# How to Divide Fractions by Canceling

When you divide one fraction containing variables by another fraction with variables in it, you should definitely factorize and cancel first. You might remember the “keep, change, flip” method, where you multiply by the flipped fraction. This method actually lets you make a division problem consisting of two fractions into a multiplication problem.

You will learn that the “keep, change, flip” method works just as well when you have variables in the numerator and denominator, as when they’re just numbers.

Rule

### DivisionbyCanceling

1.
Flip the second fraction upside down and change the division sign into a multiplication sign.
2.
Factorize the numerators.
3.
Factorize the denominators.
4.
Cancel common factors.
5.

Example 1

Evaluate $\frac{x+2}{x+1}÷\frac{2x+4}{3x+3}$

Apply the “keep, change, flip” method and cancel the common factors: $\begin{array}{llll}\hfill \frac{x+2}{x+1}÷\frac{2x+4}{3x+3}& =\frac{x+2}{x+1}\cdot \frac{3x+3}{2x+4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\left(x+2\right)}{\left(x+1\right)}\cdot \frac{3\left(x+1\right)}{2\left(x+2\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\text{(x+2)}}{\text{(x+1)}}\cdot \frac{3\text{(x+1)}}{2\text{(x+2)}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{3}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 2

Evaluate $\frac{2x+6}{x+3}÷\frac{{x}^{2}-9}{x+3}$

Apply the “keep, change, flip” method, factorize by using the third algebraic identity of quadratic expressions, and cancel the common factors: $\begin{array}{llll}\hfill \frac{2x+6}{x+3}÷\frac{{x}^{2}-9}{x+3}& =\frac{2x+6}{x+3}\cdot \frac{x+3}{{x}^{2}-9}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2\left(x+3\right)}{\left(x+3\right)}\cdot \frac{\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2\text{(x+3)}}{\text{(x+3)}}\cdot \frac{\text{(x+3)}}{\text{(x+3)}\left(x-3\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2}{x-3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 3

Evaluate $\frac{{x}^{2}-1}{x+5}÷\frac{x-1}{{x}^{2}-25}$

Apply the “keep, change, flip” method, factorize by using the third algebraic identity of quadratic expressions, cross-cancel the common factors, and multiply the remaining factors: $\begin{array}{llll}\hfill \frac{{x}^{2}-1}{x+5}÷\frac{x-1}{{x}^{2}-25}& =\frac{{x}^{2}-1}{x+5}\cdot \frac{{x}^{2}-25}{x-1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\left(x-1\right)\left(x+1\right)}{\left(x+5\right)}\cdot \frac{\left(x-5\right)\left(x+5\right)}{\left(x-1\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\text{(x−1)}\left(x+1\right)}{\text{(x+5)}}\cdot \frac{\text{(x+5)}\left(x-5\right)}{\text{(x−1)}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(x+1\right)\left(x-5\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}-5x+x-5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}-4x-5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 4

Evaluate $\frac{7x+49}{2x-2}÷\frac{x+7}{4x-4}$

Apply the “keep, change, flip” method, factorize both fractions, cross-cancel common factors, and do the remaining calculation: $\begin{array}{llll}\hfill \frac{7x+49}{2x-2}÷\frac{x+7}{4x-4}& =\frac{7x+49}{2x-2}\cdot \frac{4x-4}{x+7}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7\left(x+7\right)}{2\left(x-1\right)}\cdot \frac{4\left(x-1\right)}{\left(x+7\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7\text{(x+7)}}{2\text{(x−1)}}\cdot \frac{4\text{(x−1)}}{\text{(x+7)}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7\cdot \text{4}}{\text{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =7\cdot 2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =14\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$