# 3.5 Solve Equations with Fractions or Decimals

## Learning Objectives

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

• Solve equations with fraction coefficients
• Solve equations with decimal coefficients

Before you get started, take this readiness quiz:

## Try It

1) Multiply: $8\times\frac{3}{8}$
2) Find the LCD of $\frac{5}{6}$ and $\frac{1}{4}$
3) Multiply $4.78$ by $100$

## Solve Equations with Fraction Coefficients

Let’s use the general strategy for solving linear equations introduced earlier to solve the equation, $\frac{1}{8}x+\frac{1}{2}=\frac{1}{4}$.

\begin{alignat}{3} &\;&\;\;\;\;\;\frac18x+\frac12&=\frac14\\ &\text{To isolate the}\;x\;\text{term, subtract}\;\frac12\;\text{from both sides.}&\;\;\;\;\;\frac18x+\frac12{\color{red}{-}}{\color{red}{\frac12}}&=\frac14{\color{red}{-}}{\color{red}{\frac12}}\\ &\text{Simplify the left side.}&\;\;\;\;\;\frac18x&=\frac14-\frac12\\ &\text{Change the constants to equivalent fractions with the LCD.}&\;\;\;\;\;\frac18x&=\frac14-\frac24\\ &\text{Subtract.}&\;\;\;\;\;\frac18x&=-\frac14\\ &\text{Multiply both sides by the reciprocal of}\;\frac18&\;\;\;\;\;\;{\color{red}{\frac81}}\cdot\frac18x&={\color{red}{\frac81}}\left(-\frac14\right)\\ &\text{Simplify.}&\;\;\;\;\;x&=-2 \end{alignat}

This method worked 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 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.

Let’s solve a similar equation, but this time use the method that eliminates the fractions.

### Example 3.5.1

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

Solution

Step 1: Find the least common denominator of all the fractions in the equation.

What is the LCD of $\frac16$, $\frac13$, and $\frac56$?

$\frac16y-\frac13=\frac56 \;\;LCD=6$

Step 2: Multiply both sides of the equation by that LCD. This clears the fractions.

\begin{align*} &\text{Multiply both sides of the equation by the LCD 6.}\;&{\color{red}{6}}{\color{red}{\left({\color{black}{\frac16}}{\color{black}{y}}{\color{black}{-}}{\color{black}{\frac13}}\right)}}&{\color{black}{=}}{\color{red}{6}}{\color{red}{\left({\color{black}{\frac56}}\right)}}\\ &\text{Use the Distributive Property.}\;&{\color{black}{6}}{\color{black}{\cdot}}{\color{black}{\frac16}}{\color{black}{y}}{\color{black}{-}}{\color{black}{6}}{\color{black}{\cdot}}{\color{black}{\frac13}}&{\color{black}{=}}{\color{black}{6}}{\color{black}{\cdot}}{\color{black}{\frac56}}\\ &\text{Simplify - and notice, no more fractions!}\;&{\color{black}{y}}{\color{black}{-}}{\color{black}{2}}&{\color{black}{=}}{\color{black}{5}} \end{align*}

Step 3: Sole using the General Strategy for Solving Linear Equations.

\begin{align*} &\text{To isolate the "y" term, add 2.}\;&y-2{\color{red}{+}}{\color{red}{2}}&=5{\color{red}{+}}{\color{red}{2}}\\ &\text{Simplify.}\;&y&=7\end{align*}

### Try It

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

Solution

$x=\frac{1}{2}$

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

Solution

$x=-2$

Notice in Example 3.5.1, 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

#### Strategy to solve equations with fraction coefficients.

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

### Example 3.5.2

Solve: $6=\frac{1}{2}v+\frac{2}{5}v-\frac{3}{4}v$

Solution

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

Step 1: Find the LCD of all fractions in the equation.

$6=\frac12v+\frac25v-\frac34v$

Step 2: The LCD is $20$.

