Contents

Preface
1Foundations
- Introduction
- 1.1Use the Language of Algebra
- 1.2Integers
- 1.3Fractions
- 1.4Decimals
- 1.5Properties of Real Numbers
- Chapter Review
  Key Terms
  Key Concepts
- Exercises
  Review Exercises
  Practice Test
2Solving Linear Equations
3Graphs and Functions
4Systems of Linear Equations
5Polynomials and Polynomial Functions
6Factoring
7Rational Expressions and Functions
8Roots and Radicals
9Quadratic Equations and Functions
10Exponential and Logarithmic Functions
11Conics
12Sequences, Series and Binomial Theorem
- Introduction
- 12.1Sequences
- 12.2Arithmetic Sequences
- 12.3Geometric Sequences and Series
- 12.4Binomial Theorem
- Chapter Review
  Key Terms
Answer Key
Index

Use a General Strategy to Solve Linear Equations

Learning Objectives

By the end of this section, you will be able to:

Solve linear equations using a general strategy
Classify equations
Solve equations with fraction or decimal coefficients

Be Prepared 2.1

Before you get started, take this readiness quiz.

Simplify: $\frac{3}{2} (12 x + 20) . \frac{3}{2} (12 x + 20) .$
If you missed this problem, review Example 1.51.
Simplify: $5 - 2 (n + 1) . 5 - 2 (n + 1) .$
If you missed this problem, review Example 1.55.
Find the LCD of $\frac{5}{6} \frac{5}{6}$ and $\frac{1}{4} . \frac{1}{4} .$
If you missed this problem, review Example 1.28.

Solve Linear Equations Using a General Strategy

Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that makes it a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!

Solution of an Equation

A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

To determine whether a number is a solution to an equation, we substitute the value for the variable in the equation. If the resulting equation is a true statement, then the number is a solution of the equation.

How To

Determine Whether a Number is a Solution to an Equation.

Step 1. Substitute the number for the variable in the equation.
Step 2. Simplify the expressions on both sides of the equation.
Step 3.
Determine whether the resulting equation is true.
- If it is true, the number is a solution.
- If it is not true, the number is not a solution.

Example 2.1

Determine whether the values are solutions to the equation: $5 y + 3 = 10 y - 4 . 5 y + 3 = 10 y - 4 .$

ⓐ $y = \frac{3}{5} y = \frac{3}{5}$ ⓑ $y = \frac{7}{5} y = \frac{7}{5}$

Show solution

Solution

Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.

ⓐ



Multiply.
Simplify.

Since $y = \frac{3}{5} y = \frac{3}{5}$ does not result in a true equation, $y = \frac{3}{5} y = \frac{3}{5}$ is not a solution to the equation $5 y + 3 = 10 y - 4 . 5 y + 3 = 10 y - 4 .$

ⓑ



Multiply.
Simplify.

Since $y = \frac{7}{5} y = \frac{7}{5}$ results in a true equation, $y = \frac{7}{5} y = \frac{7}{5}$ is a solution to the equation $5 y + 3 = 10 y - 4 . 5 y + 3 = 10 y - 4 .$

Try It 2.1

Determine whether the values are solutions to the equation: $9 y + 2 = 6 y + 3 . 9 y + 2 = 6 y + 3 .$

ⓐ $y = \frac{4}{3} y = \frac{4}{3}$ ⓑ $y = \frac{1}{3} y = \frac{1}{3}$

Try It 2.2

Determine whether the values are solutions to the equation: $4 x - 2 = 2 x + 1 . 4 x - 2 = 2 x + 1 .$

ⓐ $x = \frac{3}{2} x = \frac{3}{2}$ ⓑ $x = - \frac{1}{2} x = - \frac{1}{2}$

There are many types of equations that we will learn to solve. In this section we will focus on a linear equation.

Linear Equation

A linear equation is an equation in one variable that can be written, where a and b are real numbers and $a \neq 0, a \neq 0,$ as:

a x + b = 0 a x + b = 0

To solve a linear equation it is a good idea to have an overall strategy that can be used to solve any linear equation. In the next example, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

Example 2.2

How to Solve a Linear Equation Using a General Strategy

Solve: $7 (n - 3) - 8 = −15 7 (n - 3) - 8 = −15$ .

