1 Copyright © Cengage Learning. All rights reserved

1Copyright © Cengage Learning. All rights reserved. Copyright © Cengage Learning. All rights reserved.

Equations, Inequalities, and Problem Solving

2Copyright © Cengage Learning. All rights reserved.

3.2 Equations that Reduce to Linear Form

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What You Will Learn

Solve linear equations containing symbols of grouping

Solve linear equations involving fractions

Solve linear equations involving decimals

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Equations Containing Symbols of Grouping

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In this section you will continue your study of linear equations by looking at more complicated types of linear equations.

To solve a linear equation that contains symbols of grouping, use the following guidelines.

1. Remove symbols of grouping from each side by using

the Distributive Property.

2. Combine like terms.

3. Isolate the variable using properties of equality.

4. Check your solution in the original equation.

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Example 1 – Solving Linear Equations Involving Parenthesis

Solve 4(x – 3) = 8. Then check your solution.

Solution:

4(x – 3) = 8

4 x – 4 3 = 8

4x – 12 = 8

4x – 12 + 12 = 8 + 12

4x = 20

x = 5

Write original equation.

Distributive Property

Simplify.

Add 12 to each side.

Combine like terms.

Divide each side by 4.

Simplify.

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Example 1 – Solving Linear Equations Involving Parenthesiscont’d

Check

4(5 – 3) 8

4(2) 8

8 = 8

The solution is x = 5.

Substitute 5 for x in original equation.

Simplify.

Solution checks.

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The linear equation in the next example involves both

brackets and parentheses.

Watch out for nested symbols of grouping such as these.

The innermost symbols of grouping should be removed

first.


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Example 2 – Equations Involving Symbols of Grouping (a)

a. Solve 5(x + 2) = 2(x – 1)

Solution:

5(x + 2) = 2(x – 1)

5x + 10 = 2x – 2

5x – 2x + 10 = 2x – 2x – 2

3x + 10 = – 2

3x + 10 – 10 = – 2 – 10

3x = –12

x = –4

Original equation


Subtract 2x from each side

Combine like terms

Subtract 10 from each side

Combine like terms

Divide each side by 3

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Example 2 – Equations Involving Symbols of Grouping (b)

b. Solve 2(x – 7) – 3(x + 4) = 4 – (5x – 2)

Solution:

2(x – 7) – 3(x + 4) = 4 – (5x – 2)

2x – 14 – 3x – 12 = 4 – 5x – 2

–x – 26 = –5x + 6

–x + 5x – 26 = –5x + 5x + 6

4x – 26 = 6

4x – 26 + 26 = 6 + 26

4x = 32

x = 8

Original equation


Combine like terms

Add 5x to each side

Combine like terms

Add 26 to each side

Combine like terms


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Example 2 – Equations Involving Symbols of Grouping (c)

c. Solve 5x – 2[4x + 3(x – 1)] = 8 – 3x.

Solution:

5x – 2[4x + 3(x – 1)] = 8 – 3x

5x – 2[4x + 3x – 3] = 8 – 3x

5x – 2[7x – 3] = 8 – 3x

5x – 14x + 6 = 8 – 3x

–9x + 6 = 8 – 3x

–9x + 3x + 6 = 8 – 3x + 3x



Combine like terms inside brackets.


Combine like terms.

Add 3x to each side.

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cont’d

–6x + 6 = 8

–6x + 6 – 6 = 8 – 6

–6x = 2

The solution is x = . Check this in the original equation.

Combine like terms.

Subtract 6 from each side.

Combine like terms.

Divide each side by –6.

Example 2 – Equations Involving Symbols of Grouping (c)

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Equations Involving Fractions

To solve a linear equation that contains one or more fractions, it is usually best to first clear the equation of fraction by multiplying each side by the least common multiple (LCM) of the denominators.

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Example 3 – Solving Linear Equations Involving Fractions (a)

a. Solve

Solution:

Original Equations

Multiply each side by LCM 6


Simplify

Add 2 to each side


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Example 3 – Solving Linear Equations Involving Fractions (b)

b. Solve

Solution:

Original Equations



Simplify

Add 2 to each side

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c. Solve

Solution:

Example 3 – Solving Linear Equations Involving Fractions (c)

Original Equations



Simplify

Subtract 2 to each side


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Equations Involving Fractions

A common type of linear equation is one that equates two

fractions.

To solve such an equation, consider the fractions to be

equivalent and use cross-multiplication.

That is, if

then a d = b c.

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Example 4 – Finding a Test Score

To get an A in a course, you must have an average of at least 90 points for 4 tests of 100 points each. For the first 3 tests, your scores were 87, 92, and 94. What must you score on the fourth test to earn a 90% average for the course?

Solution:

Verbal Model:

Labels: Score of 4th test = x (points)

Score of first 3 tests: 87, 82, 84 (points)

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Example 4 – Finding a Test Score

Equation:

You can solve this equation by multiplying each side by 4

You need a score of 97 on the fourth test to earn a 90% average

cont’d

Write equation


Combine like terms

Simplify

Subtract 263 from each side

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Equations Involving Decimals

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Equations Involving Decimals

Many real-life applications of linear equations involve

decimal coefficients.

To solve such an equation, you can clear it of decimals in

much the same way you clear an equation of fractions.

Multiply each side by a power of 10 that converts all

decimal coefficients to integers.

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Example 5 – Solving a Linear Equation Involving Decimals

Solve 0.3x + 0.2(10 – x) = 0.15(30). Then check your solution.

Solution:

0.3x + 0.2(10 – x) = 0.15(30)

0.3x + 2 – 0.2x = 4.5

0.1x + 2 = 4.5

10(0.1x + 2) = 10(4.5)

x + 20 = 45

x = 25



Combine like terms.

Multiply each side by 10.

Simplify.

Subtract 20 from each side.

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cont’d

Example 5 – Solving a Linear Equation Involving Decimals

Check

0.3(25) + 0.2(10 – 25) 0.15(30)

0.3(25) + 0.2(–15) 0.15(30)

7.5 – 3.0 4.5

4.5 = 4.5

The solution is x = 25.

Substitute 25 for x in original equation.

Perform subtraction withinparentheses.

Multiply.

Solution checks.

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Example 7 – Finding Your Gross Pay per Paycheck

The enrollment y (in millions) at postsecondary schools from 2000 through 2009 can be approximated by the linear model y = .035 + 9.1, where t represents the year, with t = 0 corresponding to 2000. Use the model to predict the year in which the enrollment will be 14 million students. (Source: U.S. Department of Education)

Solution:

To find the year in which the enrollment will be 14 million students, substitute 14 for y in the original equation and solve the equation for t.

14 = 0.35t + 9.1.

4.9 = 0.35t

14 = t

Because t = 0 corresponds to 2000, the enrollment at postsecondary schools will be 14 million during 2014.

Substitute 14 for y in original equation

Subtract 9.1 from each side

Divide each side by 0.35

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