Solving Linear Equations and Inequalities

21 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.

  1. Multiply: 8·\frac{3}{8}.
    If you missed this problem, review (Figure).
  2. Find the LCD of \frac{5}{6} and \frac{1}{4}.
    If you missed this problem, review (Figure).
  3. Multiply 4.78 by 100.
    If you missed this problem, review (Figure).

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}.

.
To isolate the x term, subtract \frac{1}{2} from both sides. .
Simplify the left side. .
Change the constants to equivalent fractions with the LCD. .
Subtract. .
Multiply both sides by the reciprocal of \frac{1}{8}. .
Simplify. .

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.

How to Solve Equations with Fraction Coefficients

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

Solution

This figure is a table that has three columns and three rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. On the top row of the table, the first cell on the left reads: “Step 1. Find the least common denominator of all the fractions in the equation.” The text in the second cell reads: “What is the LCD of 1/6, 1/3, and 5/6?” The third cell contains the equation one-sixth y minus 1/3 equals 5/6, with LCD equals 6 written next to it.In the second row of the table, the first cell says: “Step 2. Multiply both sides of the equation by that LCD. This clears the fractions.” In the second cell, the instructions say: “Multiply both sides of the equation by the LCD 6. Use the Distributive Property. Simplify—and notice, no more fractions!” The third cell contains the equation 6 times one-sixth y minus 1/3, with one-sixth y minus 1/3 in brackets, equals 6 times 5/6, with “6 times” written in red on both sides. Below this is the same equation with the 6 distributed on both sides: 6 times one-sixth y minus 6 times 1/3 equals 6 times 5/6. Below this is the equation y minus 2 equals 5.In the third row of the table, the first cell says: “Step 3. Solve using the General Strategy for Solving Linear Equations.” In the second cell, the instructions say: “Isolate the x term, add 2. Simplify.” The third cell contains the equation with 2 added to both sides: y minus 2 plus 2 equals 5 plus 2, with “plus 2” written in red on both sides. Below this is the equation y equals 7.

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

x=\frac{1}{2}

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

x=-2

Notice in (Figure), 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.

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.

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.

Find the LCD of all fractions in the equation. .
The LCD is 20.
Multiply both sides of the equation by 20. .
Distribute. .
Simplify—notice, no more fractions! .
Combine like terms. .
Divide by 3. .
Simplify. .
Check: .
Let v=40. .
.
.

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

x=12

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

u=-12

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

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

Solution
.
Find the LCD of all fractions in the equation.
The LCD is 8.
Multiply both sides by the LCD. .
Distribute. .
Simplify—no more fractions. .
Subtract 3a from both sides. .
Simplify. .
Subtract 6 from both sides. .
Simplify. .
Divide by 5. .
Simplify. .
Check: .
Let a=-2. .
.
.
.

Solve: x+\frac{1}{3}=\frac{1}{6}x-\frac{1}{2}.

x=-1

Solve: c+\frac{3}{4}=\frac{1}{2}c-\frac{1}{4}.

c=-2

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

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

Solution
.
Distribute. .
Simplify.
Now there are no fractions.
.
Subtract 1 from both sides. .
Simplify. .
Divide by 2. .
Simplify. .
Check: .
Let x=-3. .
.
.
.

Solve: -11=\frac{1}{2}\left(6p+2\right).

p=-4

Solve: 8=\frac{1}{3}\left(9q+6\right).

q=2

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

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

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

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

n=2

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

m=-1

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

Solution
.
Multiply by the LCD, 4. .
Simplify. .
Collect the variables to the right. .
Simplify. .
Divide. .
Simplify. .
Check: .
Let x=1. .
.
.

Solve: \frac{4y-7}{3}=\frac{y}{6}.

y=2

Solve: \frac{-2z-5}{4}=\frac{z}{8}.

z=-2

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

Solution
.
Multiply by the LCD, 12. .
Distribute. .
Simplify. .
Collect the variables to the right. .
Simplify. .
Collect the constants to the left. .
Simplify. .
Check: .
Let a=-12. .
.
.

