Adding and Subtracting Fractions

If a pizza was cut into 8 slices and you ate 2, how much of the pizza is left? The answer is 6 slices, or $\frac{6}{8}$ of the pizza. Now that you've studied what fractions are and how they work, it's time to learn what you can do with them. For example, you can add and subtract fractions just like you can with whole numbers. The first step is checking whether the denominators are the same. After that, you'll solve a simple addition or subtraction problem to get your answer. Let's get some practice in.

Adding and subtracting fractions with like denominators

If the fractions you're working with have the same denominator, adding and subtracting them is fairly simple. You ignore the denominator entirely, add or subtract the numerators, and put your answer over the denominator you started with. Let's say you wanted to add the following fractions:

$\frac{2}{13}+\frac{5}{13}$

Both fractions have the same denominator of 13, meaning you add $2+5=7$ and put the answer over 13. The answer is $\frac{7}{13}$ as illustrated below:

Subtraction works similarly. For instance, let's say that you want to subtract

$\frac{4}{5}-\frac{1}{5}$

Both fractions have "5" as the denominator, so we simply subtract 1 from 4 and put the resulting 3 over the 5. The answer is $\frac{3}{5}$ as illustrated below:

Your answer may not always be in the lowest terms even if the fractions you started with were, so you may have to reduce the fraction as a final step. Consider the following example:

$\frac{5}{12}+\frac{1}{12}$

The denominators are the same, so we simply add 5 and 1 to get an answer of $\frac{6}{12}$ . However, our fraction shares a common multiple of 6, allowing us to factor it out to reduce our answer to $\frac{1}{2}$ as illustrated below:

The procedure is rather straightforward so long as the denominators are the same. Just make sure that you always express your answer in the simplest form to avoid any mistakes.

Adding and subtracting fractions with unlike denominators

Of course, the fractions you're adding or subtracting won't always have the same denominator. If the denominators aren't the same, you'll need to use equivalent fractions which do have a common denominator. That means finding the least common multiple, or LCM, of the denominators you're working with. Here is an example problem:

$\frac{1}{11}+\frac{2}{3}$

The LCM of 3 and 11 is 33, so we need to convert both fractions to something equivalent with a denominator of 33. For $\frac{1}{11}$ , that means multiplying both the numerator and denominator by 3 to get $\frac{3}{33}$ . For $\frac{2}{3}$ , we multiply both numbers by 11 to get $\frac{22}{33}$ . Now that we have fractions with like denominators, we can add them as described above:

$\frac{3}{33}+\frac{22}{33}=\frac{25}{33}$

Once again, the procedure for subtraction is identical. Let's say we're looking at:

$\frac{3}{4}-\frac{2}{3}$

The LCM of 4 and 3 is 12, so we need two fractions equivalent to the two above with a denominator of 12. A little bit of multiplication will get us to:

$\frac{9}{12}-\frac{8}{12}=\frac{1}{12}$

All of the same rules apply if you're given more than two fractions to add or subtract, but the math is a little harder if you're working with three or more numbers. You'll want to be especially careful when you're finding the LCM of the denominators.

Adding and subtracting fractions with unlike denominators basically combines an addition or subtraction problem with an LCM problem. It might be more steps than you're used to, but it's nothing you haven't seen before.

Practice questions on adding and subtracting fractions

1. $\frac{2}{6}+\frac{2}{6}$

$\frac{2+2}{6}=\frac{4}{6}$

$\frac{4}{6}=\frac{\frac{4}{2}}{\frac{6}{2}}$

$\frac{2}{3}$

2. $\frac{5}{7}-\frac{3}{7}$

$\frac{5-3}{7}$

$\frac{2}{7}$

3. $\frac{2}{5}+\frac{2}{4}$

$\frac{2}{5}\times \frac{4}{4}+\frac{2}{4}\times \frac{5}{5}$

$\frac{8}{20}+\frac{10}{20}=\frac{18}{20}$

$\frac{9}{10}$

4. $\frac{2}{3}-\frac{4}{9}$

$\frac{2}{3}\times \frac{3}{3}-\frac{4}{9}$

$\frac{6}{9}-\frac{4}{9}$

$\frac{2}{9}$

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