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Study Guides > Mathematics for the Liberal Arts Corequisite

Solving Equations By Clearing Fractions

Learning Outcomes

  • Use the least common denominator to eliminate fractions from a linear equation before solving it
  • Solve equations with fractions that require several steps
You may feel overwhelmed when you see fractions in an equation, so we are going to show a method to solve equations with fractions where you use the common denominator to eliminate the fractions from an equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions. Pay attention to the fact that each term in the equation gets multiplied by the least common denominator. That's what makes it equal to the original!


Solve: [latex]\Large\frac{1}{8}\normalsize x+\Large\frac{1}{2}=\Large\frac{1}{4}[/latex] Solution:
[latex]\Large\frac{1}{8}\normalsize x+\Large\frac{1}{2}=\Large\frac{1}{4}\normalsize\quad{LCD=8}[/latex]
Multiply both sides of the equation by that LCD, [latex]8[/latex]. This clears the fractions.[latex]\color{red}{8(}\Large\frac{1}{8}\normalsize x+\Large\frac{1}{2}\color{red}{)}=\normalsize\color{red}{8(}\Large\frac{1}{4}\color{red}{)}[/latex]
Use the Distributive Property.[latex]8\cdot\Large\frac{1}{8}\normalsize x+8\cdot\Large\frac{1}{2}\normalsize=8\cdot\Large\frac{1}{4}[/latex]
Simplify — and notice, no more fractions![latex]x+4=2[/latex]
Solve using the General Strategy for Solving Linear Equations.[latex]x+4\color{red}{-4}=2\color{red}{-4}[/latex]
Check: Let [latex]x=-2[/latex][latex] \Large\frac{1}{8}\normalsize x+ \Large\frac{1}{2}= \Large\frac{1}{4}[/latex] [latex] \Large\frac{1}{8}\normalsize(\color{red}{-2})+ \Large\frac{1}{2}\normalsize\stackrel{\text{?}}{=} \Large\frac{1}{4}[/latex] [latex] \Large\frac{-2}{8}+ \Large\frac{1}{2}\normalsize\stackrel{\text{?}}{=} \Large\frac{1}{4}[/latex] [latex] \Large\frac{-2}{8}+ \Large\frac{4}{8}\normalsize\stackrel{\text{?}}{=} \Large\frac{1}{4}[/latex] [latex] \Large\frac{2}{8}\normalsize\stackrel{\text{?}}{=} \Large\frac{1}{4}[/latex] [latex] \Large\frac{1}{4}= \Large\frac{1}{4}\quad\checkmark[/latex]  
In the last example, the least common denominator was [latex]8[/latex]. Now it's your turn to find an LCD, and clear the fractions before you solve these linear equations.

Try it

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!

Solve equations by clearing the Denominators

  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. Isolate the variable terms on one side, and the constant terms on the other side.
  4. Simplify both sides.
  5. Use the multiplication or division property to make the coefficient on the variable equal to [latex]1[/latex].
Here's an example where you have three variable terms. After you clear fractions with the LCD, you will simplify the three variable terms, then isolate the variable.


Solve: [latex]7=\Large\frac{1}{2}\normalsize x+\Large\frac{3}{4}\normalsize x-\Large\frac{2}{3}\normalsize x[/latex]

Answer: 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 least common denominator of all the fractions in the equation.[latex]7=\Large\frac{1}{2}\normalsize x+\Large\frac{3}{4}\normalsize x-\Large\frac{2}{3}\normalsize x\quad{LCD=12}[/latex]
Multiply both sides of the equation by [latex]12[/latex].[latex]\color{red}{12}(7)=\color{red}{12}\cdot\Large(\frac{1}{2}\normalsize x+\Large\frac{3}{4}\normalsize x-\Large\frac{2}{3}\normalsize x\Large)[/latex]
Distribute.[latex]12(7)=12\cdot\Large\frac{1}{2}\normalsize x+12\cdot\Large\frac{3}{4}\normalsize x-12\cdot\Large\frac{2}{3}\normalsize x[/latex]
Simplify — and notice, no more fractions![latex]84=6x+9x-8x[/latex]
Combine like terms.[latex]84=7x[/latex]
Divide by [latex]7[/latex].[latex]\Large\frac{84}{\color{red}{7}}=\Large\frac{7x}{\color{red}{7}}[/latex]
Check: Let [latex]x=12[/latex].
[latex]7=\Large\frac{1}{2}\normalsize x+ \Large\frac{3}{4}\normalsize x- \Large\frac{2}{3}\normalsize x[/latex] [latex]7\stackrel{\text{?}}{=} \Large\frac{1}{2}\normalsize(\color{red}{12})+ \Large\frac{3}{4}\normalsize(\color{red}{12})- \Large\frac{2}{3}\normalsize(\color{red}{12})[/latex] [latex-display]7\stackrel{\text{?}}{=}6+9-8[/latex-display] [latex-display]7=7\quad\checkmark[/latex-display]  

Now here's a similar problem for you to try. Clear the fractions, simplify, then solve.

