Solve Equations with Fraction or Decimal Coefficients

By the end of this section, you will be able to: solve equations with fraction coefficients, and solve equations with decimal coefficients.

Solve equations with fraction coefficients

Let’s use the General Strategy for Solving Linear Equations introduced earlier to solve the equation $\tfrac{1}{8}x + \tfrac{1}{2} = \tfrac{1}{4}$ .

To isolate the $x$ term, subtract $\tfrac{1}{2}$ from both sides:

\tfrac{1}{8}x + \tfrac{1}{2} - \tfrac{1}{2} = \tfrac{1}{4} - \tfrac{1}{2}

Simplify the left side:

\tfrac{1}{8}x = \tfrac{1}{4} - \tfrac{1}{2}

Change the constants to equivalent fractions with the LCD:

\tfrac{1}{8}x = \tfrac{1}{4} - \tfrac{2}{4}

Subtract:

\tfrac{1}{8}x = -\tfrac{1}{4}

Multiply both sides by the reciprocal of $\tfrac{1}{8}$ :

8 \cdot \tfrac{1}{8}x = 8\left(-\tfrac{1}{4}\right)

Simplify:

x = -2

This method worked fine, but many students don’t 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 with no fractions. This process is called clearing the equation of fractions. Let’s solve the same equation again, but this time use the method that clears the fractions.

Example. Solve: $\tfrac{1}{8}x + \tfrac{1}{2} = \tfrac{1}{4}$ .

Find the least common denominator of all the fractions in the equation. Here, the LCD is $8$ .

Multiply both sides of the equation by that LCD, $8$ . This clears the fractions:

8\left(\tfrac{1}{8}x + \tfrac{1}{2}\right) = 8\left(\tfrac{1}{4}\right)

Use the Distributive Property:

8 \cdot \tfrac{1}{8}x + 8 \cdot \tfrac{1}{2} = 8 \cdot \tfrac{1}{4}

Simplify — and notice, no more fractions!

x + 4 = 2

Solve using the General Strategy for Solving Linear Equations. Subtract $4$ from both sides:

x + 4 - 4 = 2 - 4

Simplify:

x = -2

Check: substitute $x = -2$ into $\tfrac{1}{8}x + \tfrac{1}{2} = \tfrac{1}{4}$ :

\tfrac{1}{8}(-2) + \tfrac{1}{2} \stackrel{?}{=} \tfrac{1}{4} \qquad -\tfrac{2}{8} + \tfrac{4}{8} \stackrel{?}{=} \tfrac{1}{4} \qquad \tfrac{2}{4} \stackrel{?}{=} \tfrac{1}{4} \qquad \tfrac{1}{4} = \tfrac{1}{4} \checkmark

Notice in this example that 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.

Solve equations with fraction coefficients by clearing the fractions.

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

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

Solve: $\tfrac{1}{6}y - \tfrac{1}{3} = \tfrac{1}{6}$ .

Example. Solve: $7 = \tfrac{1}{2}x + \tfrac{3}{4}x - \tfrac{2}{3}x$ .

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 least common denominator of all the fractions in the equation. Here, the LCD is $12$ .

Multiply both sides of the equation by $12$ :

12(7) = 12 \cdot \left(\tfrac{1}{2}x + \tfrac{3}{4}x - \tfrac{2}{3}x\right)

Distribute:

12(7) = 12 \cdot \tfrac{1}{2}x + 12 \cdot \tfrac{3}{4}x - 12 \cdot \tfrac{2}{3}x

Simplify — and notice, no more fractions!

84 = 6x + 9x - 8x

Combine like terms:

84 = 7x

Divide by $7$ :

\tfrac{84}{7} = \tfrac{7x}{7}

Simplify:

12 = x

Check: substitute $x = 12$ into $7 = \tfrac{1}{2}x + \tfrac{3}{4}x - \tfrac{2}{3}x$ :

7 \stackrel{?}{=} \tfrac{1}{2}(12) + \tfrac{3}{4}(12) - \tfrac{2}{3}(12) \qquad 7 \stackrel{?}{=} 6 + 9 - 8 \qquad 7 = 7 \checkmark

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

In the next example, we’ll have variables and fractions on both sides of the equation.

