# 4.12: Solve Equations with Fractions (Part 1)

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##### Learning Objectives
• Determine whether a fraction is a solution of an equation
• Solve equations with fractions using the Addition, Subtraction, and Division Properties of Equality
• Solve equations using the Multiplication Property of Equality
• Translate sentences to equations and solve
##### be prepared!

Before you get started, take this readiness quiz. If you miss a problem, go back to the section listed and review the material.

1. Evaluate $$x + 4$$ when $$x = −3$$ If you missed this problem, review Example 3.2.10.
2. Solve: $$2y − 3 = 9$$. If you missed this problem, review Example 3.5.2.
3. Solve: $$y − 3 = −9$$ If you missed this problem, review Example 4.2.10.

## Determine Whether a Fraction is a Solution of an Equation

As we saw in Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers; The Division Property of Equality, a solution of an equation is a value that makes a true statement when substituted for the variable in the equation. In those sections, we found whole number and integer solutions to equations. Now that we have worked with fractions, we are ready to find fraction solutions to equations.

The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number, an integer, or a fraction.

##### 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 $$\PageIndex{1}$$: find solution

Determine whether each of the following is a solution of $$x − \dfrac{3}{10} = \dfrac{1}{2}$$.

1. $$x = 1$$
2. $$x = \dfrac{4}{5}$$
3. $$x = − \dfrac{4}{5}$$

Solution

 Substitute $$\textcolor{red}{1}$$ for x. $$\textcolor{red}{1} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{1}{2}$$ Change to fractions with a LCD of 10. $$\textcolor{red}{\dfrac{10}{10}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{5}{10}$$ Subtract. $$\dfrac{7}{10} \stackrel{?}{=} \dfrac{5}{10}$$

Since $$x = 1$$ does not result in a true equation, $$1$$ is not a solution to the equation.

 Substitute $$\textcolor{red}{\dfrac{4}{5}}$$ for x. $$\textcolor{red}{\dfrac{4}{5}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{1}{2}$$ $$\textcolor{red}{\dfrac{8}{10}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{5}{10}$$ Subtract. $$\dfrac{5}{10} = \dfrac{5}{10} \; \checkmark$$

Since $$x = \dfrac{4}{5}$$ results in a true equation, $$\dfrac{4}{5}$$ is a solution to the equation $$x − \dfrac{3}{10} = \dfrac{1}{2}$$.

 Substitute $$\textcolor{red}{- \dfrac{4}{5}}$$ for x. $$\textcolor{red}{- \dfrac{4}{5}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{1}{2}$$ $$\textcolor{red}{- \dfrac{8}{10}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{5}{10}$$ Subtract. $$\dfrac{11}{10} \neq \dfrac{5}{10}$$

Since $$x = − \dfrac{4}{5}$$ does not result in a true equation, $$− \dfrac{4}{5}$$ is not a solution to the equation.

##### Exercise $$\PageIndex{1}$$

Determine whether each number is a solution of the given equation. $$x − \dfrac{2}{3} = \dfrac{1}{6}$$

1. $$x = 1$$
2. $$x = \dfrac{5}{6}$$
3. $$x = − \dfrac{5}{6}$$

no

yes

no

##### Exercise $$\PageIndex{2}$$

Determine whether each number is a solution of the given equation. $$y − \dfrac{1}{4} = \dfrac{3}{8}$$

1. $$y = 1$$
2. $$y = - \dfrac{5}{8}$$
3. $$y = \dfrac{5}{8}$$

no

no

yes

## Solve Equations with Fractions using the Addition, Subtraction, and Division Properties of Equality

In Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers; The Division Property of Equality, we solved equations using the Addition, Subtraction, and Division Properties of Equality. We will use these same properties to solve equations with fractions.

##### Addition, Subtraction, and Division Properties of Equality

For any numbers $$a$$, $$b$$, and $$c$$,

 if $$a = b$$, then $$a + c = b + c$$. Addition Property of Equality if $$a = b$$, then $$a − c = b − c$$. Subtraction Property of Equality if $$a = b$$, then $$a c = b c$$, $$c ≠ 0$$. Division Property of Equality

In other words, when you add or subtract the same quantity from both sides of an equation, or divide both sides by the same quantity, you still have equality.

