How to Multiply Fractions by Cross-canceling

Here, you will learn a smart approach to multiplying fractions by each other. The trick is to use some initial simplification before multiplication!

Let’s have a look at how the trick works. The basic idea is to simplify the fraction as much as possible. We do this by cross-canceling factors.

When you simplify a fraction, you divide the numerator and denominator by the same number. What you do is the following:

Rule

Cross-cancelation

1.
Factorize the numerator
2.
Factorize the denominator
3.
Cancel the factors that are in both the numerator and the denominator
4.
Multiply the remaining factors

The purpose of cross-cancelation is to avoid larger multiplications with simplifications at the end. With cancelations, you’re simplifying throughout the process instead. For that reason, you often end up with a simple multiplication from the times tables.

Below you can see two examples. Study them closely. They might seem challenging at first, but when you read the calculations you’ll see that they are not too troublesome. Your most important tool to master this method is the times tables.

Earlier, we looked at which factors make up a number, as with

 $40=4\cdot 10=10\cdot 4=8\cdot 5=5\cdot 8.$

When you can see which factors make up a number, the simplification will become much easier, and the whole method turns into arranging numbers.

Example 1

Find $\frac{4}{10}\cdot \frac{25}{16}$.

You use cross-cancelation. $\begin{array}{llll}\hfill \frac{4}{10}\cdot \frac{25}{16}& =\frac{4\cdot 1}{2\cdot 5}\cdot \frac{5\cdot 5}{4\cdot 4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{4\cdot 1\cdot 5\cdot 5}{2\cdot 5\cdot 4\cdot 4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\text{4}\cdot 1\cdot \text{5}\cdot 5}{2\cdot \text{5}\cdot 4\cdot \text{4}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1\cdot 5}{2\cdot 4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{5}{8},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

because

 $\frac{4}{10}\cdot \frac{25}{16}=\frac{4\cdot 25}{10\cdot 16}=\frac{100}{160}=\frac{5}{8}.$

Example 2

Find $\frac{49}{21}\cdot \frac{42}{14}$.

You use cross-cancelation. $\begin{array}{llll}\hfill \frac{49}{21}\cdot \frac{42}{14}& =\frac{7\cdot 7}{3\cdot 7}\cdot \frac{6\cdot 7}{2\cdot 7}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7\cdot 7\cdot 6\cdot 7}{3\cdot 7\cdot 2\cdot 7}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\text{7}\cdot 7\cdot 6\cdot \text{7}}{3\cdot \text{7}\cdot 2\cdot \text{7}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7\cdot 6}{6}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7\cdot \text{6}}{\text{6}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =7,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

because

 $\frac{49}{21}\cdot \frac{42}{14}=\frac{49\cdot 42}{21\cdot 14}=\frac{2058}{294}=7.$