Step 3: Multiply both sides of the equation by $20$.

${\color{red}{20}}{\left(6\right)=}{\color{red}{20}}{\cdot\left(\frac12v+\frac25v-\frac34v\right)}$

Step 4: Distribute.

\begin{align*}&\;&20(6)&=20\cdot\frac12v+20\cdot\frac25v-20\cdot\frac34v\\ &\text{Simplify - notice, no more fractions!.}\;&120&=10v+8v-15v\end{align*}

Step 5: Combine like terms.

$120=3v$

Step 6: Divide by $3$.

\begin{align*}&\;&\frac{120}{\color{red}{3}}&=\frac{3v}{\color{red}{3}}\\ &\text{Simplify.}\;&40&=v \end{align*}

Step 7: Check:

$6=\frac12v+\frac25v-\frac34v$

Step 8: Let $v=40$.

\begin{align*} 6&\overset?=\frac12\left({\color{red}{40}}\right)+\frac25\left({\color{red}{40}}\right)-\frac34\left({\color{red}{40}}\right)\\ 6&\overset?=20+16-30\\ 6&=6\checkmark \end{align*}

### Try It

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

Solution

$x=12$

7) Solve: $-1=\frac{1}{2}u+\frac{1}{4}u-\frac{2}{3}u$

Solution

$u=-12$

In the next example, we again have variables on both sides of the equation.

### Example 3.5.3

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

Solution

Step 1: Find the LCD of all fractions in the equation.

The LCD is $8$.

Step 2: Multiply both sides by the LCD.

${\color{red}{8}}{\left(a+\frac34\right)=}{\color{red}{8}}{\left(\frac38a-\frac12\right)}$

Step 3: Distribute.

\begin{align*}&\;&8\cdot a+8\cdot\frac34&=8\cdot\frac38a-8\cdot\frac12\\ &\text{Simplify - no more fractions.}\;&8a+6&=3a-4\end{align*}

Step 4: Subtract $3a$ from both sides.

\begin{align*}&\;&8a{\color{red}{-}}{\color{red}{3}}{\color{red}{a}}+6&=3a{\color{red}{-}}{\color{red}{3}}{\color{red}{a}}-4\\ &\text{Simplify.}\;&5a+6&=-4 \end{align*}

Step 5: Subtract $6$ from both sides.

\begin{align*}&\;&5a+6{\color{red}{-}}{\color{red}{6}}&=-4{\color{red}{-}}{\color{red}{6}}\\ &\text{Simplify.}\;&5a&=-10 \end{align*}

Step 6: Divide by $5$.

\begin{align*}&\;&\frac{5a}{\color{red}{5}}&=\frac{-10}{\color{red}{5}}\\ &\text{Simplify.}\;&a&=-2 \end{align*}

Step 7: Check:

$a+\frac34=\frac38a-\frac12$

Step 8: Let $a=-2$.

\begin{align*} {\color{red}{-}}{\color{red}{2}}+\frac34&\overset?=\frac38\left({\color{red}{-}}{\color{red}{2}}\right)-\frac12\\ -\frac84+\frac34&\overset?=-\frac{16}8-\frac48\\ -\frac54&=-\frac{10}8\\ -\frac54&=-\frac54\checkmark \end{align*}

### Try It

8) Solve: $x+\frac{1}{3}=\frac{1}{6}x-\frac{1}{2}$

Solution

$x=-1$

9) Solve: $c+\frac{3}{4}=\frac{1}{2}c-\frac{1}{4}$

Solution

$c=-2$

In the next example, we start by using the Distributive Property. This step clears the fractions right away.

### Example 3.5.4

Solve: $-5=\frac{1}{4}(8x+4)$

Solution

Step 1: Distribute.

\begin{align*}&\;&-5&=\frac14\cdot8x+\frac14\cdot4\\ &\text{Simplify.}\;&-5&=2x+1 \end{align*}

Now there are no fractions.

