Subtraction of Fractions

Subtracting fractions works just like adding them. When the parts are the same size, you subtract how many there are. When they are different sizes, you first rewrite the fractions with a common denominator. This page shows both cases, then mixed numbers (with borrowing), negative answers and fractions with letters.

Fractions with the same denominator

Subtract the numerators and keep the denominator. $$\frac{a}{c}-\frac{b}{c}=\frac{a-b}{c}$$

Five eighths take away two eighths leaves three eighths. The parts are still eighths, so the denominator stays 8.

$$\frac{5}{8}-\frac{2}{8}=\frac{5-2}{8}=\frac{3}{8}$$

If the answer can be simplified, simplify it:

$$\frac{7}{10}-\frac{3}{10}=\frac{4}{10}=\frac{2}{5}$$

Never subtract the denominators. $\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$. Subtracting the denominators would give $\frac{2}{0}$, which isn't even a number.

Fractions with different denominators

Rewrite the fractions with a common denominator first, exactly as in addition:

Find the least common denominator: the least common multiple of the denominators.
Rewrite each fraction with that denominator, multiplying its numerator and denominator by the same number.
Subtract the numerators and keep the common denominator.
Simplify the result.

Example 1: $\frac{5}{6}-\frac{1}{4}$

The least common multiple of 6 and 4 is 12, and $12=6\cdot 2=4\cdot 3$:

$$\begin{aligned}\frac{5}{6}-\frac{1}{4}&=\frac{5\cdot 2}{6\cdot 2}-\frac{1\cdot 3}{4\cdot 3}\\&=\frac{10}{12}-\frac{3}{12}=\frac{7}{12}\end{aligned}$$

Example 2: $\frac{7}{10}-\frac{8}{15}$

The least common denominator is 30. The answer can be simplified at the end:

$$\frac{7}{10}-\frac{8}{15}=\frac{21}{30}-\frac{16}{30}=\frac{5}{30}=\frac{1}{6}$$

Shortcut for two fractions

The product of the denominators is always a common denominator:

$$\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}$$

With Example 1: $\frac{5}{6}-\frac{1}{4}=\frac{20-6}{24}=\frac{14}{24}=\frac{7}{12}$. It always works, but it may leave a fraction to simplify.

When the answer is negative

If you take a bigger fraction from a smaller one, the answer is negative. Subtract as usual and keep the sign:

$$\begin{aligned}\frac{3}{4}-\frac{5}{6}&=\frac{9}{12}-\frac{10}{12}\\&=\frac{9-10}{12}=-\frac{1}{12}\end{aligned}$$

Whole numbers and mixed numbers

A whole number minus a fraction

Write the whole number as a fraction with the same denominator, for example $1=\frac{8}{8}$ and $3=\frac{15}{5}$:

$$1-\frac{3}{8}=\frac{8}{8}-\frac{3}{8}=\frac{5}{8}$$ $$3-\frac{2}{5}=\frac{15}{5}-\frac{2}{5}=\frac{13}{5}=2\frac{3}{5}$$

Subtracting mixed numbers

Method 1: subtract the whole parts and the fractions separately. First give the fractions a common denominator:

$$5\frac{1}{4}-2\frac{2}{3}=5\frac{3}{12}-2\frac{8}{12}$$

$\frac{3}{12}$ is smaller than $\frac{8}{12}$, so borrow 1 from the 5 and write it as $\frac{12}{12}$:

$$5\frac{3}{12}=4+\frac{12}{12}+\frac{3}{12}=4\frac{15}{12}$$

Now subtract the whole parts and the fractions:

$$4\frac{15}{12}-2\frac{8}{12}=2\frac{7}{12}$$

Method 2: change to improper fractions first. No borrowing is needed:

$$\begin{aligned}5\frac{1}{4}-2\frac{2}{3}&=\frac{21}{4}-\frac{8}{3}=\frac{63}{12}-\frac{32}{12}\\&=\frac{31}{12}=2\frac{7}{12}\end{aligned}$$

The borrowed 1 is a whole, in the fraction's own parts. Here it is $\frac{12}{12}$, so $5\frac{3}{12}$ becomes $4\frac{15}{12}$, not $4\frac{13}{12}$.

