# Find the Common Denominator by Expanding or Simplifying

## Expanding One Fraction

If one fraction can be expanded such that the denominator becomes equal to the other, then you only need to expand that fraction before adding or subtracting.

Example 1

Find $\frac{2}{5}+\frac{7}{15}$.

$\begin{array}{llll}\hfill \frac{2}{5}+\frac{7}{15}& =\frac{2\cdot 3}{5\cdot 3}+\frac{7}{15}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{6}{15}+\frac{7}{15}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{13}{15}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Rule

### Expandingonefraction

When expanding one of the fractions to get the common denominator, you multiply by the missing factor such that the denominators become equal.

In Example 1, you can see that the first denominator is $5$ and the other denominator is $15$. From the multiplication table you know that $5$ is a factor in $15$, because $15=5\cdot 3$. So you can expand the first fraction by $3$ to get equal denominators.

Example 2

Find $\frac{2}{9}-\frac{11}{27}$.

We know that $9\cdot 3=27$, which means we can expand the first fraction by $3$. The calculation looks like this: $\begin{array}{llll}\hfill \frac{2}{9}-\frac{11}{27}& =\frac{2\cdot 3}{9\cdot 3}-\frac{11}{27}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{6}{27}-\frac{11}{27}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\frac{5}{27}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 3

Find $\frac{19}{81}-\frac{2}{9}$.

We know that $9\cdot 9=81$, which means we can expand the second fraction by $9$. The calculation looks like this: $\begin{array}{llll}\hfill \frac{19}{81}-\frac{2}{9}& =\frac{19}{81}-\frac{2\cdot 9}{9\cdot 9}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{19}{81}-\frac{18}{81}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{81}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

## Simplify One Fraction

If one of the fractions can be simplified in such a way that the denominator becomes equal to the other denominators, it’s enough to cancel those factors before adding or subtracting.

Example 4

Find $\frac{2}{3}-\frac{36}{27}$.

$\begin{array}{llll}\hfill \frac{2}{3}-\frac{36}{27}& =\frac{2}{3}-\frac{36:9}{27:9}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2}{3}-\frac{4}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\frac{2}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Rule

### SimplifyingOneFraction

When you simplify one of the fractions to get the common denominator, you divide by factors such that the denominators become equal.

In Example 4, you can see that the first denominator is $3$ and that the other denominator is $27$. From the multiplication table you know that $3$ is a factor in $27$, because $27=3\cdot 9$. Simultaneously, $9$ is a factor in the numerator $36=9\cdot 4$. That means you can simplify the second fraction by dividing by $9$ in both the numerator and denominator to get equal denominators.

Example 5

Find $\frac{11}{5}+\frac{15}{25}$.

You know that $25:5=5$, which is the denominator of the first fraction, and $15:5=3$. This means that you can reduce the second fraction by dividing both the numerator and denominator by $5$. The calculation looks like this: $\begin{array}{llll}\hfill \frac{11}{5}+\frac{15}{25}& =\frac{11}{5}+\frac{15:5}{25:5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{11}{5}+\frac{3}{5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{14}{5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 6

Find $\frac{21}{49}-\frac{2}{7}$.

You know that $49:7=7$, which is the denominator of the second fraction, and $21:7=3$. This means that you can reduce the first fraction by dividing both the numerator and denominator by $7$. The calculation looks like this: $\begin{array}{llll}\hfill \frac{21}{49}-\frac{2}{7}& =\frac{21:7}{49:7}-\frac{2}{7}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{3}{7}-\frac{2}{7}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{7}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$