##### Example $$\PageIndex{2}$$: solve

Solve: $$y + \dfrac{9}{16} = \dfrac{5}{16}$$.

Solution

 Subtract $$\dfrac{9}{16}$$ from each side to undo the addition. $$y + \dfrac{9}{16} \textcolor{red}{- \dfrac{9}{16}} = \dfrac{5}{16} \textcolor{red}{- \dfrac{9}{16}}$$ Simplify on each side of the equation. $$y + 0 = - \dfrac{4}{16}$$ Simplify the fraction. $$y = - \dfrac{1}{4}$$

Check:

 Substitute y = (− \dfrac{1}{4}\). $$\textcolor{red}{- \dfrac{1}{4}} + \dfrac{9}{16} \stackrel{?}{=} \dfrac{5}{16}$$ Rewrite as fractions with the LCD. \textcolor{red}{- \dfrac{4}{16}} + \dfrac{9}{16} \stackrel{?}{=} \dfrac{5}{16}\) Add. $$\dfrac{5}{16} \stackrel{?}{=} \dfrac{5}{16} \; \checkmark$$

Since $$y = − \dfrac{1}{4}$$ makes $$y + \dfrac{9}{16} = \dfrac{5}{16}$$ a true statement, we know we have found the solution to this equation.

##### Exercise $$\PageIndex{3}$$

Solve: $$y + \dfrac{11}{12} = \dfrac{5}{12}$$.

$$-\dfrac{1}{2}$$

##### Exercise $$\PageIndex{4}$$

Solve: $$y + \dfrac{8}{15} = \dfrac{4}{15}$$.

$$-\dfrac{4}{15}$$

We used the Subtraction Property of Equality in Example $$\PageIndex{2}$$. Now we’ll use the Addition Property of Equality.

##### Example $$\PageIndex{3}$$: solve

Solve: a − $$\dfrac{5}{9}$$ = $$− \dfrac{8}{9}$$.

Solution

 Add $$\dfrac{5}{9}$$ from each side to undo the addition. $$a - \dfrac{5}{9} \textcolor{red}{+ \dfrac{5}{9}} = - \dfrac{8}{9} \textcolor{red}{+ \dfrac{5}{9}}$$ Simplify on each side of the equation. $$a + 0 = - \dfrac{3}{9}$$ Simplify the fraction. $$a = - \dfrac{1}{3}$$

Check:

 Substitute a = $$− \dfrac{1}{3}$$. $$\textcolor{red}{- \dfrac{1}{3}} - \dfrac{5}{9} \stackrel{?}{=} - \dfrac{8}{9}$$ Change to common denominator. $$\textcolor{red}{- \dfrac{3}{9}} - \dfrac{5}{9} \stackrel{?}{=} - \dfrac{8}{9}$$ Subtract. $$- \dfrac{8}{9} = - \dfrac{8}{9} \; \checkmark$$

Since $$a = − \dfrac{1}{3}$$ makes the equation true, we know that $$a = − \dfrac{1}{3}$$ is the solution to the equation.

##### Exercise $$\PageIndex{5}$$

Solve: $$a − \dfrac{3}{5} = − \dfrac{8}{5}$$.

$$-1$$

##### Exercise $$\PageIndex{6}$$

Solve: $$n − \dfrac{3}{7} = − \dfrac{9}{7}$$.

$$-\dfrac{6}{7}$$

The next example may not seem to have a fraction, but let’s see what happens when we solve it.

##### Example $$\PageIndex{4}$$: solve

Solve: $$10q = 44$$.

Solution

 Divide both sides by 10 to undo the multiplication. $$\dfrac{10q}{10} = \dfrac{44}{10}$$ Simplify. $$q =\dfrac{22}{5}$$

Check:

 Substitute $$q = \dfrac{22}{5}$$ into the original equation. $$10 \left(\dfrac{22}{5}\right) \stackrel{?}{=} 44$$ Simplify. $$\stackrel{2}{\cancel{10}} \left(\dfrac{22}{\cancel{5}}\right) \stackrel{?}{=} 44$$ Multiply. $$44 = 44 \; \checkmark$$

The solution to the equation was the fraction $$\dfrac{22}{5}$$. We leave it as an improper fraction.

