Here, you’ll learn how to solve equations that contain fractions. The only major difference from what you’ve done so far is that you must find a common denominator and multiply all the terms by it. This will allow you to cancel out the denominators and isolate the variable. The whole point of this method is to get rid of the fractions, which simplifies the calculations!

Rule

- 1.
- Find the common denominator.
- 2.
- Multiply all terms by the common denominator and simplify the denominators. If there’s anything left of the common denominator, you have to multiply the rest by the numerator. Notice how I put parentheses around the numerators.
- 3.
- Think PEMDAS! Solve the parentheses first.
- 4.
- Move all the terms with only numbers to one side. Remember to change signs.
- 5.
- Move all the terms containing $x$ to the other side. Remember to change signs.
- 6.
- Simplify both sides.
- 7.
- Multiply or divide both sides by whatever number is preventing $x$ from standing by itself.

Example 1

**Solve the equation $\frac{x}{3}+1=4-\frac{2x}{3}$ **

$$\begin{array}{llll}\hfill \frac{x}{3}+1& =4-\frac{2x}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3\phantom{\rule{-0.17em}{0ex}}\left(\frac{x}{3}+1\right)& =3\phantom{\rule{-0.17em}{0ex}}\left(4-\frac{2x}{3}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{\text{3}x}{\text{3}}+3& =12-\frac{2x\cdot \text{3}}{\text{3}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x+3& =12-2x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x+2x& =12-3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3x& =9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

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