Solve Equations with Fractions or Decimals

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, we subtract $\tfrac{1}{2}$ from both sides, giving $\tfrac{1}{8}x = \tfrac{1}{4} - \tfrac{1}{2}$ . Changing the constants to equivalent fractions with the LCD $4$ and subtracting gives $\tfrac{1}{8}x = -\tfrac{1}{4}$ . Multiplying both sides by the reciprocal of $\tfrac{1}{8}$ , which is $8$ , gives $x = -2$ .

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 — one that eliminates the fractions.

We 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 is a new equation, equivalent to the first, but without fractions. This process is called “clearing” the equation of fractions.

Strategy to solve equations with fraction coefficients.

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.

Example. Solve: $\tfrac{1}{6}y - \tfrac{1}{3} = \tfrac{5}{6}$ .

The LCD of $\tfrac{1}{6}, \tfrac{1}{3},$ and $\tfrac{5}{6}$ is $6$ . We multiply both sides of the equation by $6$ , then distribute:

\begin{array}{lrcl} & \tfrac{1}{6}y - \tfrac{1}{3} &=& \tfrac{5}{6} \\[4pt] \text{Multiply both sides by the LCD, 6.} & 6\left(\tfrac{1}{6}y - \tfrac{1}{3}\right) &=& 6\left(\tfrac{5}{6}\right) \\[4pt] \text{Distribute.} & 6 \cdot \tfrac{1}{6}y - 6 \cdot \tfrac{1}{3} &=& 6 \cdot \tfrac{5}{6} \\[4pt] \text{Simplify — notice, no more fractions!} & y - 2 &=& 5 \\[4pt] \text{Add 2 to both sides.} & y - 2 + 2 &=& 5 + 2 \\[4pt] \text{Simplify.} & y &=& 7 \end{array}

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

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

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

We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation. The LCD of $\tfrac{1}{2}, \tfrac{2}{5},$ and $\tfrac{3}{4}$ is $20$ .

\begin{array}{lrcl} & 6 &=& \tfrac{1}{2}v + \tfrac{2}{5}v - \tfrac{3}{4}v \\[4pt] \text{Multiply both sides by the LCD, 20.} & 20(6) &=& 20\left(\tfrac{1}{2}v + \tfrac{2}{5}v - \tfrac{3}{4}v\right) \\[4pt] \text{Distribute.} & 20(6) &=& 20 \cdot \tfrac{1}{2}v + 20 \cdot \tfrac{2}{5}v - 20 \cdot \tfrac{3}{4}v \\[4pt] \text{Simplify — notice, no more fractions!} & 120 &=& 10v + 8v - 15v \\[4pt] \text{Combine like terms.} & 120 &=& 3v \\[4pt] \text{Divide by 3.} & \tfrac{120}{3} &=& \tfrac{3v}{3} \\[4pt] \text{Simplify.} & 40 &=& v \end{array}

Check by substituting $v = 40$ into the original equation: $6 = \tfrac{1}{2}(40) + \tfrac{2}{5}(40) - \tfrac{3}{4}(40)$ simplifies to $6 = 20 + 16 - 30$ , and indeed $6 = 6$ . ✓

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

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

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

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

The LCD of all the fractions in the equation is $8$ .

\begin{array}{lrcl} & a + \tfrac{3}{4} &=& \tfrac{3}{8}a - \tfrac{1}{2} \\[4pt] \text{Multiply both sides by the LCD, 8.} & 8\left(a + \tfrac{3}{4}\right) &=& 8\left(\tfrac{3}{8}a - \tfrac{1}{2}\right) \\[4pt] \text{Distribute.} & 8 \cdot a + 8 \cdot \tfrac{3}{4} &=& 8 \cdot \tfrac{3}{8}a - 8 \cdot \tfrac{1}{2} \\[4pt] \text{Simplify — no more fractions.} & 8a + 6 &=& 3a - 4 \\[4pt] \text{Subtract } 3a \text{ from both sides.} & 8a - 3a + 6 &=& 3a - 3a - 4 \\[4pt] \text{Simplify.} & 5a + 6 &=& -4 \\[4pt] \text{Subtract 6 from both sides.} & 5a + 6 - 6 &=& -4 - 6 \\[4pt] \text{Simplify.} & 5a &=& -10 \\[4pt] \text{Divide by 5.} & \tfrac{5a}{5} &=& \tfrac{-10}{5} \\[4pt] \text{Simplify.} & a &=& -2 \end{array}