Subtracting a negative fraction

Subtracting a negative number is the same as adding the positive number:

$$\begin{aligned}\frac{1}{2}-\left(-\frac{1}{3}\right)&=\frac{1}{2}+\frac{1}{3}\\&=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}\end{aligned}$$

It works the other way too: subtracting a positive fraction is adding a negative one, see negative fractions on the addition page.

Algebraic fractions

Fractions with letters are subtracted by the same rules. One new thing to watch: put the numerator you subtract in parentheses, because the minus sign applies to all of it.

Example 3: $\frac{a}{b}-\frac{c}{d}$

$$\frac{a}{b}-\frac{c}{d}=\frac{ad}{bd}-\frac{bc}{bd}=\frac{ad-bc}{bd}$$

Example 4: $\frac{2x+1}{x+3}-\frac{x-2}{x+3}$

The denominators are equal, so subtract the numerators, the second one in parentheses:

$$\begin{aligned}&\frac{2x+1}{x+3}-\frac{x-2}{x+3}\\&=\frac{2x+1-(x-2)}{x+3}\\&=\frac{2x+1-x+2}{x+3}=\frac{x+3}{x+3}=1\end{aligned}$$

This holds for every $x$ except $x=-3$. Without the parentheses you would get $2x+1-x-2=x-1$ in the numerator, which is wrong.

Example 5: $\frac{1}{x-2}-\frac{1}{x+2}$

The common denominator is $(x-2)(x+2)=x^2-4$:

$$\begin{aligned}&\frac{1}{x-2}-\frac{1}{x+2}\\&=\frac{(x+2)-(x-2)}{(x-2)(x+2)}=\frac{4}{x^2-4}\end{aligned}$$

for $x\ne 2$ and $x\ne -2$.

Example 6: $\frac{h}{y}-m$

Write $m$ as $\frac{my}{y}$:

$$\frac{h}{y}-m=\frac{h}{y}-\frac{my}{y}=\frac{h-my}{y}$$

Common mistakes

Subtracting the denominators: $\frac{3}{4}-\frac{1}{2}\ne\frac{2}{2}$. The correct difference is $\frac{3}{4}-\frac{2}{4}=\frac{1}{4}$.
Borrowing 10 instead of a whole: in $5\frac{3}{12}$ the borrowed 1 is $\frac{12}{12}$, so it becomes $4\frac{15}{12}$.
Losing the sign: $\frac{1}{4}-\frac{3}{4}=-\frac{2}{4}=-\frac{1}{2}$, not $\frac{1}{2}$.
Forgetting the parentheses: $\frac{x}{3}-\frac{x-1}{3}=\frac{x-(x-1)}{3}=\frac{1}{3}$, not $-\frac{1}{3}$.

Practice

Try each problem, then open it to check your answer.

1. $\frac{7}{9}-\frac{4}{9}$

$\frac{7-4}{9}=\frac{3}{9}=\frac{1}{3}$

2. $\frac{3}{4}-\frac{1}{6}$

The least common denominator is 12: $\frac{9}{12}-\frac{2}{12}=\frac{7}{12}$

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

$\frac{8}{20}-\frac{15}{20}=-\frac{7}{20}$

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

$\frac{12}{3}-\frac{2}{3}=\frac{10}{3}=3\frac{1}{3}$

5. $6\frac{1}{3}-2\frac{3}{4}$

$6\frac{4}{12}-2\frac{9}{12}$. Borrow 1: $5\frac{16}{12}-2\frac{9}{12}=3\frac{7}{12}$

6. $\frac{5}{x}-\frac{2}{3x}$

The common denominator is $3x$: $\frac{15}{3x}-\frac{2}{3x}=\frac{13}{3x}$

7. $\frac{3}{x}-\frac{x-3}{x}$

$\frac{3-(x-3)}{x}=\frac{6-x}{x}$

You can check any difference of numeric fractions, step by step, with the fraction calculator below.