Step 2: Subtract $1$ from both sides.

\begin{align*}&\;&-5{\color{red}{-}}{\color{red}{1}}&=2+1{\color{red}{-}}{\color{red}{1}}\\ &\text{Simplify.}\;&-6&=2x \end{align*}

Step 3: Divide by $2$.

\begin{align*}&\;&\frac{-6}{\color{red}{2}}&=\frac{2x}{\color{red}{2}}\\ &\text{Simplify.}\;&-3&=x \end{align*}

Step 4: Check:

$-5=\frac14(8x+4)$

Step 5: Let $x=-3$.

\begin{align*} -5&\overset?=\frac12\left(4\left({\color{red}{-}}{\color{red}{3}}\right)+2\right)\\ -5&\overset?=\frac12\left(-12+2\right)\\ -5&\overset?=\frac12\left(-10\right)\\ -5&=-5\checkmark \end{align*}

### Try It

10) Solve: $-11=\frac{1}{2}(6p+2)$

Solution

$p=-4$

11) Solve: $8=\frac{1}{3}(9q+6)$

Solution

$q=2$

In the next example, even after distributing, we still have fractions to clear.

### Example 3.5.5

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

Solution

Step 1: Distribute.

\begin{align*}&\;&\frac12\cdot y-\frac12\cdot5&=\frac14\cdot y-\frac14\cdot1\\ &\text{Simplify.}\;&\frac12y-\frac52&=\frac14y-\frac14 \end{align*}

Step 2: Multiply by the LCD, $4$.

\begin{align*}&\;&{\color{red}{4}}\left(\frac12y-\frac52\right)&={\color{red}{4}}\left(\frac14y-\frac14\right)\\ &\text{Distribute.}\;&4\cdot\frac12y-4\cdot\frac52&=4\cdot\frac14y-4\cdot\frac14\\ &\text{Simplify.}\;&2y-10&=y-1 \end{align*}

Step 3: Collect the variables to the left.

\begin{align*}&\;&2y{\color{red}{-}}{\color{red}{y}}-10&=y{\color{red}{-}}{\color{red}{y}}-1\\ &\text{Simplify.}\;&y-10&=-1 \end{align*}

Step 4: Collect the constants to the right.

\begin{align*}&\;&y-10{\color{red}{+}}{\color{red}{10}}&=-1{\color{red}{+}}{\color{red}{10}}\\ &\text{Simplify.}\;&y&=9 \end{align*}

Step 5: Check:

$\frac12(y-5)=\frac14(y-1)$

Step 6: Let $y=9$.

$\frac12\left({\color{red}{9}}-5\right)\overset?=\frac14\left({\color{red}{9}}-1\right)$

Finish the check on your own.

### Try It

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

Solution

$n=2$

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

Solution

$m=-1$

### Example 3.5.6

Solve: $\frac{5x-3}{4}=\frac{x}{2}$

Solution

Step 1: Multiply by the LCD, $4$.

\begin{align*}&\;&{\color{red}{4}}\left(\frac{5x-3}4\right)&={\color{red}{4}}\left(\frac x2\right)\\ &\text{Simplify.}\;&5x-3&=2x \end{align*}

Step 2: Collect the variables to the right.

\begin{align*}&\;&5x{\color{red}{-}}{\color{red}{5}}{\color{red}{x}}-3&=2x{\color{red}{-}}{\color{red}{5}}{\color{red}{x}}\\ &\text{Simplify.}\;&-3&=-3x \end{align*}

Step 3: Divide.

\begin{align*}&\;&\frac{-3}{{\color{red}{-}}{\color{red}{3}}}&=\frac{-3x}{{\color{red}{-}}{\color{red}{3}}}\\ &\text{Simplify.}\;&1&=x \end{align*}

Step 4: Check:

$\frac{5x-3}{4}=\frac{x}{2}$

Step 5: Let $x =1$.

\begin{align*} \frac{5\left({\color{red}{1}}\right)-3}4&\overset?=\frac{\color{red}{1}}2\\ \frac24&\overset?=\frac12\\ \frac12&=\frac12\checkmark \end{align*}

### Try It

14) Solve: $\frac{4y-7}{3}=\frac{y}{6}$

Solution

$y=2$

15) Solve: $\frac{-2z-5}{4}=\frac{z}{8}$

Solution

$z=-2$

### Example 3.5.7

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

Solution

Step 1: Multiply by the LCD, $12$.

