Lesson 9.8 (word problems)

Chapter 9

Basic Algebra

© 2010 Pearson Education, Inc.All rights reserved.

9.8 Using Equations to Solve Application Problems

Objectives

Slide 9.8- 2

1. Translate word phrases into expressions

with variables.

2. Translate sentences into equations.

3. Solve application problems.

Copyright © 2010 Pearson Education, Inc. All rights reserved.

ParallelExample 1

Translating Word Phrases into Expressions with Variables

Slide 9.8- 3

Write each word phrase in symbols, using x as the variable.

Words Algebraic Expression

A number plus nine

7 more than a number

−12 added to a number

3 less than a number

A number decreased by 1

14 minus a number

x + 9 or 9 + x

x + 7 or 7 + x

−12 + x or x + (−12)

x – 3

x – 1

14 – x


ParallelExample 2

Translating Word Phrases into Expressions with Variables

Slide 9.8- 4

Write each word phrase in symbols, using x as the variable.

Words Algebraic Expression

3 times a number

Twice a number

The quotient of 8 and a number

A number divided by 15

The result is

3x

2x

8

x

15

x

=


ParallelExample 2

Translating a Sentence into an Equation

Slide 9.8- 5

If 8 times a number is added to 13, the result is 45. Find the number.Let x represent the unknown number.

Check:8x + 13 − 13 = 45 − 13

8 times a number

Next, solve the equation.8x

added to

+

13

13

is

=

45

45

8x = 328 32

8 8

x

4x

8x + 13 = 45

8(4) + 13 = 45

45 = 45

The solution is 4. Copyright © 2010 Pearson Education, Inc. All rights reserved.

Slide 9.8- 6 Copyright © 2010 Pearson Education, Inc. All rights reserved.

Frankie has washed 6 less than twice as many windows washed as Rita. If Frankie has washed 14 windows, how many windows has Rita washed?

ParallelExample 5

Solving an Application Problem with One Unknown

Step 1

Step 2

Slide 9.8- 7

Read. The problem asks for the number of windows that Rita has washed.

Assign a variable. There is only one unknown, Rita’s number of windows washed.

Step 3 Write an equation.

14 = 2x – 6

The number Frankie washed.

6 less than twice Rita’s number.



ParallelExample 5continued


Step 4

Step 5 State the answer. Rita washed 20 windows.

Slide 9.8- 8

Solve. 14 = 2x – 614 + 6 = 2x – 6 + 6

20 = 2x20

2

2

2

x

10 x




Step 6

So 10 is the correct solution because it “works” in the original problem.

Slide 9.8- 9

Check. 14 = 2x – 6

14 = 2(10) – 6

14 = 14



On a shopping spree, Yoshi spent $54 more than Lowell. The total spent by them both was $276. Find the amount spent by each person.

ParallelExample 6

Solving an Application Problem with Two Unknowns

Step 1

Step 2

Slide 9.8- 10

Read. The problem asks for the amount spent by each person.

Assign a variable. There are two unknowns. Let x represent the amounts spent by Lowell and x + 54 be the amount spent by Yoshi.


x + x + 54 = 276Amount spent by Lowell

Amount spent by Yoshi.



Solving an Application Problem with Two Unknowns

Step 4

Slide 9.8- 11

Solve.

2 2

2

22

2

x

On a shopping spree, Yoshi spent $54 more than Lowell. The total spent by them both was $276. Find the amount spent by each person.

x + x + 54 = 276

2x + 54 = 276

2x + 54 − 54 = 276 − 54

2x = 2221

1

111x




Step 6

Yoshi’s $165 is $54 more dollars than Lowell’s $111, so that checks. The total spent is $111 + $165 = $276 which also checks.

Slide 9.8- 12

Check. Use the words in the original problem.

Step 5 State the answer. The amount Lowell spent is x, so Lowell spent $111. The amount Yoshi spent is x + 54, so Yoshi spent $165.


The length of a rectangle is 3 inches more than the width. The perimeter is 78 inches. Find the length and width.

ParallelExample 7

Solving a Geometry Application Problem

Step 1

Step 2

Slide 9.8- 13

Read. The problem asks for the length and width of the rectangle.

Assign a variable. There are two unknowns, length and width. Let x represent the width and x + 3 represent the length.


P = 2l + 2wUse the formula for perimeter of a rectangle.



Step 4

Slide 9.8- 14

Solve.

72

4

4

4

x

1

1


The length of a rectangle is 3 inches more than the width. The perimeter is 78 inches. Find the length and width.

P = 2l + 2w

78 = 2(x + 3) + 2 ∙ x

78 = 2x + 6 + 2x

78 = 4x + 6

78 – 6 = 4x + 6 – 672 = 4x

18 = x



Step 6

The original problem says that the perimeter is 78 inches.

Slide 9.8- 15

Check. Use the words in the original problem.

Step 5 State the answer.


The width is x, so the width is 18 inches.

The length is x + 3, so the length is 21 inches.

P = 2 ∙ 18 in. + 2 ∙21in.

P = 36 in. + 42 in.P = 78 in. checks Copyright © 2010 Pearson Education, Inc. All rights reserved.