Solve: \frac{b}{10}+2=\frac{b}{4}+5.

b=-20

Solve: \frac{c}{6}+3=\frac{c}{3}+4.

c=-6

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

Solution
.
Multiply by the LCD, 4. .
Distribute. .
Simplify. .
.
.
Collect the variables to the left. .
Simplify. .
Collect the constants to the right. .
Simplify. .
Divide by 5. .
Simplify. .
Check: .
Let q=-5. .
Finish the check on your own.

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

r=1

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

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.

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}\phantom{\rule{2em}{0ex}}0.02=\frac{2}{100}\phantom{\rule{2em}{0ex}}0.25=\frac{25}{100}\phantom{\rule{2em}{0ex}}1.5=1\frac{5}{10}

Notice, the LCD is 100.

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

.
Multiply both sides by 100. .
Distribute. .
Multiply, and now we have no more decimals. .
Collect the variables to the right. .
Simplify. .
Collect the constants to the left. .
Simplify. .
Divide by 19. .
Simplify. .
Check: Let x=8.
.

Solve: 0.14h+0.12=0.35h-2.4.

h=12

Solve: 0.65k-0.1=0.4k-0.35.

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.

Solve: 0.25x+0.05\left(x+3\right)=2.85.

Solution
.
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.

Solve: 0.25n+0.05\left(n+5\right)=2.95.

n=9

Solve: 0.10d+0.05\left(d-5\right)=2.15.

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.

Practice Makes Perfect

Solve Equations with Fraction Coefficients

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

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

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

x=1

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

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

y=-1

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

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

b=-2

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

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

x=3

\frac{1}{4}m-\frac{4}{5}m+\frac{1}{2}m=-1

\frac{5}{6}n-\frac{1}{4}n-\frac{1}{2}n=-2

n=-24

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

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

x=-4

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

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

z=-2

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

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

a=1

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

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

x=-6

1=\frac{1}{6}\left(12x-6\right)

1=\frac{1}{5}\left(15x-10\right)

x=1

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

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

q=7

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

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

x=-\frac{5}{2}

\frac{5q-8}{5}=\frac{2q}{10}

\frac{4m+2}{6}=\frac{m}{3}

m=-1

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

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

p=-4

\frac{u}{3}-4=\frac{u}{2}-3

\frac{v}{10}+1=\frac{v}{4}-2

v=20

\frac{c}{15}+1=\frac{c}{10}-1

\frac{d}{6}+3=\frac{d}{8}+2

d=-24

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

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

y=-1

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

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

v=4

Solve Equations with Decimal Coefficients

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

0.6y+3=9

0.4y-4=2

y=15

3.6j-2=5.2

2.1k+3=7.2

k=2

0.4x+0.6=0.5x-1.2

0.7x+0.4=0.6x+2.4

x=20

0.23x+1.47=0.37x-1.05

0.48x+1.56=0.58x-0.64

x=22

0.9x-1.25=0.75x+1.75

1.2x-0.91=0.8x+2.29

x=8

0.05n+0.10\left(n+8\right)=2.15

0.05n+0.10\left(n+7\right)=3.55

n=19

0.10d+0.25\left(d+5\right)=4.05

0.10d+0.25\left(d+7\right)=5.25

d=10

0.05\left(q-5\right)+0.25q=3.05

0.05\left(q-8\right)+0.25q=4.10

q=15

Everyday Math

Coins Taylor has ?2.00 in dimes and pennies. The number of pennies is 2 more than the number of dimes. Solve the equation 0.10d+0.01\left(d+2\right)=2 for d, the number of dimes.

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

s=35

Writing Exercises

Explain how you find the least common denominator of \frac{3}{8}, \frac{1}{6}, and \frac{2}{3}.

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

Answers will vary.

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

In the equation 0.35x+2.1=3.85 what is the LCD? How do you know?

100. Justifications will vary.

Self Check

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

This is a table that has three rows and four columns. In the first row, which is a header row, the cells read from left to right: “I can…,” “confidently,” “with some help,” and “no-I don’t get it!” The first column below “I can…” reads: “solve equations with fraction coefficients,” and “solve equations with decimal coefficients.” The rest of the cells are blank.

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

License

Icon for the Creative Commons Attribution 4.0 International License

Elementary Algebra by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Share This Book