##### Exercise $$\PageIndex{7}$$

Solve: $$12u = −76$$.

$$-\dfrac{19}{3}$$

##### Exercise $$\PageIndex{8}$$

Solve: $$8m = 92$$.

$$\dfrac{23}{2}$$

## Solve Equations with Fractions Using the Multiplication Property of Equality

Consider the equation $$\dfrac{x}{4} = 3$$. We want to know what number divided by $$4$$ gives $$3$$. So to “undo” the division, we will need to multiply by $$4$$. The Multiplication Property of Equality will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal.

##### Definition: The Multiplication Property of Equality

For any numbers $$a$$, $$b$$, and $$c$$, if $$a = b$$, then $$ac = bc$$. If you multiply both sides of an equation by the same quantity, you still have equality.

Let’s use the Multiplication Property of Equality to solve the equation $$\dfrac{x}{7} = −9$$.

##### Example $$\PageIndex{5}$$: solve

Solve: $$\dfrac{x}{7} = −9$$.

Solution

 Use the Multiplication Property of Equality to multiply both sides by 7 . This will isolate the variable. $$\textcolor{red}{7} \cdot \dfrac{x}{7} = \textcolor{red}{7} (-9)$$ Multiply. $$\dfrac{7x}{7} = -63$$ Simplify. $$x = -63$$ Check. Substitute $$\textcolor{red}{-63}$$ for x in the original equation. $$\dfrac{\textcolor{red}{-63}}{7} \stackrel{?}{=} -9$$ The equation is true. $$-9 = -9 ; \checkmark$$
##### Exercise $$\PageIndex{9}$$

Solve: $$\dfrac{f}{5} = −25$$.

$$-125$$

##### Exercise $$\PageIndex{10}$$

Solve: $$\dfrac{h}{9} = −27$$.

$$-243$$

##### Example $$\PageIndex{6}$$:solve

Solve: $$\dfrac{p}{−8} = −40$$.

Solution

Here, $$p$$ is divided by $$−8$$. We must multiply by $$−8$$ to isolate $$p$$.

 Multiply both sides by −8. $$\textcolor{red}{-8} \left(\dfrac{p}{-8}\right) = \textcolor{red}{-8} (-40)$$ Multiply. $$\dfrac{-8p}{-8} = 320$$ Simplify. $$p = 320$$

Check:

 Substitute p = 320. $$\dfrac{\textcolor{red}{320}}{-8} \stackrel{?}{=} -40$$ The equation is true. $$-40 = -40 \; \checkmark$$
##### Exercise $$\PageIndex{11}$$

Solve: $$\dfrac{c}{−7} = −35$$.

$$245$$

##### Exercise $$\PageIndex{12}$$

Solve: $$\dfrac{x}{−11} = −12$$.

$$132$$

### Solve Equations with a Coefficient of $$−1$$

Look at the equation $$−y = 15$$. Does it look as if y is already isolated? But there is a negative sign in front of $$y$$, so it is not isolated.

There are three different ways to isolate the variable in this type of equation. We will show all three ways in Example $$\PageIndex{7}$$.

##### Example $$\PageIndex{7}$$: solve

Solve: $$−y = 15$$.

Solution

One way to solve the equation is to rewrite $$−y$$ as $$−1y$$, and then use the Division Property of Equality to isolate $$y$$.

 Rewrite −y as −1y. $$-1y = 15$$ Divide both sides by −1. $$\dfrac{-1y}{\textcolor{red}{-1}} = \dfrac{15}{\textcolor{red}{-1}}$$ Simplify each side. $$y = -15$$

Another way to solve this equation is to multiply both sides of the equation by −1.

 Multiply both sides by −1. $$\textcolor{red}{-1} (-y) = \textcolor{red}{-1} (15)$$ Simplify each side. $$y = -15$$

The third way to solve the equation is to read $$−y$$ as “the opposite of $$y$$.” What number has $$15$$ as its opposite? The opposite of $$15$$ is $$−15$$. So $$y = −15$$.

For all three methods, we isolated $$y$$ is isolated and solved the equation.

Check:

 Substitute y = −15. $$-(\textcolor{red}{15}) \stackrel{?}{=} (15)$$ Simplify. The equation is true. $$15 = 15 \; \checkmark$$
##### Exercise $$\PageIndex{13}$$

Solve: $$−y = 48$$.

$$-48$$

##### Exercise $$\PageIndex{14}$$

Solve: $$−c = −23$$.

$$23$$