Show solution

Solution

Step 1 is to simplify each side of the equation, the product of 7 and the quantity n minus 3 minus 8 is equal to negative 15. Use the Distributive Property. The equation first simplifies to 7 n minus 21 minus 8 is equal to negative 15. Then it simplifies to 7 n minus 29 is equal to negative 15. Notice that each side of the equation is now simplified as much as possible.

Step 2 is to collect all variable terms on the left side of the equation, 7 n minus 29 is equal to negative 15. Notice there is nothing to do because all n’s are on the left side.

Step 3 is to collect all constant terms on the other side of the equation, 7 n minus 29 is equal to negative 15. To get constants only on the right, add 29 to each side. The result is 7 n minus 29 plus 29 is equal to negative 15 plus 29. Simplify. The result is 7 n is equal to 14.

Step 4 is to make the coefficient of the equation, 7 n is equal to 14, 1. Divide each side of the equation by 7. The result is 7 n divided by 7 is equal to 14 divided by 7. Simplify. The result is n is equal to 2.

Step 5 is to check the solution, n is equal to 2, by substituting into the equation, the product of 7 and the quantity n minus 3 minus 8 is equal to negative 15. Is the product of 7 and the quantity 2 minus 3 minus 8 equal to negative 15? Subtract. Is 7 times negative 1 minus 8 equal to negative 15? Is negative 7 minus 8 equal to negative 15. Negative 15 is equal to negative 15. The solution checks.

Try It 2.3

Solve: $2 (m - 4) + 3 = −1 . 2 (m - 4) + 3 = −1 .$

Try It 2.4

Solve: $5 (a - 3) + 5 = −10 . 5 (a - 3) + 5 = −10 .$

These steps are summarized in the General Strategy for Solving Linear Equations below.

How To

Solve linear equations using a general strategy.

Step 1. Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms.
Step 2. Collect all the variable terms on one side of the equation.
Use the Addition or Subtraction Property of Equality.
Step 3. Collect all the constant terms on the other side of the equation.
Use the Addition or Subtraction Property of Equality.
Step 4. Make the coefficient of the variable term equal to 1.
Use the Multiplication or Division Property of Equality.
State the solution to the equation.
Step 5. Check the solution.
Substitute the solution into the original equation to make sure the result is a true statement.

Example 2.3

Solve: $\frac{2}{3} (3 m - 6) = 5 - m . \frac{2}{3} (3 m - 6) = 5 - m .$

Show solution

Solution


Distribute.
Add m to both sides to get the variables only on the left.
Simplify.
Add 4 to both sides to get constants only on the right.
Simplify.
Divide both sides by three.
Simplify.
Check:
Let $m = 3. m = 3.$

Try It 2.5

Solve: $\frac{1}{3} (6 u + 3) = 7 - u . \frac{1}{3} (6 u + 3) = 7 - u .$

Try It 2.6

Solve: $\frac{2}{3} (9 x - 12) = 8 + 2 x . \frac{2}{3} (9 x - 12) = 8 + 2 x .$

We can solve equations by getting all the variable terms to either side of the equal sign. By collecting the variable terms on the side where the coefficient of the variable is larger, we avoid working with some negatives. This will be a good strategy when we solve inequalities later in this chapter. It also helps us prevent errors with negatives.

Example 2.4

Solve: $4 (x - 1) - 2 = 5 (2 x + 3) + 6 . 4 (x - 1) - 2 = 5 (2 x + 3) + 6 .$

Show solution

Solution


Distribute.
Combine like terms.
Subtract $4 x 4 x$ from each side to get the variables only on the right since $10 > 4 . 10 > 4 .$
Simplify.
Subtract 21 from each side to get the constants on left.
Simplify.
Divide both sides by 6.
Simplify.
Check:
Let $x = - \frac{9}{2} . x = - \frac{9}{2} .$

Try It 2.7

Solve: $6 (p - 3) - 7 = 5 (4 p + 3) - 12 . 6 (p - 3) - 7 = 5 (4 p + 3) - 12 .$

Try It 2.8

Solve: $8 (q + 1) - 5 = 3 (2 q - 4) - 1 . 8 (q + 1) - 5 = 3 (2 q - 4) - 1 .$

Example 2.5

Solve: $10 [3 - 8 (2 s - 5)] = 15 (40 - 5 s) . 10 [3 - 8 (2 s - 5)] = 15 (40 - 5 s) .$

