# How to Solve Equations with Fractions

You will now expand your knowledge of equations further by learning to solve equations with fractions. These kinds of equations are also called rational equations.

The only new thing is that you have to find the common denominator, multiply all terms by that common denominator, and then simplify. The whole purpose of this method is to get rid of the fractions!

Rule

1.
Find the common denominator.
2.
Multiply all the terms by the common denominator and simplify. You’ll notice below that I’ve put parentheses around the numerators.
3.
Think PEMDAS! Expand the parentheses first.
4.
Move all the terms that only consist of numbers to one side of the equation. Remember to change the sign when you change sides.
5.
Move all the terms with $x$ in them to the other side. Remember to change the sign when you change sides.
6.
Combine and simplify the terms on both sides.
7.
Multiply or divide on both sides by the number that is preventing $x$ from standing alone.

To be able to master equations, you should be comfortable with fraction calculations and algebra. If you experience a gap in your knowledge, you can go back and study these entries.

Example 1

Solve the equation

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

with respect to $x$

You multiply the equation by the common denominator, which is 2: $\begin{array}{llll}\hfill 2\cdot \frac{\left(x+1\right)}{2}+2\cdot 2& =2\cdot \frac{1}{2}-2x\cdot 2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{2}\cdot \frac{\left(x+1\right)}{\text{2}}+4& =\text{2}\cdot \frac{1}{\text{2}}-4x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x+1+4& =1-4x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x+4x& =1-1-4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 5x& =-4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{\text{5}x}{\text{5}}& =-\frac{4}{5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =-\frac{4}{5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 2

Solve the equation

 $\frac{-x+2}{5}-4x=\frac{5-2x}{10}-5x$

with respect to $x$

You multiply the equation by the common denominator, which is 10:

$\begin{array}{llll}\hfill 10\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(-x+2\right)}{5}-4x\cdot 10& =10\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(5-2x\right)}{10}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-5x\cdot 10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 10\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(-x+2\right)}{5}-40x& =10\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(5-2x\right)}{10}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-50x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{10}\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(-x+2\right)}{\text{5}}-40x& =\text{10}\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(5-2x\right)}{\text{10}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-50x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2\phantom{\rule{-0.17em}{0ex}}\left(-x+2\right)-40x& =\phantom{\rule{-0.17em}{0ex}}\left(5-2x\right)-50x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x+4-40x& =5-2x-50x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x-40x+50x+2x& =5-4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 10x& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{\text{10}x}{\text{10}}& =\frac{1}{10}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =\frac{1}{10}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill 10\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(-x+2\right)}{5}-4x\cdot 10& =10\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(5-2x\right)}{10}-5x\cdot 10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 10\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(-x+2\right)}{5}-40x& =10\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(5-2x\right)}{10}-50x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{10}\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(-x+2\right)}{\text{5}}-40x& =\text{10}\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(5-2x\right)}{\text{10}}-50x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2\phantom{\rule{-0.17em}{0ex}}\left(-x+2\right)-40x& =\phantom{\rule{-0.17em}{0ex}}\left(5-2x\right)-50x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x+4-40x& =5-2x-50x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x-40x+50x+2x& =5-4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 10x& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{\text{10}x}{\text{10}}& =\frac{1}{10}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =\frac{1}{10}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$