How to Add and Subtract Fractions

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Fractions describe equal parts of a whole. The top number is the numerator, and the bottom number is the denominator. When we add or subtract fractions, the denominator matters because it tells us the size of the pieces.

Think about adding slices of pizza. If both slices are the same size, you can count them right away. For example, \(\frac{2}{7}+\frac{3}{7}=\frac{5}{7}\). But if the pieces have different sizes, like fourths and sixths, you first rewrite them as the same-size pieces. That shared denominator is called a common denominator.

Types of fractions:
Proper Fractions: The numerator is less than the denominator, such as \(\frac{3}{4}\) or \(\frac{1}{3}\).
Improper Fractions: The numerator is greater than or equal to the denominator, such as \(\frac{7}{4}\) or \(\frac{5}{2}\).
Mixed Numbers: A whole number and a fraction written together, such as \(1\frac{1}{3}\) or \(2\frac{3}{7}\).

Addition & Subtraction with Like Fractions

Like fractions have the same denominator. To add or subtract like fractions, keep the denominator and add or subtract only the numerators.

\(\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}\) and \(\frac{a}{b}-\frac{c}{b}=\frac{a-c}{b}\)

Example: \(\frac{2}{9}+\frac{5}{9}=\frac{7}{9}\). The denominator stays \(9\) because the pieces are already the same size.

Addition and Subtraction with Unlike Fractions

Unlike fractions have different denominators. First, rewrite both fractions with a common denominator. The least common denominator is usually the easiest choice because it keeps the numbers smaller.

\(\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\) and \(\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}\)

This cross-multiply formula works, but you should still simplify the final answer whenever possible.

Example: \(\frac{2}{5}-\frac{1}{6}\). A common denominator is \(30\), so \(\frac{2}{5}=\frac{12}{30}\) and \(\frac{1}{6}=\frac{5}{30}\). Then \(\frac{12}{30}-\frac{5}{30}=\frac{7}{30}\).

Finally, always check whether your answer can be reduced to simplest form. Learn more

Adding and Subtracting Fractions

Think of this lesson as more than a rule to memorize. Adding and Subtracting Fractions is about number sense, equivalent forms, and careful arithmetic. A strong student does not rush to the first formula on the page; they pause, identify the structure of the problem, and then choose the tool that matches that structure. That pause is what prevents most mistakes.

Fractions compare a part to a whole. Keep track of the numerator, denominator, and whether the pieces are the same size before adding, subtracting, or simplifying.

Here is the teacher way to approach the topic. First, read the problem slowly and underline the information that is actually given. Next, name the target: are you finding a value, simplifying an expression, comparing two quantities, solving for a variable, or interpreting a graph? Once the target is clear, the calculation becomes much less mysterious because every step has a job.

Check whether denominators already match.
If they do not match, build a common denominator.
Add or subtract only the numerators.
Simplify the result and check reasonableness.

A helpful habit is to say what each number represents before using it. For example, if a number is a denominator, a radius, a slope, a common difference, or a coefficient, it should not be treated like an ordinary loose number. Its role tells you where it belongs in the formula. This is especially important on ACT-style questions because many wrong answer choices come from using the right numbers in the wrong places.

Another good habit is to keep the work organized vertically. Write one clean line for substitution, one line for simplifying, and one line for the final answer. If the problem has units, keep the units attached. If the problem has signs, exponents, or parentheses, copy them carefully from one line to the next. Most errors in this topic are not caused by a hard idea; they are caused by dropping a negative sign, combining unlike terms, using the wrong denominator, or skipping a check.

When you finish, ask a quick reasonableness question. Should the answer be positive or negative? Should it be larger or smaller than the original number? Does it fit the graph, table, shape, or equation? Can you plug it back into the original problem? This final check turns the lesson from a procedure into understanding.

On a test, the goal is not to write the longest solution. The goal is to write enough clear work that you can see the structure, avoid traps, and recover quickly if one line goes wrong. Practice the examples below with that mindset: identify the type of problem, choose the matching rule, show the substitution, simplify carefully, and check the answer before moving on.