${\color{red}{12}}{\left(\frac a6+2\right)=}{\color{red}{12}}{\left(\frac a4+3\right)}$

Step 2: Distribute.

\begin{align*}&\;&12\cdot\frac{a}{6}+12\cdot2&=12\cdot\frac{a}{4}+12\cdot3\\ &\text{Simplify.}\;&2a+24&=3a+36\\ \end{align*}

Step 3: Collect the variables to the right.

\begin{align*}&\;&2a{\color{red}{-}}{\color{red}{2}}{\color{red}{a}}+24&=3a{\color{red}{-}}{\color{red}{2}}{\color{red}{a}}+36\\ &\text{Simplify.}\;&24&=a+36\\ \end{align*}

Step 4: Collect the constants to the left.

\begin{align*}&\;&24{\color{red}{-}}{\color{red}{36}}&=a+36{\color{red}{-}}{\color{red}{36}}\\ &\text{Simplify.}\;&a&=-12\\ \end{align*}

Step 5: Check:

$\frac{a}{6}+2=\frac{a}{4}+3$

Step 6: Let $a=-12$.

\begin{align*} \frac{{\color{red}{-}}{\color{red}{12}}}6+2&\overset?=\frac{{\color{red}{-}}{\color{red}{12}}}4+3\\ -2+2&\overset?=-3+3\\ 0&=0\checkmark \end{align*}

### Try It

16) Solve: $\frac{b}{10}+2=\frac{b}{4}+5$

Solution

$b=-20$

17) Solve: $\frac{c}{6}+3=\frac{c}{3}+4$

Solution

$c=-6$

### Example 3.5.8

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

Solution

Step 1: Multiply by the LCD, $4$.

${\color{red}{4}}{\left(\frac{4q+3}2+6\right)=}{\color{red}{4}}{\left(\frac{3q+5}4\right)}$

Step 2: Distribute.

\begin{align*}&\;&4\left(\frac{4q+3}2\right)+4\cdot6&=4\cdot\left(\frac{3q+5}4\right)\\ &\text{Simplify.}\;&2(4q+3)+24&=3q+5\\ &\;&8q+6+24&=3q+5\\ &\;&8q+30&=3q+5 \end{align*}

Step 3: Collect the variables to the left.

\begin{align*}&\;&8q{\color{red}{-}}{\color{red}{3}}{\color{red}{q}}+30&=3q{\color{red}{-}}{\color{red}{3}}{\color{red}{q}}+5\\ &\text{Simplify.}\;&5q+30&=5\\ \end{align*}

Step 4: Collect the constants to the right.

\begin{align*}&\;&5q+30{\color{red}{-}}{\color{red}{30}}&=5{\color{red}{-}}{\color{red}{30}}\\ &\text{Simplify.}\;&5q&=-25\\ \end{align*}

Step 5: Divide by $5$.

\begin{align*}&\;&\frac{5q}{\color{red}{5}}&=\frac{-25}{\color{red}{5}}\\ &\text{Simplify.}\;&q&=-5\\ \end{align*}

Step 6: Check:

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

Step 7: Let $q=-5$.

$\frac{4\left({\color{red}{-}}{\color{red}{5}}\right)+3}2+6\overset?=\frac{3\left({\color{red}{-}}{\color{red}{5}}\right)+5}4$

Finish the check on your own.

