–5

3.

5–5

5

–5

4.



1 GRAPHS OF LINES GRAPHS OF LINES

The graph of Ax + By = C is a straight line, and every straight line has an equation that can be written in this form. (A, B and C are real numbers and A and B are not both 0.) When A = 0, and B = 1, y = C is a horizontal line. When B = 0 and A = 1, x = C is a ve

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ALGEBRA - CengageIC4a: Solve linear equations IC4b: Solve linear inequalities IC5: Use given formulas to compute results, when geometric measurements are not involved IC6: Find particular