Free printable Worksheets

Exercises for Add or Subtract Fractions

1) \({3 \over 8}+{1 \over 8}=\)

2) \({7 \over 10}-{3 \over 10}=\)

3) \({5 \over 12}+{1 \over 6}=\)

4) \({3 \over 4}-{1 \over 6}=\)

5) \({2 \over 9}+{5 \over 12}=\)

6) \({7 \over 15}-{2 \over 5}=\)

7) \({5 \over 6}+{3 \over 10}=\)

8) \({11 \over 14}-{2 \over 7}=\)

9) \({4 \over 5}+{7 \over 15}=\)

10) \({9 \over 16}-{1 \over 4}=\)

11) \(1{1 \over 3}+2{1 \over 6}=\)

12) \(4{3 \over 5}-1{7 \over 10}=\)

13) A recipe uses \({2 \over 3}\) cup of flour and \({3 \over 8}\) cup of oats. How much dry ingredient is used altogether?

14) A tank is \({5 \over 6}\) full. After water is used, it is \({1 \over 4}\) full. What fraction of the tank was used?

15) Simplify \({13 \over 18}+{5 \over 24}\).

16) Simplify \({7 \over 9}-{5 \over 12}\).

17) Find the missing numerator: \({x \over 12}+{1 \over 3}={3 \over 4}\).

18) Which is greater: \({5 \over 6}-{1 \over 3}\) or \({3 \over 4}-{1 \over 8}\)?

19) A runner jogs \(2{3 \over 4}\) miles on Monday and \(1{5 \over 8}\) miles on Tuesday. How many miles did the runner jog in all?

20) A board is \(6{1 \over 2}\) feet long. If \(2{5 \over 6}\) feet are cut off, how much board remains?

Answers

1) The denominators are the same, so add the numerators: \({3 \over 8}+{1 \over 8}={3+1 \over 8}={4 \over 8}\). Simplify using GCF(4,8) = 4: \({4 \over 8}={1 \over 2}\). Answer: \(\color{red}{{1 \over 2}}\)

2) The denominators are the same: \({7 \over 10}-{3 \over 10}={7-3 \over 10}={4 \over 10}\). Simplify using GCF(4,10) = 2: \({4 \over 10}={2 \over 5}\). Answer: \(\color{red}{{2 \over 5}}\)

3) Use denominator \(12\): \({1 \over 6}={2 \over 12}\). Then \({5 \over 12}+{2 \over 12}={7 \over 12}\). The answer is already simplest. Answer: \(\color{red}{{7 \over 12}}\)

4) Use denominator \(12\): \({3 \over 4}={9 \over 12}\) and \({1 \over 6}={2 \over 12}\). Subtract: \({9 \over 12}-{2 \over 12}={7 \over 12}\). Answer: \(\color{red}{{7 \over 12}}\)

5) A common denominator for \(9\) and \(12\) is \(36\). \({2 \over 9}={8 \over 36}\) and \({5 \over 12}={15 \over 36}\). Add: \({8 \over 36}+{15 \over 36}={23 \over 36}\). Answer: \(\color{red}{{23 \over 36}}\)

6) Rewrite \({2 \over 5}\) with denominator \(15\): \({2 \over 5}={6 \over 15}\). Then \({7 \over 15}-{6 \over 15}={1 \over 15}\). Answer: \(\color{red}{{1 \over 15}}\)

7) Use denominator \(30\): \({5 \over 6}={25 \over 30}\) and \({3 \over 10}={9 \over 30}\). Add: \({25 \over 30}+{9 \over 30}={34 \over 30}\). Simplify using GCF(34,30) = 2: \({34 \over 30}={17 \over 15}\). Answer: \(\color{red}{{17 \over 15}}\)

8) Rewrite \({2 \over 7}\) as \({4 \over 14}\). Then \({11 \over 14}-{4 \over 14}={7 \over 14}\). Simplify using GCF(7,14) = 7: \({7 \over 14}={1 \over 2}\). Answer: \(\color{red}{{1 \over 2}}\)