### Try It

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

Solution

$r=1$

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

Solution

$s=-8$

## Solve Equations with Decimal Coefficients

Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money or percentages. But decimals can also be expressed as fractions. For example, $0.3=\frac{3}{10}$ and $0.17=\frac{17}{100}$. 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.

### Example 3.5.9

Solve: $0.06x+0.02=0.25x-1.5$

Solution

Look at the decimals and think of the equivalent fractions.

$0.06=\frac{6}{100}$        $0.02=\frac{2}{100}$        $0.25=\frac{25}{100}$          $1.5=1\frac{5}{10}$

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

$0.06x+0.02=0.25x-1.5$

Step 1: Multiply both sides by $100$.

${\color{red}{100}}{(0.06x+0.02)=}{\color{red}{100}}{(0.25x-1.5)}$

Step 2: Distribute.

$100(0.06x)+100(0.02)=100(0.25x)-100(1.5)$

Step 3: Multiply, and now we have no more decimals.

$6x+2=25x-150$

Step 4: Collect the variables to the right.

\begin{align*}&\;&6x{\color{red}{-}}{\color{red}{6}}{\color{red}{x}}+2&=25x{\color{red}{-}}{\color{red}{6}}{\color{red}{x}}-150\\ &\text{Simplify.}\;&2&=19x-150\\ \end{align*}

Step 5: Collect the constants to the left.

\begin{align*}&\;&2{\color{red}{+}}{\color{red}{150}}&=19x-150{\color{red}{+}}{\color{red}{150}}\\ &\text{Simplify.}\;&152&=19x\\ \end{align*}

Step 6: Divide by $19$.

\begin{align*}&\;&\frac{152}{\color{red}{19}}&=\frac{19x}{\color{red}{19}}\\ &\text{Simplify.}\;&8&=x\\ \end{align*}

Step 7: Check: Let $x=8$.

\begin{align*} 0.06\left({\color{red}{8}}\right)+0.02&\overset?=0.25\left({\color{red}{8}}\right)-1.5\\ 0.48+0.02&\overset?=2.00-1.5\\ 0.50&=0.50\checkmark \end{align*}

### Try It

20) Solve: $0.14h+0.12=0.35h-2.4$

Solution

$h=12$

21) Solve: $0.65k-0.1=0.4k-0.35$

Solution

$k=-1$

The next example uses an equation that is typical of the money applications in the next chapter. Notice that we distribute the decimal before we clear all the decimals.

### Example 3.5.10

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

Solution

Step 1: Distribute first.

$0.25x+0.05x+0.15=2.85$

Step 2: Combine like terms.

$0.30x+0.15=2.85$

Step 3: To clear decimals, multiply by $100$.

${\color{red}{100}}{\left(0.30x+0.15\right)=}{\color{red}{100}}{\left(2.8\right)}$

Step 4: Distribute.

$30x+15=285$

Step 5: Subtract $15$ from both sides.

\begin{align*}&\;&30x+15{\color{red}{-}}{\color{red}{15}}&=285{\color{red}{-}}{\color{red}{15}}\\ &\text{Simplify.}\;&30x&=270\\ \end{align*}

Step 6: Divide by $30$.

\begin{align*}&\;&\frac{30x}{\color{red}{30}}&=\frac{270}{\color{red}{30}}\\ &\text{Simplify.}\;&x&=9\\ \end{align*}

Step 7: Check it yourself by substituting $x=9$ into the original equation.

### Try It

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

Solution

$n=9$

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

Solution

$d=16$

### Key Concepts

• Strategy to Solve an Equation with Fraction Coefficients
1. Find the least common denominator of all the fractions in the equation.
2. Multiply both sides of the equation by that LCD. This clears the fractions.
3. Solve using the General Strategy for Solving Linear Equations.

### Self Check

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

Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

## License

Fanshawe Pre-Health Sciences Mathematics 1 Copyright © 2022 by Domenic Spilotro, MSc is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, except where otherwise noted.