Chapter 2 Section 4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Chapter Chapter 22Section Section 44

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley


An Introduction to Applications of Linear Equations

Learn the six steps for solving applied problems.Solve problems involving unknown numbers.Solve problems involving sums of quantities.Solve problems involving supplementary and complementary angles.Solve problems involving consecutive integers.

11

44

33

22

55

2.42.42.42.4


Objective 11

Slide 2.4 - 3

Learn the six steps for solving applied problems.


Learn the six steps for solving applied problems.We now look at how algebra is used to solve applied problems.

While there is no one specific method that enables you to solve all kinds of applied problems, the following six-step method is often applicable.

Step 1: Read the problem carefully until you understand what is given and what is to be found.

Step 2: Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the variable represents. If necessary, express any other unknown values in terms of the variable.

Step 3: Write an equation using the variable expression(s).Step 4: Solve the equation.Step 5: State the answer. Does it seem reasonable?Step 6: Check the answer in the words of the original problem.

Slide 2.4 - 4


The third step in solving an applied problem is often the hardest. To translate the problem into an equation, write the given phrases as mathematical expressions. Replace any words that mean equals or same with an = sign.

Slide 2.4 - 5

Other forms of the verb “to be,” such as is, are, was, and were, also translate this way. The = sign leads to an equation to be solved.

Learn the six steps for solving applied problems. (cont’d)


Objective 22

Slide 2.4 - 6

Solve problems involving unknown numbers.


EXAMPLE 1

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When solving an equation, use solution set notation to write the answer. When solving an application, state the answer in a sentence.

Solution:

Let x = the number.

Finding the Value of an Unknown Number

If 5 is added to the product of 9 and a number, the result is 19 less than the number. Find the number.

9 5 19x x

9 5 19x xx x 8 5 195 5x

8 2

8 8

4x

3x The number is −3.


Objective 33

Slide 2.4 - 8

Solve problems involving sums of quantities.


A common type of problem in elementary algebra involves finding two quantities when the sum of the quantities is known. In Example 9 of Section 2.3, we prepared for this type of problem, by writing mathematical expressions for two related unknown quantities.

Solve problems involving sums of quantities.

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PROBLEM-SOLVING HINTPROBLEM-SOLVING HINT

To solve problems involving sums of quantities, choose a variable to represent one of the unknowns. Represent the other quantity in terms of the same variable, using information from the problem. Write an equation based on the words of the problem.


EXAMPLE 2

Slide 2.4 - 10Slide 2.4 - 10

Solution:Let x = the number of medals Norway won.

Finding Numbers of Olympics Medals

In the 2006 Winter Olympics in Torino, Italy, the United States won 6 more medals than Norway. The two countries won a total of 44 medals. How many medals did each country win? (Source: U.S. Olympic Committee.)

44 6x x 44 26 66x

38

2 2

2x

19x

Norway won 19 medals and the U.S. won 25 medals.

Let x + 6 = the number of medals the U.S. won.

19 6 25


Solve problems involving sums of quantities. (cont’d)

Slide 2.4 - 11

The problem in example 2 could also be solved by letting x represent the number of medals the United States won. Then x − 6 would represent the number of medals Norway won. The equation would be

44 6x x The solution to this equation is 25, which is the number of

medals the U.S. won. The number of Norwegian medals would be 25 − 6 = 19. The answers are the same, whichever approach is used.

The nature of the applied problem restricts the set of possible solutions. For example, an answer such as −33 medals or 25 medals should be recognized as inappropriate.

1

2


EXAMPLE 3

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Finding the Number of Orders for Croissants

On that same day, the owner of Terry’s Coffeehouse found that the number of orders for croissants was the number of muffins. If the total number for the two breakfast rolls was 56, how many orders were placed for croissants?Solution:Let x = the number of muffins.Then x = the number of croissants.

1

61

566

x x

6 6 61

566

x x

336 6x x 336 7

7 7

x

48x

1

648 8

8 croissants were ordered.

1

6


Finding the Number of Orders for Croissants (cont’d)

PROBLEM SOLVING HINTPROBLEM SOLVING HINTIn Example 3, it was easier to let the variable represent the

quantity that was not specified. This required extra work in Step 5 to find the number of orders for croissants. In some cases, this approach is easier than letting the variable represent the quantity that we are asked to find.