Example. Solve: $x + \tfrac{1}{3} = \tfrac{1}{6}x - \tfrac{1}{2}$ .

Find the LCD of all the fractions in the equation. Here, the LCD is $6$ .

Multiply both sides by the LCD:

6\left(x + \tfrac{1}{3}\right) = 6\left(\tfrac{1}{6}x - \tfrac{1}{2}\right)

Distribute:

6 \cdot x + 6 \cdot \tfrac{1}{3} = 6 \cdot \tfrac{1}{6}x - 6 \cdot \tfrac{1}{2}

Simplify — no more fractions!

6x + 2 = x - 3

Subtract $x$ from both sides:

6x - x + 2 = x - x - 3

Simplify:

5x + 2 = -3

Subtract $2$ from both sides:

5x + 2 - 2 = -3 - 2

Simplify:

5x = -5

Divide by $5$ :

\tfrac{5x}{5} = \tfrac{-5}{5}

Simplify:

x = -1

Check: substitute $x = -1$ into $x + \tfrac{1}{3} = \tfrac{1}{6}x - \tfrac{1}{2}$ :

(-1) + \tfrac{1}{3} \stackrel{?}{=} \tfrac{1}{6}(-1) - \tfrac{1}{2} \qquad -\tfrac{2}{3} \stackrel{?}{=} -\tfrac{1}{6} - \tfrac{3}{6} \qquad -\tfrac{2}{3} = -\tfrac{2}{3} \checkmark

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

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

In the next example, we’ll start by using the Distributive Property. This step will clear the fractions right away.

Example. Solve: $1 = \tfrac{1}{2}(4x + 2)$ .

Distribute:

1 = \tfrac{1}{2} \cdot 4x + \tfrac{1}{2} \cdot 2

Simplify. Now there are no fractions to clear!

1 = 2x + 1

Subtract $1$ from both sides:

1 - 1 = 2x + 1 - 1

Simplify:

0 = 2x

Divide by $2$ :

\tfrac{0}{2} = \tfrac{2x}{2}

Simplify:

0 = x

Check: substitute $x = 0$ into $1 = \tfrac{1}{2}(4x + 2)$ :

1 \stackrel{?}{=} \tfrac{1}{2}(4(0) + 2) \qquad 1 \stackrel{?}{=} \tfrac{1}{2}(2) \qquad 1 \stackrel{?}{=} \tfrac{2}{2} \qquad 1 = 1 \checkmark

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

Many times, there will still be fractions, even after distributing.

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

Distribute:

\tfrac{1}{2} \cdot y - \tfrac{1}{2} \cdot 5 = \tfrac{1}{4} \cdot y - \tfrac{1}{4} \cdot 1

Simplify:

\tfrac{1}{2}y - \tfrac{5}{2} = \tfrac{1}{4}y - \tfrac{1}{4}

Multiply by the LCD, $4$ :

4\left(\tfrac{1}{2}y - \tfrac{5}{2}\right) = 4\left(\tfrac{1}{4}y - \tfrac{1}{4}\right)

Distribute:

4 \cdot \tfrac{1}{2}y - 4 \cdot \tfrac{5}{2} = 4 \cdot \tfrac{1}{4}y - 4 \cdot \tfrac{1}{4}

Simplify:

2y - 10 = y - 1

Collect the $y$ terms to the left:

2y - 10 - y = y - 1 - y

Simplify:

y - 10 = -1

Collect the constants to the right:

y - 10 + 10 = -1 + 10

Simplify:

y = 9

Check: substitute $y = 9$ into $\tfrac{1}{2}(y - 5) = \tfrac{1}{4}(y - 1)$ :

\tfrac{1}{2}(9 - 5) \stackrel{?}{=} \tfrac{1}{4}(9 - 1) \qquad \tfrac{1}{2}(4) \stackrel{?}{=} \tfrac{1}{4}(8) \qquad 2 = 2 \checkmark

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

Solve equations with decimal coefficients

Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money and percent. But decimals are really another way to represent fractions. For example, $0.3 = \tfrac{3}{10}$ and $0.17 = \tfrac{17}{100}$ . So, when we have an equation with decimals, we can use the same process we used to clear fractions — multiply both sides of the equation by the least common denominator.