9) Rewrite \({4 \over 5}\) as \({12 \over 15}\). Then \({12 \over 15}+{7 \over 15}={19 \over 15}\). Answer: \(\color{red}{{19 \over 15}}\)

10) Rewrite \({1 \over 4}\) as \({4 \over 16}\). Then \({9 \over 16}-{4 \over 16}={5 \over 16}\). Answer: \(\color{red}{{5 \over 16}}\)

11) Add whole numbers and fractions: \(1{1 \over 3}+2{1 \over 6}=3+({1 \over 3}+{1 \over 6})\). Rewrite \({1 \over 3}={2 \over 6}\), so \({2 \over 6}+{1 \over 6}={3 \over 6}\). Simplify using GCF(3,6) = 3: \({3 \over 6}={1 \over 2}\). Answer: \(\color{red}{3{1 \over 2}}\)

12) Convert to tenths: \(4{3 \over 5}=4{6 \over 10}\). Then \(4{6 \over 10}-1{7 \over 10}\) requires borrowing: \(4{6 \over 10}=3{16 \over 10}\). Now \(3{16 \over 10}-1{7 \over 10}=2{9 \over 10}\). Answer: \(\color{red}{2{9 \over 10}}\)

13) Add the amounts: \({2 \over 3}+{3 \over 8}\). Use denominator \(24\): \({2 \over 3}={16 \over 24}\), \({3 \over 8}={9 \over 24}\). So \({16 \over 24}+{9 \over 24}={25 \over 24}\). Answer: \(\color{red}{{25 \over 24}\text{ cups}}\)

14) Subtract the remaining amount from the starting amount: \({5 \over 6}-{1 \over 4}\). Use denominator \(12\): \({5 \over 6}={10 \over 12}\), \({1 \over 4}={3 \over 12}\). Then \({10 \over 12}-{3 \over 12}={7 \over 12}\). Answer: \(\color{red}{{7 \over 12}}\)

15) Use denominator \(72\): \({13 \over 18}={52 \over 72}\) and \({5 \over 24}={15 \over 72}\). Add: \({52 \over 72}+{15 \over 72}={67 \over 72}\). Answer: \(\color{red}{{67 \over 72}}\)

16) Use denominator \(36\): \({7 \over 9}={28 \over 36}\) and \({5 \over 12}={15 \over 36}\). Subtract: \({28 \over 36}-{15 \over 36}={13 \over 36}\). Answer: \(\color{red}{{13 \over 36}}\)

17) Rewrite \({1 \over 3}\) as \({4 \over 12}\) and \({3 \over 4}\) as \({9 \over 12}\). Then \({x \over 12}+{4 \over 12}={9 \over 12}\), so \(x+4=9\), and \(x=5\). Answer: \(\color{red}{5}\)

18) First expression: \({5 \over 6}-{1 \over 3}={5 \over 6}-{2 \over 6}={3 \over 6}={1 \over 2}\). Second expression: \({3 \over 4}-{1 \over 8}={6 \over 8}-{1 \over 8}={5 \over 8}\). Since \({5 \over 8}>{1 \over 2}\), the second expression is greater. Answer: \(\color{red}{{3 \over 4}-{1 \over 8}}\)

19) Add the mixed numbers: \(2{3 \over 4}+1{5 \over 8}=3+({3 \over 4}+{5 \over 8})\). Rewrite \({3 \over 4}={6 \over 8}\). Then \({6 \over 8}+{5 \over 8}={11 \over 8}=1{3 \over 8}\). So the total is \(3+1{3 \over 8}=4{3 \over 8}\). Answer: \(\color{red}{4{3 \over 8}\text{ miles}}\)

20) Subtract \(6{1 \over 2}-2{5 \over 6}\). Use sixths: \(6{1 \over 2}=6{3 \over 6}\). Borrow from \(6\): \(6{3 \over 6}=5{9 \over 6}\). Now \(5{9 \over 6}-2{5 \over 6}=3{4 \over 6}\). Simplify using GCF(4,6) = 2: \(3{4 \over 6}=3{2 \over 3}\). Answer: \(\color{red}{3{2 \over 3}\text{ feet}}\)