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© 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 6 Algebra: Equations and Inequalities

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CHAPTER 6. Algebra: Equations and Inequalities. 6.3. Applications of Linear Equations. Objectives Use linear equations to solve problems. Solve a formula for a variable. Strategy for Solving Word Problems. - PowerPoint PPT Presentation

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Page 1: CHAPTER 6

© 2010 Pearson Prentice Hall. All rights reserved.

CHAPTER 6

Algebra: Equations and Inequalities

Page 2: CHAPTER 6

© 2010 Pearson Prentice Hall. All rights reserved. 2

6.3

Applications of Linear Equations

Page 3: CHAPTER 6

© 2010 Pearson Prentice Hall. All rights reserved.3

Objectives

1. Use linear equations to solve problems.

2. Solve a formula for a variable.

Page 4: CHAPTER 6

© 2010 Pearson Prentice Hall. All rights reserved.4

Strategy for Solving Word Problems

Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the quantities in the problem.

Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

Step 3 Write an equation in x that models the verbal conditions of the problem.

Step 4 Solve the equation and answer the problem’s question.

Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Page 5: CHAPTER 6

© 2010 Pearson Prentice Hall. All rights reserved. 5

Algebraic Translations of English Phrases

Addition Subtraction Multiplication Divisionsum

more than

increased by

minus

decreased by

subtracted from

difference between

less than

fewer than

times

product of

percent of a number

multiplied by

twice

divided by

quotient

reciprocal

Page 6: CHAPTER 6

© 2010 Pearson Prentice Hall. All rights reserved.6

Example 1: Education Pays Off

This graph shows the average yearly earnings in the United States by highest educational attainment.The average yearly salary of a man with an associate degree exceeds that of a man with some college by $3 thousand. The average yearly salary of a man with a bachelor’s degree or more exceeds that of a man with some college by $41 thousand. Combined, three men with these educational attainments earn $188 thousand. Find the average yearly salary of men with each of these levels of education.

Page 7: CHAPTER 6

© 2010 Pearson Prentice Hall. All rights reserved.7

Example 2: continued

Step 1: Let x represent one of the unknown quantities.Let x = the average yearly salary of a man with some college.

Step 2: Represent the other unknown quantities in terms of x.x + 3 = the average yearly salary of a man with an associate degreex + 41 = the average yearly salary of a man with a bachelor’s degree or more.

Page 8: CHAPTER 6

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Example 2: continued

Step 3: Write an equation in x that models the conditions.x + (x + 3) + (x + 41) = 188

Step 4: Solve the equation and answer the question.

( 3) ( 41) 188

3 44 188

3 144

48

x x x

x

x

x

Page 9: CHAPTER 6

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Example 2: continued

The average salary with some college = 48The average salary with an associate degree = x + 3

= 48 + 3 = 51The average salary with a bachelor’s degree or more

= x + 41 = 48 + 41 = 89.

Some college = $48 thousand per yearAssociate degree = $51 thousandBachelor’s degree = $89 thousandStep 5: Check the proposed solution in the wording

of the problem. The solution checks.

Page 10: CHAPTER 6

© 2010 Pearson Prentice Hall. All rights reserved.10

Example 6: Solving a Formula for One of its Variables

The total price of an article purchased on a monthly deferred payment plan isdescribed by the following formula:

T is the total price, D is the down payment, p is the monthly payment, and m is the number of months onepays. Solve the formula for p.

T – D = D – D + pm

T – D = pm

T – D = pm m m

T – D = p m