Example. Solve: $0.8x - 5 = 7$ .

The only decimal in the equation is $0.8$ . Since $0.8 = \tfrac{8}{10}$ , the LCD is $10$ . We can multiply both sides by $10$ to clear the decimal.

Multiply both sides by the LCD:

10(0.8x - 5) = 10(7)

Distribute:

10(0.8x) - 10(5) = 10(7)

Multiply, and notice, no more decimals!

8x - 50 = 70

Add $50$ to get all constants to the right:

8x - 50 + 50 = 70 + 50

Simplify:

8x = 120

Divide both sides by $8$ :

\tfrac{8x}{8} = \tfrac{120}{8}

Simplify:

x = 15

Check: substitute $x = 15$ into $0.8x - 5 = 7$ :

0.8(15) - 5 \stackrel{?}{=} 7 \qquad 12 - 5 \stackrel{?}{=} 7 \qquad 7 = 7 \checkmark

Solve: $0.6x - 1 = 11$ .

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

Look at the decimals and think of the equivalent fractions: $0.06 = \tfrac{6}{100}$ , $0.02 = \tfrac{2}{100}$ , $0.25 = \tfrac{25}{100}$ , $1.5 = 1\tfrac{5}{10}$ . Notice the LCD is $100$ .

By multiplying by the LCD we will clear the decimals. Multiply both sides by $100$ :

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

Distribute:

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

Multiply, and now no more decimals:

6x + 2 = 25x - 150

Collect the variables to the right by subtracting $6x$ from both sides:

6x - 6x + 2 = 25x - 6x - 150

Simplify:

2 = 19x - 150

Collect the constants to the left by adding $150$ to both sides:

2 + 150 = 19x - 150 + 150

Simplify:

152 = 19x

Divide by $19$ :

\tfrac{152}{19} = \tfrac{19x}{19}

Simplify:

8 = x

Check: substitute $x = 8$ into $0.06x + 0.02 = 0.25x - 1.5$ :

0.06(8) + 0.02 \stackrel{?}{=} 0.25(8) - 1.5 \qquad 0.48 + 0.02 \stackrel{?}{=} 2.00 - 1.5 \qquad 0.50 = 0.50 \checkmark

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

The next example uses an equation that is typical of the ones we will see in money applications later in this course. Notice that we will distribute the decimal first before we clear all decimals in the equation.

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

Distribute first:

0.25x + 0.05x + 0.15 = 2.85

Combine like terms:

0.30x + 0.15 = 2.85

To clear decimals, multiply by $100$ :

100(0.30x + 0.15) = 100(2.85)

Distribute:

30x + 15 = 285

Subtract $15$ from both sides:

30x + 15 - 15 = 285 - 15

Simplify:

30x = 270

Divide by $30$ :

\tfrac{30x}{30} = \tfrac{270}{30}

Simplify:

x = 9

Check: substitute $x = 9$ into $0.25x + 0.05(x + 3) = 2.85$ :

0.25(9) + 0.05(9 + 3) \stackrel{?}{=} 2.85 \qquad 2.25 + 0.05(12) \stackrel{?}{=} 2.85 \qquad 2.25 + 0.60 \stackrel{?}{=} 2.85 \qquad 2.85 = 2.85 \checkmark

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

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

Key terms

clearing the equation of fractions — multiplying both sides of an equation by the least common denominator of all the fractions in the equation, so the resulting equivalent equation has no fractions. clearing the equation of decimals — the same idea applied to decimal coefficients: multiplying both sides by a power of ten equal to the LCD of the decimals’ equivalent fractions removes the decimal points.

This section is adapted from Prealgebra 2e, Section 8.4: Solve Equations with Fraction or Decimal Coefficients by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the Be Prepared quiz, Media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

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