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

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

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

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

\begin{array}{lrcl} & -5 &=& \tfrac{1}{4}(8x+4) \\[4pt] \text{Distribute.} & -5 &=& \tfrac{1}{4}\cdot 8x + \tfrac{1}{4}\cdot 4 \\[4pt] \text{Simplify — now there are no fractions.} & -5 &=& 2x + 1 \\[4pt] \text{Subtract 1 from both sides.} & -5 - 1 &=& 2x + 1 - 1 \\[4pt] \text{Simplify.} & -6 &=& 2x \\[4pt] \text{Divide by 2.} & \tfrac{-6}{2} &=& \tfrac{2x}{2} \\[4pt] \text{Simplify.} & -3 &=& x \end{array}

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

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

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

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

\begin{array}{lrcl} & \tfrac{1}{2}(y-5) &=& \tfrac{1}{4}(y-1) \\[4pt] \text{Distribute.} & \tfrac{1}{2}y - \tfrac{5}{2} &=& \tfrac{1}{4}y - \tfrac{1}{4} \\[4pt] \text{Multiply both sides by the LCD, 4.} & 4\left(\tfrac{1}{2}y - \tfrac{5}{2}\right) &=& 4\left(\tfrac{1}{4}y - \tfrac{1}{4}\right) \\[4pt] \text{Distribute.} & 4\cdot\tfrac{1}{2}y - 4\cdot\tfrac{5}{2} &=& 4\cdot\tfrac{1}{4}y - 4\cdot\tfrac{1}{4} \\[4pt] \text{Simplify.} & 2y - 10 &=& y - 1 \\[4pt] \text{Collect the variables to the left.} & 2y - y - 10 &=& y - y - 1 \\[4pt] \text{Simplify.} & y - 10 &=& -1 \\[4pt] \text{Collect the constants to the right.} & y - 10 + 10 &=& -1 + 10 \\[4pt] \text{Simplify.} & y &=& 9 \end{array}

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

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

Some equations have fractions written as a single quotient, such as $\tfrac{5x-3}{4}$ . Since a fraction bar is a grouping symbol, we treat the numerator as if it were in parentheses, even without seeing them.

Example. Solve: $\tfrac{5x-3}{4} = \tfrac{x}{2}$ .

\begin{array}{lrcl} & \tfrac{5x-3}{4} &=& \tfrac{x}{2} \\[4pt] \text{Multiply both sides by the LCD, 4.} & 4\left(\tfrac{5x-3}{4}\right) &=& 4\left(\tfrac{x}{2}\right) \\[4pt] \text{Simplify.} & 5x - 3 &=& 2x \\[4pt] \text{Collect the variables to the right.} & 5x - 5x - 3 &=& 2x - 5x \\[4pt] \text{Simplify.} & -3 &=& -3x \\[4pt] \text{Divide by }{-3}. & \tfrac{-3}{-3} &=& \tfrac{-3x}{-3} \\[4pt] \text{Simplify.} & 1 &=& x \end{array}

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

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

Example. Solve: $\tfrac{a}{6} + 2 = \tfrac{a}{4} + 3$ .

\begin{array}{lrcl} & \tfrac{a}{6} + 2 &=& \tfrac{a}{4} + 3 \\[4pt] \text{Multiply both sides by the LCD, 12.} & 12\left(\tfrac{a}{6} + 2\right) &=& 12\left(\tfrac{a}{4} + 3\right) \\[4pt] \text{Distribute.} & 12\cdot\tfrac{a}{6} + 12\cdot 2 &=& 12\cdot\tfrac{a}{4} + 12\cdot 3 \\[4pt] \text{Simplify.} & 2a + 24 &=& 3a + 36 \\[4pt] \text{Collect the variables to the right.} & 2a - 2a + 24 &=& 3a - 2a + 36 \\[4pt] \text{Simplify.} & 24 &=& a + 36 \\[4pt] \text{Collect the constants to the left.} & 24 - 36 &=& a + 36 - 36 \\[4pt] \text{Simplify.} & a &=& -12 \end{array}

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

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

Example. Solve: $\tfrac{4q+3}{2} + 6 = \tfrac{3q+5}{4}$ .

