5.5 Solving Linear Equations with Fractions or Decimals

Solving Linear Equations with Fractions or Decimals

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Solve Equations with Fraction Coefficients

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

	$A mathematical equation is displayed: 1/8x + 1/2 = 1/4. The equation involves fractions, a variable 'x', addition, and an equality sign, set against a plain white background.$
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.

Notice in Example 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.

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

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

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

Example 5

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

Show solution


Distribute.	$The image shows a linear equation involving fractions and the variable 'y': (1/2) * y - (1/2) * 5 = (1/4) * y - (1/4) * 1.$
Simplify.
Multiply by the LCD, 4.
Distribute.	$A linear equation involving fractions and the variable 'y' is presented as 4(1/2)y - 4(5/2) = 4(1/4)y - 4(1/4).$
Simplify.
Collect the variables to the left.
Simplify.
Collect the constants to the right.
Simplify.
Check:
Let $y equals 9$ .
Finish the check on your own.

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 9

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

Show 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.


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 $x equals 8$ .

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

Example 10

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

Show 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 $x equals 9$ into the original equation.

Practice Makes Perfect

Solve Equations with Fraction Coefficients

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

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

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

Show solution

$x equals 1$

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

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

Show solution

$y equals −1$

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

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

Show solution

$b equals −2$

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

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

Show solution

$x equals 3$

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

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

Show solution

$n equals −24$

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

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

Show solution

$x equals −4$

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

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

Show solution

$z equals −2$

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

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

Show solution

$a equals 1$

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

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

Show solution

$x equals −6$

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

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

Show solution

$x equals 1$

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

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

Show solution

$q equals 7$

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

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

Show solution

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

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

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

Show solution

$m equals −1$

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

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

Show solution

$p equals −4$

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

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

Show solution

$v equals 20$

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

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

Show solution

$d equals −24$

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

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

Show solution

$y equals −1$

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

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

Show solution

$v equals 4$

Solve Equations with Decimal Coefficients

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

$0.6 y + 3 = 9$

$0.4 y - 4 = 2$

Show solution

$y equals 15$

$3.6 j - 2 = 5.2$

$2.1 k + 3 = 7.2$

Show solution

$k equals 2$

$0.4 x + 0.6 = 0.5 x - 1.2$

$0.7 x + 0.4 = 0.6 x + 2.4$

Show solution

$x equals 20$

$0.23 x + 1.47 = 0.37 x - 1.05$

$0.48 x + 1.56 = 0.58 x - 0.64$

Show solution

$x equals 22$

$0.9 x - 1.25 = 0.75 x + 1.75$

$1.2 x - 0.91 = 0.8 x + 2.29$

Show solution

$x equals 8$

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

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

Show solution

$n equals 19$

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

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

Show solution

$d equals 10$

$0.05 (q - 5) + 0.25 q = 3.05$

$0.05 (q - 8) + 0.25 q = 4.10$

Show solution

$q equals 15$

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 $0.10 d + 0.01 (d + 2) = 2$ 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 $0.49 s + 0.21 (s - 8) = 22.82$ for s, to find the number of 49-cent stamps Paula bought.

Show solution

$s equals 35$

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?

Show solution

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

Show solution

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?

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