Show solution

Solution


Simplify from the innermost parentheses first.
Combine like terms in the brackets.
Distribute.
Add $160 s 160 s$ to both sides to get the variables to the right.
Simplify.
Subtract 600 from both sides to get the constants to the left.
Simplify.
Divide both sides by 85.
Simplify.
Check:
Let $s = - 2. s = - 2.$

Try It 2.9

Solve: $6 [4 - 2 (7 y - 1)] = 8 (13 - 8 y) . 6 [4 - 2 (7 y - 1)] = 8 (13 - 8 y) .$

Try It 2.10

Solve: $12 [1 - 5 (4 z - 1)] = 3 (24 + 11 z) . 12 [1 - 5 (4 z - 1)] = 3 (24 + 11 z) .$

Classify Equations

Whether or not an equation is true depends on the value of the variable. The equation $7 x + 8 = −13 7 x + 8 = −13$ is true when we replace the variable, x, with the value $−3, −3,$ but not true when we replace x with any other value. An equation like this is called a conditional equation. All the equations we have solved so far are conditional equations.

Conditional Equation

An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.

Now let’s consider the equation $7 y + 14 = 7 (y + 2) . 7 y + 14 = 7 (y + 2) .$ Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.

Solve:


Distribute.
Subtract $7 y 7 y$ to each side to get the $y ’ s y ’ s$ to one side.
Simplify—the y’s are eliminated.
	But $14 = 14 14 = 14$ is true.

This means that the equation $7 y + 14 = 7 (y + 2) 7 y + 14 = 7 (y + 2)$ is true for any value of y. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable is called an identity.

Identity

An equation that is true for any value of the variable is called an identity.

The solution of an identity is all real numbers.

What happens when we solve the equation $−8 z = −8 z + 9 ? −8 z = −8 z + 9 ?$

Solve:


Add $8 z 8 z$ to both sides to leave the constant alone on the right.
Simplify—the $z ’ s z ’ s$ are eliminated.
	But $0 \neq 9 . 0 \neq 9 .$

Solving the equation $−8 z = −8 z + 9 −8 z = −8 z + 9$ led to the false statement $0 = 9 . 0 = 9 .$ The equation $−8 z = −8 z + 9 −8 z = −8 z + 9$ will not be true for any value of z. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction.

Contradiction

An equation that is false for all values of the variable is called a contradiction.

A contradiction has no solution.

The next few examples will ask us to classify an equation as conditional, an identity, or as a contradiction.

Example 2.6

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $6 (2 n - 1) + 3 = 2 n - 8 + 5 (2 n + 1) . 6 (2 n - 1) + 3 = 2 n - 8 + 5 (2 n + 1) .$

Show solution

Solution


Distribute.
Combine like terms.
Subtract $12 n 12 n$ from each side to get the n’s to one side.
Simplify.
This is a true statement.	The equation is an identity.
	The solution is all real numbers.

Try It 2.11

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $4 + 9 (3 x - 7) = −42 x - 13 + 23 (3 x - 2) . 4 + 9 (3 x - 7) = −42 x - 13 + 23 (3 x - 2) .$

Try It 2.12

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $8 (1 - 3 x) + 15 (2 x + 7) = 2 (x + 50) + 4 (x + 3) + 1 . 8 (1 - 3 x) + 15 (2 x + 7) = 2 (x + 50) + 4 (x + 3) + 1 .$

Example 2.7

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $8 + 3 (a - 4) = 0 . 8 + 3 (a - 4) = 0 .$

Show solution

Solution


Distribute.
Combine like terms.
Add 4 to both sides.
Simplify.
Divide.
Simplify.
The equation is true when $a = \frac{4}{3} . a = \frac{4}{3} .$	This is a conditional equation.
	The solution is $a = \frac{4}{3} . a = \frac{4}{3} .$

Try It 2.13

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $11 (q + 3) - 5 = 19 . 11 (q + 3) - 5 = 19 .$

Try It 2.14

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $6+14 (k - 8) = 95 . 6+14 (k - 8) = 95 .$

Example 2.8

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $5 m + 3 (9 + 3 m) = 2 (7 m - 11) . 5 m + 3 (9 + 3 m) = 2 (7 m - 11) .$

Show solution

Solution


Distribute.
Combine like terms.
Subtract $14 m 14 m$ from both sides.
Simplify.
But $27 \neq − 22 . 27 \neq − 22 .$	The equation is a contradiction.
	It has no solution.

Try It 2.15

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $12 c + 5 (5 + 3 c) = 3 (9 c - 4) . 12 c + 5 (5 + 3 c) = 3 (9 c - 4) .$

Try It 2.16

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $4 (7 d + 18) = 13 (3 d - 2) - 11 d . 4 (7 d + 18) = 13 (3 d - 2) - 11 d .$

We summarize the methods for classifying equations in the table.

Type of equation	What happens when you solve it?	Solution
Conditional Equation	True for one or more values of the variables and false for all other values	One or more values
Identity	True for any value of the variable	All real numbers
Contradiction	False for all values of the variable	No solution

Table 2.1

Solve Equations with Fraction or Decimal Coefficients

We could use the General Strategy to solve the next example. This method would work fine, but many students do not feel very confident when they see all those fractions. So, we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator (LCD) of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but without fractions. This process is called clearing the equation of fractions.

To clear an equation of decimals, we think of all the decimals in their fraction form and then find the LCD of those denominators.

Example 2.9

How to Solve Equations with Fraction or Decimal Coefficients

Solve: $\frac{1}{12} x + \frac{5}{6} = \frac{3}{4} . \frac{1}{12} x + \frac{5}{6} = \frac{3}{4} .$

Show solution

Solution

$Step 1 is to find the least common denominator of all the fractions and decimals in the equation, one-twelfth x plus five-sixth is equal to three-fourths. What is the L C D of one-twelfth, five-sixths, and three-fourths? The L C D is equal to 12.$ $Step 2 is multiply both sides of the equation by the L C D. This clears the fractions and decimals. Multiply both sides of the equation by the L C D, 12. The result is 12 times the quantity one-twelfth x plus five-sixths is equal to 12 times three-fourths. Use the Distributive Property. The result is 12 times one-twelfth x plus 12 times five-sixths is equal to 12 times three-fourths. Simplify. The result is x plus 10 is equal to 9. Notice there are no more fractions.$

Step 3 is to solve using the General Strategy for Solving Linear Equations. To isolate the variable term, subtract 10. The result is x plus 10 minus 10 is equal to 9 minus 10. Simplify. The result is x is equal to negative 1. Check the solution. Substitute negative into the original equation one-twelfth x plus five-sixths is equal to three-fourth. Is one-twelfth times negative 1 plus five-sixths equal to three-fourths? Is negative one-twelfth plus five-sixths equal to three-fourths? Is negative one-twelfth plus ten-twelfths equal to nine-twelfths? Is nine-twelfths equal to nine-twelfths? Yes. The solution checks.

Try It 2.17

Solve: $\frac{1}{4} x + \frac{1}{2} = \frac{5}{8} . \frac{1}{4} x + \frac{1}{2} = \frac{5}{8} .$

Try It 2.18

Solve: $\frac{1}{8} x + \frac{1}{2} = \frac{1}{4} . \frac{1}{8} x + \frac{1}{2} = \frac{1}{4} .$

Notice in the previous example, once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve. We then used the General Strategy for Solving Linear Equations.

How To

Solve Equations with Fraction or Decimal Coefficients.

Step 1. Find the least common denominator (LCD) of all the fractions and decimals (in fraction form) in the equation.
Step 2. Multiply both sides of the equation by that LCD. This clears the fractions and decimals.
Step 3. Solve using the General Strategy for Solving Linear Equations.

Example 2.10

Solve: $5 = \frac{1}{2} y + \frac{2}{3} y - \frac{3}{4} y . 5 = \frac{1}{2} y + \frac{2}{3} y - \frac{3}{4} y .$

Show solution

Solution

We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

Find the LCD of all fractions in the equation.
The LCD is 12.
Multiply both sides of the equation by 12.
Distribute.
Simplify—notice, no more fractions.
Combine like terms.
Divide by five.
Simplify.
Check:
Let $y = 12. y = 12.$

Try It 2.19

Solve: $7 = \frac{1}{2} x + \frac{3}{4} x - \frac{2}{3} x . 7 = \frac{1}{2} x + \frac{3}{4} x - \frac{2}{3} x .$

Try It 2.20

Solve: $−1 = \frac{1}{2} u + \frac{1}{4} u - \frac{2}{3} u . −1 = \frac{1}{2} u + \frac{1}{4} u - \frac{2}{3} u .$

In the next example, we’ll distribute before we clear the fractions.

Example 2.11

Solve: $\frac{1}{2} (y - 5) = \frac{1}{4} (y - 1) . \frac{1}{2} (y - 5) = \frac{1}{4} (y - 1) .$

Show solution

Solution


Distribute.
Simplify.
Multiply by the LCD, four.
Distribute.
Simplify.
Collect the variables to the left.
Simplify.
Collect the constants to the right.
Simplify.
An alternate way to solve this equation is to clear the fractions without distributing first. If you multiply the factors correctly, this method will be easier.

Multiply by the LCD, 4.
Multiply four times the fractions.
Distribute.
Collect the variables to the left.
Simplify.
Collect the constants to the right.
Simplify.
Check:
Let $y = 9. y = 9.$
Finish the check on your own.

Try It 2.21

Solve: $\frac{1}{5} (n + 3) = \frac{1}{4} (n + 2) . \frac{1}{5} (n + 3) = \frac{1}{4} (n + 2) .$

Try It 2.22

Solve: $\frac{1}{2} (m - 3) = \frac{1}{4} (m - 7) . \frac{1}{2} (m - 3) = \frac{1}{4} (m - 7) .$

When you multiply both sides of an equation by the LCD of the fractions, make sure you multiply each term by the LCD—even if it does not contain a fraction.

Example 2.12

Solve: $\frac{4 q + 3}{2} + 6 = \frac{3 q + 5}{4} \frac{4 q + 3}{2} + 6 = \frac{3 q + 5}{4}$

Show solution

Solution


Multiply both sides by the LCD, 4.
Distribute.
Simplify.


Collect the variables to the left.
Simplify.
Collect the constants to the right.
Simplify.
Divide both sides by five.
Simplify.
Check:
Let $q = - 5. q = - 5.$
Finish the check on your own.

Try It 2.23

Solve: $\frac{3 r + 5}{6} + 1 = \frac{4 r + 3}{3} . \frac{3 r + 5}{6} + 1 = \frac{4 r + 3}{3} .$

Try It 2.24

Solve: $\frac{2 s + 3}{2} + 1 = \frac{3 s + 2}{4} . \frac{2 s + 3}{2} + 1 = \frac{3 s + 2}{4} .$

Some equations have decimals in them. This kind of equation may occur when we solve problems dealing with money or percentages. But decimals can also be expressed as fractions. For example, $0.7 = \frac{7}{10} 0.7 = \frac{7}{10}$ and $0.29 = \frac{29}{100} . 0.29 = \frac{29}{100} .$ So, with an equation with decimals, we can use the same method we used to clear fractions—multiply both sides of the equation by the least common denominator.

The next example uses an equation that is typical of the ones we will see in the money applications in a later section. Notice that we will clear all decimals by multiplying by the LCD of their fraction form.

Example 2.13

Solve: $0.25 x + 0.05 (x + 3) = 2.85 . 0.25 x + 0.05 (x + 3) = 2.85 .$

Show solution

Solution

Look at the decimals and think of the equivalent fractions:

0.25 = \frac{25}{100}, 0.05 = \frac{5}{100}, 2.85 = 2 \frac{85}{100} . 0.25 = \frac{25}{100},

Notice, the LCD is 100. By multiplying by the LCD we will clear the decimals from the equation.


Distribute first.
Combine like terms.
To clear decimals, multiply by 100.
Distribute.
Subtract 15 from both sides.
Simplify.
Divide by 30.
Simplify.
Check it yourself by substituting $x = 9 x = 9$ into the original equation.

Try It 2.25

Solve: $0.25 n + 0.05 (n + 5) = 2.95 . 0.25 n + 0.05 (n + 5) = 2.95 .$

Try It 2.26

Solve: $0.10 d + 0.05 (d - 5) = 2.15 . 0.10 d + 0.05 (d - 5) = 2.15 .$

Practice Makes Perfect

Solve Equations Using the General Strategy

In the following exercises, determine whether the given values are solutions to the equation.

$6 y + 10 = 12 y 6 y + 10 = 12 y$

ⓐ $y = \frac{5}{3} y = \frac{5}{3}$ ⓑ $y = - \frac{1}{2} y = - \frac{1}{2}$

$4 x + 9 = 8 x 4 x + 9 = 8 x$

ⓐ $x = - \frac{7}{8} x = - \frac{7}{8}$ ⓑ $x = \frac{9}{4} x = \frac{9}{4}$

$8 u - 1 = 6 u 8 u - 1 = 6 u$

ⓐ $u = - \frac{1}{2} u = - \frac{1}{2}$ ⓑ $u = \frac{1}{2} u = \frac{1}{2}$

$9 v - 2 = 3 v 9 v - 2 = 3 v$

ⓐ $v = - \frac{1}{3} v = - \frac{1}{3}$ ⓑ $v = \frac{1}{3} v = \frac{1}{3}$

In the following exercises, solve each linear equation.

$15 (y - 9) = −60 15 (y - 9) = −60$

$−16 (3 n + 4) = 32 −16 (3 n + 4) = 32$

$− (w - 12) = 30 − (w - 12) = 30$

$− (t - 19) = 28 − (t - 19) = 28$

$51 + 5 (4 - q) = 56 51 + 5 (4 - q) = 56$

10.

$−6 + 6 (5 - k) = 15 −6 + 6 (5 - k) = 15$

11.

$3 (10 - 2 x) + 54 = 0 3 (10 - 2 x) + 54 = 0$

12.

$−2 (11 - 7 x) + 54 = 4 −2 (11 - 7 x) + 54 = 4$

13.

$\frac{2}{3} (9 c - 3) = 22 \frac{2}{3} (9 c - 3) = 22$

14.

$\frac{3}{5} (10 x - 5) = 27 \frac{3}{5} (10 x - 5) = 27$

15.

$\frac{1}{5} (15 c + 10) = c + 7 \frac{1}{5} (15 c + 10) = c + 7$

16.

$\frac{1}{4} (20 d + 12) = d + 7 \frac{1}{4} (20 d + 12) = d + 7$

17.

$3 (4 n - 1) - 2 = 8 n + 3 3 (4 n - 1) - 2 = 8 n + 3$

18.

$9 (2 m - 3) - 8 = 4 m + 7 9 (2 m - 3) - 8 = 4 m + 7$

19.

$12 + 2 (5 - 3 y) = −9 (y - 1) - 2 12 + 2 (5 - 3 y) = −9 (y - 1) - 2$

20.

$−15 + 4 (2 - 5 y) = −7 (y - 4) + 4 −15 + 4 (2 - 5 y) = −7 (y - 4) + 4$

21.

$5 + 6 (3 s - 5) = −3 + 2 (8 s - 1) 5 + 6 (3 s - 5) = −3 + 2 (8 s - 1)$

22.

$−12 + 8 (x - 5) = −4 + 3 (5 x - 2) −12 + 8 (x - 5) = −4 + 3 (5 x - 2)$

23.

$4 (p - 4) - (p + 7) = 5 (p - 3) 4 (p - 4) - (p + 7) = 5 (p - 3)$

24.

$3 (a - 2) - (a + 6) = 4 (a - 1) 3 (a - 2) - (a + 6) = 4 (a - 1)$

25.

$4 [5 - 8 (4 c - 3)] = 12 (1 - 13 c) - 8 4 [5 - 8 (4 c - 3)] = 12 (1 - 13 c) - 8$

26.

$5 [9 - 2 (6 d - 1)] = 11 (4 - 10 d) - 139 5 [9 - 2 (6 d - 1)] = 11 (4 - 10 d) - 139$

27.

$3 [−9 + 8 (4 h - 3)] = 2 (5 - 12 h) - 19 3 [−9 + 8 (4 h - 3)] = 2 (5 - 12 h) - 19$

28.

$3 [−14 + 2 (15 k - 6)] = 8 (3 - 5 k) - 24 3 [−14 + 2 (15 k - 6)] = 8 (3 - 5 k) - 24$

29.

$5 [2 (m + 4) + 8 (m - 7)] = 2 [3 (5 + m) - (21 - 3 m)] 5 [2 (m + 4) + 8 (m - 7)] = 2 [3 (5 + m) - (21 - 3 m)]$

30.

$10 [5 (n + 1) + 4 (n - 1)] = 11 [7 (5 + n) - (25 - 3 n)] 10 [5 (n + 1) + 4 (n - 1)] = 11 [7 (5 + n) - (25 - 3 n)]$

Classify Equations

In the following exercises, classify each equation as a conditional equation, an identity, or a contradiction and then state the solution.

31.

$23 z + 19 = 3 (5 z - 9) + 8 z + 46 23 z + 19 = 3 (5 z - 9) + 8 z + 46$

32.

$15 y + 32 = 2 (10 y - 7) - 5 y + 46 15 y + 32 = 2 (10 y - 7) - 5 y + 46$

33.

$18 (5 j - 1) + 29 = 47 18 (5 j - 1) + 29 = 47$

34.

$24 (3 d - 4) + 100 = 52 24 (3 d - 4) + 100 = 52$

35.

$22 (3 m - 4) = 8 (2 m + 9) 22 (3 m - 4) = 8 (2 m + 9)$

36.

$30 (2 n - 1) = 5 (10 n + 8) 30 (2 n - 1) = 5 (10 n + 8)$

37.

$7 v + 42 = 11 (3 v + 8) - 2 (13 v - 1) 7 v + 42 = 11 (3 v + 8) - 2 (13 v - 1)$

38.

$18 u - 51 = 9 (4 u + 5) - 6 (3 u - 10) 18 u - 51 = 9 (4 u + 5) - 6 (3 u - 10)$

39.

$45 (3 y - 2) = 9 (15 y - 6) 45 (3 y - 2) = 9 (15 y - 6)$

40.

$60 (2 x - 1) = 15 (8 x + 5) 60 (2 x - 1) = 15 (8 x + 5)$

41.

$9 (14 d + 9) + 4 d = 13 (10 d + 6) + 3 9 (14 d + 9) + 4 d = 13 (10 d + 6) + 3$

42.

$11 (8 c + 5) - 8 c = 2 (40 c + 25) + 5 11 (8 c + 5) - 8 c = 2 (40 c + 25) + 5$

Solve Equations with Fraction or Decimal Coefficients

In the following exercises, solve each equation with fraction coefficients.

43.

$\frac{1}{4} x - \frac{1}{2} = - \frac{3}{4} \frac{1}{4} x - \frac{1}{2} = - \frac{3}{4}$

44.

$\frac{3}{4} x - \frac{1}{2} = \frac{1}{4} \frac{3}{4} x - \frac{1}{2} = \frac{1}{4}$

45.

$\frac{5}{6} y - \frac{2}{3} = - \frac{3}{2} \frac{5}{6} y - \frac{2}{3} = - \frac{3}{2}$

46.

$\frac{5}{6} y - \frac{1}{3} = - \frac{7}{6} \frac{5}{6} y - \frac{1}{3} = - \frac{7}{6}$

47.

$\frac{1}{2} a + \frac{3}{8} = \frac{3}{4} \frac{1}{2} a + \frac{3}{8} = \frac{3}{4}$

48.

$\frac{5}{8} b + \frac{1}{2} = - \frac{3}{4} \frac{5}{8} b + \frac{1}{2} = - \frac{3}{4}$

49.

$2 = \frac{1}{3} x - \frac{1}{2} x + \frac{2}{3} x 2 = \frac{1}{3} x - \frac{1}{2} x + \frac{2}{3} x$

50.

$2 = \frac{3}{5} x - \frac{1}{3} x + \frac{2}{5} x 2 = \frac{3}{5} x - \frac{1}{3} x + \frac{2}{5} x$

51.

$\frac{1}{3} w + \frac{5}{4} = w - \frac{1}{4} \frac{1}{3} w + \frac{5}{4} = w - \frac{1}{4}$

52.

$\frac{1}{2} a - \frac{1}{4} = \frac{1}{6} a + \frac{1}{12} \frac{1}{2} a - \frac{1}{4} = \frac{1}{6} a + \frac{1}{12}$

53.

$\frac{1}{3} b + \frac{1}{5} = \frac{2}{5} b - \frac{3}{5} \frac{1}{3} b + \frac{1}{5} = \frac{2}{5} b - \frac{3}{5}$

54.

$\frac{1}{3} x + \frac{2}{5} = \frac{1}{5} x - \frac{2}{5} \frac{1}{3} x + \frac{2}{5} = \frac{1}{5} x - \frac{2}{5}$

55.

$\frac{1}{4} (p - 7) = \frac{1}{3} (p + 5) \frac{1}{4} (p - 7) = \frac{1}{3} (p + 5)$

56.

$\frac{1}{5} (q + 3) = \frac{1}{2} (q - 3) \frac{1}{5} (q + 3) = \frac{1}{2} (q - 3)$

57.

$\frac{1}{2} (x + 4) = \frac{3}{4} \frac{1}{2} (x + 4) = \frac{3}{4}$

58.

$\frac{1}{3} (x + 5) = \frac{5}{6} \frac{1}{3} (x + 5) = \frac{5}{6}$

59.

$\frac{4 n + 8}{4} = \frac{n}{3} \frac{4 n + 8}{4} = \frac{n}{3}$

60.

$\frac{3 p + 6}{3} = \frac{p}{2} \frac{3 p + 6}{3} = \frac{p}{2}$

61.

$\frac{3 x + 4}{2} + 1 = \frac{5 x + 10}{8} \frac{3 x + 4}{2} + 1 = \frac{5 x + 10}{8}$

62.

$\frac{10 y - 2}{3} + 3 = \frac{10 y + 1}{9} \frac{10 y - 2}{3} + 3 = \frac{10 y + 1}{9}$

63.

$\frac{7 u - 1}{4} - 1 = \frac{4 u + 8}{5} \frac{7 u - 1}{4} - 1 = \frac{4 u + 8}{5}$

64.

$\frac{3 v - 6}{2} + 5 = \frac{11 v - 4}{5} \frac{3 v - 6}{2} + 5 = \frac{11 v - 4}{5}$

In the following exercises, solve each equation with decimal coefficients.

65.

$0.4 x + 0.6 = 0.5 x - 1.2 0.4 x + 0.6 = 0.5 x - 1.2$

66.

$0.7 x + 0.4 = 0.6 x + 2.4 0.7 x + 0.4 = 0.6 x + 2.4$

67.

$0.9 x - 1.25 = 0.75 x + 1.75 0.9 x - 1.25 = 0.75 x + 1.75$

68.

$1.2 x - 0.91 = 0.8 x + 2.29 1.2 x - 0.91 = 0.8 x + 2.29$

69.

$0.05 n + 0.10 (n + 8) = 2.15 0.05 n + 0.10 (n + 8) = 2.15$

70.

$0.05 n + 0.10 (n + 7) = 3.55 0.05 n + 0.10 (n + 7) = 3.55$

71.

$0.10 d + 0.25 (d + 5) = 4.05 0.10 d + 0.25 (d + 5) = 4.05$

72.

$0.10 d + 0.25 (d + 7) = 5.25 0.10 d + 0.25 (d + 7) = 5.25$

Everyday Math

73.

Fencing Micah has 74 feet of fencing to make a dog run in his yard. He wants the length to be 2.5 feet more than the width. Find the length, L, by solving the equation $2 L + 2 (L - 2.5) = 74 . 2 L + 2 (L - 2.5) = 74 .$

74.

Stamps Paula bought $22.82 worth of 49-cent stamps and 21-cent stamps. The number of 21-cent stamps was eight less than the number of
49-cent stamps. Solve the equation
$0.49 s + 0.21 (s - 8) = 22.82 0.49 s + 0.21 (s - 8) = 22.82$ for s, to find the number of 49-cent stamps Paula bought.

Writing Exercises

75.

Using your own words, list the steps in the general strategy for solving linear equations.

76.

Explain why you should simplify both sides of an equation as much as possible before collecting the variable terms to one side and the constant terms to the other side.

77.

What is the first step you take when solving the equation $3 - 7 (y - 4) = 38 ? 3 - 7 (y - 4) = 38 ?$ Why is this your first step?

78.

If an equation has several fractions, how does multiplying both sides by the LCD make it easier to solve?

79.

If an equation has fractions only on one side, why do you have to multiply both sides of the equation by the LCD?

80.

For the equation $0.35 x + 2.1 = 3.85, 0.35 x + 2.1 = 3.85,$ how do you clear the decimal?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

$This table has four columns and four rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve linear equations using a general strategy. In row 3, the I can was classify equations. In row 4, the I can was solve equations with fraction or decimal coefficients.$

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.