\begin{array}{lrcl} & \tfrac{4q+3}{2} + 6 &=& \tfrac{3q+5}{4} \\[4pt] \text{Multiply both sides by the LCD, 4.} & 4\left(\tfrac{4q+3}{2} + 6\right) &=& 4\left(\tfrac{3q+5}{4}\right) \\[4pt] \text{Distribute.} & 4\left(\tfrac{4q+3}{2}\right) + 4\cdot 6 &=& 4\left(\tfrac{3q+5}{4}\right) \\[4pt] \text{Simplify.} & 2(4q+3) + 24 &=& 3q+5 \\[4pt] & 8q + 6 + 24 &=& 3q + 5 \\[4pt] & 8q + 30 &=& 3q + 5 \\[4pt] \text{Collect the variables to the left.} & 8q - 3q + 30 &=& 3q - 3q + 5 \\[4pt] \text{Simplify.} & 5q + 30 &=& 5 \\[4pt] \text{Collect the constants to the right.} & 5q + 30 - 30 &=& 5 - 30 \\[4pt] \text{Simplify.} & 5q &=& -25 \\[4pt] \text{Divide by 5.} & \tfrac{5q}{5} &=& \tfrac{-25}{5} \\[4pt] \text{Simplify.} & q &=& -5 \end{array}

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

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

Solve Equations with Decimal Coefficients

Some equations have decimals in them. This kind of equation occurs when we solve problems dealing with money or percentages. But decimals can also be expressed as fractions — for example, $0.3 = \tfrac{3}{10}$ and $0.17 = \tfrac{17}{100}$ . 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. Solve: $0.06x + 0.02 = 0.25x - 1.5$ .

Looking at the decimals, $0.06 = \tfrac{6}{100}$ , $0.02 = \tfrac{2}{100}$ , $0.25 = \tfrac{25}{100}$ , and $1.5 = 1\tfrac{5}{10}$ , so the LCD is $100$ . By multiplying by the LCD, we clear the decimals from the equation:

\begin{array}{lrcl} & 0.06x + 0.02 &=& 0.25x - 1.5 \\[4pt] \text{Multiply both sides by 100.} & 100(0.06x + 0.02) &=& 100(0.25x - 1.5) \\[4pt] \text{Distribute.} & 100(0.06x) + 100(0.02) &=& 100(0.25x) - 100(1.5) \\[4pt] \text{Multiply — now there are no more decimals.} & 6x + 2 &=& 25x - 150 \\[4pt] \text{Collect the variables to the right.} & 6x - 6x + 2 &=& 25x - 6x - 150 \\[4pt] \text{Simplify.} & 2 &=& 19x - 150 \\[4pt] \text{Collect the constants to the left.} & 2 + 150 &=& 19x - 150 + 150 \\[4pt] \text{Simplify.} & 152 &=& 19x \\[4pt] \text{Divide by 19.} & \tfrac{152}{19} &=& \tfrac{19x}{19} \\[4pt] \text{Simplify.} & 8 &=& x \end{array}

Check: with $x=8$ , $0.06(8) + 0.02 = 0.48 + 0.02 = 0.50$ , and $0.25(8) - 1.5 = 2.00 - 1.5 = 0.50$ . Both sides equal $0.50$ . ✓

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

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

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

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

\begin{array}{lrcl} & 0.25x + 0.05(x+3) &=& 2.85 \\[4pt] \text{Distribute first.} & 0.25x + 0.05x + 0.15 &=& 2.85 \\[4pt] \text{Combine like terms.} & 0.30x + 0.15 &=& 2.85 \\[4pt] \text{To clear decimals, multiply by 100.} & 100(0.30x + 0.15) &=& 100(2.85) \\[4pt] \text{Distribute.} & 30x + 15 &=& 285 \\[4pt] \text{Subtract 15 from both sides.} & 30x + 15 - 15 &=& 285 - 15 \\[4pt] \text{Simplify.} & 30x &=& 270 \\[4pt] \text{Divide by 30.} & \tfrac{30x}{30} &=& \tfrac{270}{30} \\[4pt] \text{Simplify.} & x &=& 9 \end{array}

Check this yourself by substituting $x = 9$ into the original equation.

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 (LCD) of all its fractions so the resulting equivalent equation has no fractions. clearing the equation of decimals — multiplying both sides of an equation by an appropriate power of ten so the resulting equivalent equation has no decimals.

This section is adapted from Elementary Algebra 2e, Section 2.5: Solve Equations with Fractions or Decimals 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: condensed the worked-example step tables into annotated derivation displays; omitted the Be Prepared quiz, media links, Self Check checklist, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

Use a General Strategy to Solve Linear Equations Solve a Formula for a Specific Variable