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You do not always have to expand both fractions you’re adding or subtracting to get a common denominator. If it is enough to expand only one of them, then you can just do that, saving yourself a step. The only goal is for the fractions to have the same denominator, so they can be added or subtracted. Let’s look at an example:

Example 1

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

$$\frac{2}{5}+\frac{7}{15}=\frac{2\times 3}{5\times 3}+\frac{7}{15}=\frac{6}{15}+\frac{7}{15}=\frac{13}{15}$$ |

Likewise, if one fraction can be simplified so that the denominator is equal to the other, it’s fine to simplify just that one before you add or subtract. Below you see an example:

Example 2

$$\begin{array}{llll}\hfill \frac{2}{3}-\frac{36}{27}& =\frac{2}{3}-\frac{36\xf79}{27\xf79}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2}{3}-\frac{4}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\frac{2}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

$$\frac{2}{3}-\frac{36}{27}=\frac{2}{3}-\frac{36\xf79}{27\xf79}=\frac{2}{3}-\frac{4}{3}=-\frac{2}{3}$$ |

It’s good to know, that if you are unsure of which method you should use, then you can always expand each fraction by the denominator of the other fraction.

So if you’re not sure what to do, you can always expand the first fraction by the denominator of the second fraction, and expand the second fraction with the denominator of the first fraction, like in this example.

Example 3

$$\begin{array}{llll}\hfill \frac{7}{12}-\frac{1}{3}& =\frac{3\times 7}{3\times 12}-\frac{12\times 1}{12\times 3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{21}{36}-\frac{12}{36}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{21-12}{36}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{9}{36}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ You can simplify the fraction $\frac{9}{36}$ using prime factorization:

$$\frac{9}{36}=\frac{\text{3}\times \text{3}}{2\times 2\times \text{3}\times \text{3}}=\frac{1}{4}$$ |

But, if you want to be even smarter about it, take a moment to see that $12=3\times 4$, and only expand the second fraction, so that its denominator becomes 12, like the first fraction. The calculation is then

$$\frac{7}{12}-\frac{1\times 4}{3\times 4}=\frac{7}{12}-\frac{4}{12}=\frac{3}{12}=\frac{1}{4}$$ |

The answers are the same! But the second calculation was much easier and quicker than the first.

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