Try it



One of the most common mistakes when you clear fractions is forgetting to multiply BOTH sides of the equation by the LCD. If your answer doesn't check, make sure you have multiplied both sides of the equation by the LCD.
In the next example, we’ll have variables and fractions on both sides of the equation. After you clear the fractions using the LCD, you will see that this equation is similar to ones with variables on both sides that we solved previously. Remember to choose a variable side and a constant side to help you organize your work.


Solve: [latex]x+\Large\frac{1}{3}=\Large\frac{1}{6}\normalsize x-\Large\frac{1}{2}[/latex]

Answer: Solution:

Find the LCD of all the fractions in the equation.[latex]x+\Large\frac{1}{3}=\Large\frac{1}{6}\normalsize x-\Large\frac{1}{2}\normalsize,\quad{LCD=6}[/latex]
Multiply both sides by the LCD.[latex]\color{red}{6}(x+\Large\frac{1}{3}\normalsize)=\color{red}{6}(\Large\frac{1}{6}\normalsize x-\Large\frac{1}{2})[/latex]
Distribute.[latex]6\cdot{x}+6\cdot\Large\frac{1}{3}\normalsize=6\cdot\Large\frac{1}{6}\normalsize x-6\cdot\Large\frac{1}{2}[/latex]
Simplify — no more fractions![latex]6x+2=x-3[/latex]
Subtract [latex]x[/latex] from both sides.[latex]6x-\color{red}{x}+2=x-\color{red}{x}-3[/latex]
Subtract 2 from both sides.[latex]5x+2\color{red}{-2}=-3\color{red}{-2}[/latex]
Divide by [latex]5[/latex].[latex]\Large\frac{5x}{\color{red}{5}}=\Large\frac{-5}{\color{red}{5}}[/latex]
Check: Substitute [latex]x=-1[/latex].
[latex]x+\Large\frac{1}{3}= \Large\frac{1}{6}\normalsize x- \Large\frac{1}{2}[/latex] [latex](\color{red}{-1})+ \Large\frac{1}{3}\normalsize\stackrel{\text{?}}{=} \Large\frac{1}{6}\normalsize(\color{red}{-1})- \Large\frac{1}{2}[/latex] [latex](-1)+ \Large\frac{1}{3}\normalsize\stackrel{\text{?}}{=}- \Large\frac{1}{6}- \Large\frac{1}{2}[/latex] [latex]- \Large\frac{3}{3}+ \Large\frac{1}{3}\normalsize\stackrel{\text{?}}{=}- \Large\frac{1}{6}- \Large\frac{3}{6}[/latex] [latex]- \Large\frac{2}{3}\normalsize\stackrel{\text{?}}{=}- \Large\frac{4}{6}[/latex] [latex]- \Large\frac{2}{3}=- \Large\frac{2}{3}\quad\checkmark[/latex]  

Now you can try solving an equation with fractions that has variables on both sides of the equal sign. The answer may be a fraction.

Try it

In the following video we show another example of how to solve an equation that contains fractions and variables on both sides of the equal sign. https://youtu.be/G5R9jySFMpw In the next example, we start with an equation where the variable term is locked up in some parentheses and multiplied by a fraction. You can clear the fraction, or if you use the distributive property it will eliminate the fraction.  Can you see why?


Solve: [latex]1=\Large\frac{1}{2}\normalsize\left(4x+2\right)[/latex]

Answer: Solution:

Simplify. Now there are no fractions to clear![latex]1=2x+1[/latex]
Subtract 1 from both sides.[latex]1\color{red}{-1}=2x+1\color{red}{-1}[/latex]
Divide by [latex]2[/latex].[latex]\Large\frac{0}{\color{red}{2}}=\Large\frac{2x}{\color{red}{2}}[/latex]
Check: Let [latex]x=0[/latex].
[latex]1=\Large\frac{1}{2}\normalsize(4x+2)[/latex] [latex]1\stackrel{\text{?}}{=} \Large\frac{1}{2}\normalsize(4(\color{red}{0})+2)[/latex] [latex]1\stackrel{\text{?}}{=} \Large\frac{1}{2}\normalsize(2)[/latex] [latex]1\stackrel{\text{?}}{=} \Large\frac{2}{2}[/latex] [latex-display]1=1\quad\checkmark[/latex-display]  

Now you can try solving an equation that has the variable term in parentheses that are multiplied by a fraction.

Try it


Licenses & Attributions

CC licensed content, Original

  • Question ID 142514, 142542. Authored by: Lumen Learning. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
  • Solve a Linear Equation with Parentheses and a Fraction 2/3(9x-12)=8+2x. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.

CC licensed content, Shared previously

  • Ex 1: Solve an Equation with Fractions with Variable Terms on Both Sides. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
  • Question ID 71948. Authored by: Alyson Day. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.

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