This document provides examples of how to translate word phrases and sentences into algebraic expressions and equations and then solve application problems involving one or more unknowns. It demonstrates translating word phrases into expressions using variables, writing sentences as equations, and solving word problems by setting up and solving equations. Sample word problems include finding an unknown number, the number of windows two people washed, and the dimensions of a rectangle given its perimeter. The examples show each step of the problem-solving process.

1. Chapter 9
Basic Algebra
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All rights reserved.

2. 9.8 Using Equations to Solve Application Problems
Objectives
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. Slide 9.8- 2

3. Parallel
Example 1
Translating Word Phrases into
Expressions with Variables
Write each word phrase in symbols, using x as the
variable.
Words Algebraic Expression
A number plus nine x + 9 or 9 + x
7 more than a number x + 7 or 7 + x
−12 added to a number −12 + x or x + (−12)
3 less than a number x–3
A number decreased by 1 x–1
14 minus a number 14 – x
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4. Parallel
Example 2
Translating Word Phrases into
Expressions with Variables
Write each word phrase in symbols, using x as the
variable.
Words Algebraic Expression
3 times a number 3x
Twice a number 2x
The quotient of 8 and a 8
number x
A number divided by 15 x
15
The result is =
Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 4

5. Parallel
Example 2
Translating a Sentence into an
Equation
If 8 times a number is added to 13, the result is 45.
Find the number.
Let x represent the unknown number.
8 times a number added to 13 is 45
8x + 13 = 45
Next, solve the equation. Check:
8x + 13 − 13 = 45 − 13 8x + 13 = 45
8x = 32 8(4) + 13 = 45
8 x 32
= 45 = 45
8 8
x=4 The solution is 4.
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6. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 6

7. Parallel
Example 5
Solving an Application Problem with
One Unknown
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?
Step 1 Read. The problem asks for the number of
windows that Rita has washed.
Step 2 Assign a variable. There is only one
unknown, Rita’s number of windows washed.
Step 3 Write an equation.
The number Frankie 6 less than twice
washed. Rita’s number.
14 = 2x – 6
Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 7

8. Parallel
Example 5
Solving an Application Problem with
continued One Unknown
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?
Step 4 Solve. 14 = 2x – 6
14 + 6 = 2x – 6 + 6
20 = 2x
20 2 ×x
=
2 2
10 = x
Step 5 State the answer. Rita washed 20
windows.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 8

9. Parallel
Example 5
Solving an Application Problem with
continued One Unknown
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?
Step 6 Check. 14 = 2x – 6
14 = 2(10) – 6
14 = 14
So 10 is the correct solution because it
“works” in the original problem.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 9

10. Parallel
Example 6
Solving an Application Problem with
Two Unknowns
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.
Step 1 Read. The problem asks for the amount
spent by each person.
Step 2 Assign a variable. There are two unknowns.
Let x represent the amounts spent by Lowell
and x + 54 be the amount spent by Yoshi.
Step 3 Write an equation. Amount spent by
Amount spent Yoshi.
by Lowell
x + x + 54 = 276
Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 10

11. Parallel
Example 6
Solving an Application Problem with
continued Two Unknowns
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.
Step 4 Solve. x + x + 54 = 276
2x + 54 = 276
2x + 54 − 54 = 276 − 54
2x = 222
1
2 x 222
=
2 2
1
x = 111
Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 11

12. Parallel
Example 6
Solving an Application Problem with
continued One Unknown
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.
Step 6 Check. Use the words in the original
problem.
Yoshi’s $165 is $54 more dollars than Lowell’s
$111, so that checks.
The total spent is $111 + $165 = $276 which
also checks.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 12

13. Parallel
Example 7
Solving a Geometry Application
Problem
The length of a rectangle is 3 inches more than
the width. The perimeter is 78 inches. Find the
length and width.
Step 1 Read. The problem asks for the length and
width of the rectangle.
Step 2 Assign a variable. There are two unknowns,
length and width. Let x represent the width
and x + 3 represent the length.
Step 3 Write an equation.
Use the formula for
P = 2l + 2w perimeter of a rectangle.
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14. Parallel
Example 7
Solving a Geometry Application
continued Problem
The length of a rectangle is 3 inches more than the width. The perimeter is 78
inches. Find the length and width.
Step 4 Solve. P = 2l + 2w
78 = 2(x + 3) + 2 ∙ x
78 = 2x + 6 + 2x
78 = 4x + 6
78 – 6 = 4x + 6 – 6
72 = 4x
72 14 ×x
=
4 41 18 = x
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15. Parallel
Example 7
Solving a Geometry Application
continued 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.
Step 6 Check. Use the words in the original
problem.
The original problem says that the perimeter
is 78 inches.
P = 2 ∙ 18 in. + 2 ∙21in.
P = 36 in. + 42 in.
P = 78 in. checks
Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 15