Slide 2.4 - 13


EXAMPLE 4

Solution:

Let x = number of members.

Then 2x = number of nonmembers.

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3 27

3 3

x

2 27x x

9x

2 9 18

Analyzing the Mixture of a Computer User Group

At a meeting of the local computer user group, each member brought two nonmembers. If a total of 27 people attended, how many were nonmembers?

There were 9 members and 18 nonmembers at the meeting.


Dividing a Board into Pieces

PROBLEM SOLVING HINTPROBLEM SOLVING HINTSometimes it is necessary to find three unknown quantities in

an applied problem. Frequently, the three unknowns are compared in pairs. When this happens it is usually easiest to let the variable represent the unknown found in both pairs.

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EXAMPLE 5

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Solution:Let x = the length of the middle-sized piece, then x +10 = longest piece,and x − 5 = shortest piece.

10 5 50x x x 55 50 53x

3 5

3 3

4x

Dividing a Pipe into Pieces

A piece of pipe is 50 in. long. It is cut into three pieces. The longest piece is 10 in. more than the middle-sized piece, and the shortest piece measures 5 in. less than the middle-sized piece. Find the lengths of the three pieces.

15x

15 10 25

15 5 10

The shortest piece is 10 in., the middle-size piece is 15 in., and the longest is 25 in.


Objective 44

Slide 2.4 - 17

Solve problems involving supplementary and complementary angles.


Solve problems involving supplementary and complementary angles.

An angle can be measured by a unit called the degree (°), which is of a complete rotation.

Two angles whose sum is 90° are said to be complementary, or complements of each other. An angle that measures 90° is a right angle.

Two angles who sum is 180° are said to be supplementary, or supplements of each other. One angle supplements the other to form a straight angle of 180°.

Slide 2.4 - 18

1

360


Solve problems involving supplementary and complementary angles. (cont’d)

If x represents the degree measure of an angle, then

90 − x represents the degree measure of its complement,

and 180 − x represents the degree measure of is supplement.

Slide 2.4 - 19


EXAMPLE 6

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Solution:Let x = the degree measure of the angle.Then 90 − x = the degree measure of its

complement,and 180 − x = the degree measure of its supplement.

90 180 174x x 2702 270 2 70174x 2 96

2 2

x

Finding the Measure of an Angle

Find the measure of an angle such that the sum of the measures of its complement and its supplement is 174°.

48x The measure of the angle is 48°.


Objective 55

Slide 2.4 - 21

Solve problems involving consecutive integers.


Solve problems involving consecutive integers.Two integers that differ by 1 are called consecutive integers.

For example, 3 and 4, 6 and 7, and −2 and −1 are pairs of consecutive integers. In general, if x represents an integer, x + 1 represents the next larger consecutive integer.

Slide 2.4 - 22

Consecutive even integers, such as 8 and 10, differ by 2. Similarly, consecutive odd integers, such as 9 and 11, also differ by 2. In general if x represents an even integer, x + 2 represents the larger consecutive integer. The same holds true for odd integers; that is if x is an odd integer, x + 2 is the larger odd integer.


Solve problems involving consecutive integers. (cont’d)

PROBLEM SOLVING HINTPROBLEM SOLVING HINTIn solving consecutive integer problems, if x = the first integer,

then, for any

two consecutive integers, use

two consecutive even integers, use

two consecutive odd integers, use

Slide 2.4 - 23

, 1;x x , 2;x x , 2.x x


EXAMPLE 7

Slide 2.4 - 24

Solution:Let x = the lesser page number.Then x + 1= the greater page number.

1 569x x 1 12 1 569x

2 8

2 2

56x

Finding Consecutive Integers

Two back-to-back page numbers in this book have a sum of 569. What are the page numbers?

284x

The lesser page number is 284, and the greater page number is 285.

184 82 2 5

It is a good idea to use parentheses around x + 1, (even though they are not necessary here).


6 2 86x x

EXAMPLE 8

Slide 2.4 - 25

Find two consecutive even integers such that six times the lesser added to the greater gives a sum of 86.Solution:

Let x = the lesser integer.Then x + 2 = the greater integer.

The lesser integer is 12 and the greater integer is 14.

7 4

7 7

8x

22 86 27x

12 2 14

12x

Finding Consecutive Even Integers

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Chapter 2 Section 4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley