Use a General Strategy to Solve Linear Equations

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!

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.

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

ⓐ $y = \frac{3}{5}$

ⓑ $y = \frac{7}{5}$

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.

ⓐ* * *

| | | {: valign=”top”}| | | {: valign=”top”}| Multiply. | | {: valign=”top”}| Simplify. | | {: valign=”top”}{: .unnumbered .unstyled summary=”5 y plus 3 is equal to 10 y minus 4. Substitute three-fifths for y. Is the product of 5 and three-fifths plus 3 equal to the product of 10 and three-fifths minus 4. Multiply on each side of the equation. Is 3 plus 3 equal to 6 minus 4? Simplify on each side. 6 is not equal to 2. Since y is equal to three-fifths does not result in a true equation, y is equal to three-fifths is not a solution to the equation 5 y plus 3 is equal to 10 y minus 4.”}

Since $y = \frac{3}{5}$

does not result in a true equation, $y = \frac{3}{5}$

is not a solution to the equation $5 y + 3 = 10 y - 4 .$

ⓑ* * *

| | | {: valign=”top”}| | | {: valign=”top”}| Multiply. | | {: valign=”top”}| Simplify. | | {: valign=”top”}{: .unnumbered .unstyled summary=”Substitute seven-fifths for y. Is the product of 5 and seven-fifths plus 3 equal to the product of 10 and seven-fifths minus 4. Multiply on each side of the equation. Is 7 plus 3 equal to 14 minus 4? Simplify on each side. 10 is equal to 10. Since y is equal to seven-fifths results in a true equation, y is equal to seven-fifths is a solution to the equation 5 y plus 3 is equal to 10 y minus 4.”}

Since $y = \frac{7}{5}$

results in a true equation, $y = \frac{7}{5}$

is a solution to the equation $5 y + 3 = 10 y - 4 .$

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

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.

How to Solve a Linear Equation Using a General Strategy

Solve: $7 (n - 3) - 8 = −15$

![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.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_02_01_003a_img.jpg) ![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.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_02_01_003b_img.jpg) ![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.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_02_01_003c_img.jpg) ![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.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_02_01_003d_img.jpg) ![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.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_02_01_003e_img.jpg)

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

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.

Classify Equations

Whether or not an equation is true depends on the value of the variable. The equation

7 x + 8 = −13

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.

Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.

is true. | {: valign=”top”}{: .unnumbered .unstyled summary=”7 y plus 14 is equal to 7 times the quantity y plus 2. Distribute. The result is 7 y plus 14 is equal to 7 y plus 14. Subtract 7 y from each side to get the y’s on one side. The result is 7 y minus 7 y plus 14 is equal to 7 y minus 7 y plus 14. When simplified, 14 is equal to 14. The y’s are eliminated. But 14 is equal to 14 is true.”}

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.

| {: valign=”top”}{: .unnumbered .unstyled summary=”Negative 8 z is equal to negative 8 z plus 9. Add 8 z to both sides to leave the constant alone on the right. The result is negative 8 z plus 8 z is equal to negative 8 z plus 8 z plus 9. When you simplify, the z’s are eliminated. The result is 0 is equal to 9. But 0 is not equal to 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.

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

| Type of equation | What happens when you solve it? | Solution | {: valign=”top”}|———- | Conditional Equation | True for one or more values of the variables and false for all other values | One or more values | {: valign=”top”}| Identity | True for any value of the variable | All real numbers | {: valign=”top”}| Contradiction | False for all values of the variable | No solution | {: valign=”top”}{: summary=”This table has three columns and four rows. The first row is a header row and it labels each column, “Type of equation,” “What happens when you solve it?” and “Solution.” The second column is a header column and it labels each row “Conditional Equations,” Identity,” “Contradiction”. In row two, the Conditional Equation is True for one or more values of the variables and false for all other values, and the Solution is One or more values. In row three, the Identity is True for any value of the variable, and the Solution is All real numbers. In row four, the Contradiction is False for all values of the variable, and the Solution is No Solution.”}

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.

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.

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.

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}

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.

Key Concepts

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

ⓐ $y = \frac{5}{3}$

ⓑ $y = - \frac{1}{2}$

ⓐ yes ⓑ no

4 x + 9 = 8 x

ⓐ $x = - \frac{7}{8}$

ⓑ $x = \frac{9}{4}$

8 u - 1 = 6 u

ⓐ $u = - \frac{1}{2}$

ⓑ $u = \frac{1}{2}$

ⓐ no ⓑ yes

9 v - 2 = 3 v

ⓐ $v = - \frac{1}{3}$

ⓑ $v = \frac{1}{3}$

In the following exercises, solve each linear equation.

15 (y - 9) = −60

y = 5

−16 (3 n + 4) = 32

− (w - 12) = 30

w = −18

− (t - 19) = 28

51 + 5 (4 - q) = 56

q = 3

−6 + 6 (5 - k) = 15

3 (10 - 2 x) + 54 = 0

x = 14

−2 (11 - 7 x) + 54 = 4

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

c = 4

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

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

c = \frac{5}{2}

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

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

n = 2

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

12 + 2 (5 - 3 y) = −9 (y - 1) - 2

y = −5

−15 + 4 (2 - 5 y) = −7 (y - 4) + 4

5 + 6 (3 s - 5) = −3 + 2 (8 s - 1)

s = 10

−12 + 8 (x - 5) = −4 + 3 (5 x - 2)

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

p = −4

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

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

c = −4

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

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

h = \frac{3}{4}

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

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

m = 6

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.

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

identity; all real numbers

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

18 (5 j - 1) + 29 = 47

conditional equation; $j = \frac{2}{5}$

24 (3 d - 4) + 100 = 52

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

conditional equation; $m = \frac{16}{5}$

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

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

contradiction; no solution

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

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

contradiction; no solution

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

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

identity; all real numbers

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.

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

x = −1

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

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

y = −1

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

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

a = \frac{3}{4}

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

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

x = 4

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

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

w = \frac{9}{4}

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

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

b = 12

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

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

p = −41

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

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

x = - \frac{5}{2}

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

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

n = −3

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

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

x = −2

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

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

u = 3

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

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

0.4 x + 0.6 = 0.5 x - 1.2

x = 18

0.7 x + 0.4 = 0.6 x + 2.4

0.9 x - 1.25 = 0.75 x + 1.75

x = 20

1.2 x - 0.91 = 0.8 x + 2.29

0.05 n + 0.10 (n + 8) = 2.15

n = 9

0.05 n + 0.10 (n + 7) = 3.55

0.10 d + 0.25 (d + 5) = 4.05

d = 8

0.10 d + 0.25 (d + 7) = 5.25

Everyday Math

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 $2 L + 2 (L - 2.5) = 74 .$

L = 19.75

feet

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

for s, to find the number of 49-cent stamps Paula bought.

Writing Exercises

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

Answers will vary.

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.

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

Why is this your first step?

Answers will vary.

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

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

Answers will vary.

For the equation $0.35 x + 2.1 = 3.85,$

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.


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 $m = 3.$


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


Simplify from the innermost parentheses first.
Combine like terms in the brackets.
Distribute.
Add $160 s$ 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 $s = - 2.$

to one side.
{: valign=”top”}	Simplify—the y’s are eliminated.
{: valign=”top”}		But $14 = 14$


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

Use a General Strategy to Solve Linear Equations

Solve Linear Equations Using a General Strategy

Classify Equations

Solve Equations with Fraction or Decimal Coefficients

Key Concepts

Practice Makes Perfect

Everyday Math

Writing Exercises

Self Check

Glossary

to both sides to leave the constant alone on the right.
{: valign=”top”}	Simplify—the $z ’ s$

from both sides.
{: valign=”top”}	Simplify.
{: valign=”top”}	But $27 \neq − 22 .$

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 $y = 12.$


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 $y = 9.$
Finish the check on your own.


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 $q = - 5.$
Finish the check on your own.


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 $x = 9$ into the original equation.