# Plus and Minus (Different Denominators)

To be able to add and subtract fractions with different denominators, you must expand or simplify the fractions so they have equal denominators. Expanding a fraction means that you multiply the numerator and denominator by the same number. To simplify means that you divide the numerator and denominator by the same number. This does not change the value of the fraction. Expanding or simplifying fractions so that they have equal denominators is called finding a common denominator.

Rule

### FindingtheCommonDenominator

Below you see two fractions with different denominators. You will find the common denominator, and change the fractions so that they both have equal denominators.

The fractions $\frac{a}{b}$ and $\frac{c}{d}$ have different denominators. Therefore, you multiply both fractions by the denominator of the other fraction—both above and below the fraction bar—and get

 $\frac{a×d}{b×d}=\frac{ad}{bd}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{c×b}{d×b}=\frac{cb}{db}$

As you know from earlier, multiplying by the same number in the numerator and denominator does not change the value of the fraction. But now you have two fractions with equal denominators! The common denominator in this case is $bd=db$.

Example 1

Compute $\frac{5}{8}-\frac{1}{4}$ and $\frac{1}{6}+\frac{3}{8}$

You start with the first expression. Here you need to factorize both denominators to find the common denominator:

Since 2 occurs at most three times (when factoring 8), the common denominator is

 $2×2×2=8.$

Thus you get that $\begin{array}{llll}\hfill \frac{5}{8}-\frac{1}{4}& =\frac{5}{8}-\frac{1×2}{4×2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{5-2}{8}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{3}{8}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Repeat this when working with the next fraction:

Here the common denominator is

 $2×2×2×3=24.$

Then you get that

$\begin{array}{llll}\hfill \frac{1}{6}+\frac{3}{8}& =\frac{1×4}{6×4}+\frac{3×3}{8×3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{4+9}{24}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{13}{24}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\frac{1}{6}+\frac{3}{8}=\frac{1×4}{6×4}+\frac{3×3}{8×3}=\frac{4+9}{24}=\frac{13}{24}.$

Example 2

Compute $\frac{7}{12}-\frac{1}{3}$

Since the denominators are different ($12\ne 3$), you need to find the common denominator. You do this by factoring both denominators, and see how many times each prime number occurs at most:

 $12=2×2×3\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}3=3.$

Here you see that 2 occurs at most twice (when factoring 12), and 3 occurs at most once (when factoring 3 and 12). It does not matter that 3 exists in both factorizations—you only count the maximum number of times each of the prime numbers occur. Therefore, the common denominator is equal to $2×2×3=12$. So, the fraction with 12 in denominator can stay as it is, and the fraction with 3 in the denominator must be expanded by 4, so that 12 is the denominator in that fraction as well:

 $\frac{7}{12}-\frac{1×4}{3×4}=\frac{7-4}{12}=\frac{3}{12}=\frac{1}{4}.$

Note! When finding common denominators, do not count the number of times each prime number occurs in total. You should only include it as many times as it maximally occurs in one of the factorizations. Above you see that 2 is also found in the factorization of 6, but you do not include this 2 in the common denominator. In the factorization of 6, the number 2 is only occurs once, while it occurs three times in the factorization of 8. Therefore, three 2s should be included in the common denominator, not four. Since 3 occurs once (when factoring 6), there should be one 3 in the common denominator.