Adding & Subtracting Fractions

You can only add or subtract fractions when the pieces are the same size — that is, when the denominators match. If they already match, add or subtract the numerators and keep the denominator. If they do not, rewrite both fractions over a common denominator first, then combine.

That second case is where the real work lives, and it is one skill: finding the least common denominator (LCD) and scaling each fraction up to it. Master that and every fraction addition or subtraction problem is the same three steps.

Same denominator: combine the tops

When the denominators match, the denominator just names the size of the pieces — eighths, sixths, twelfths. Adding $\dfrac{3}{8} + \dfrac{1}{8}$ is adding $3$ eighths to $1$ eighth: four eighths, or $\dfrac{4}{8}$ . Add or subtract the numerators; the denominator does not change.

The most common error on this skill is adding the denominators too, turning $\dfrac{3}{8} + \dfrac{1}{8}$ into $\dfrac{4}{16}$ . That would mean the pieces changed size just because you counted them — they did not.

Different denominators: find the LCD

When the denominators differ, rewrite both fractions with the least common denominator — the smallest number both denominators divide into. For $4$ and $3$ that is $12$ ; for $6$ and $8$ it is $24$ .

To rewrite a fraction, multiply its numerator and denominator by the same number. For $\dfrac{1}{4}$ with an LCD of $12$ : since $4 \cdot 3 = 12$ , multiply top and bottom by $3$ to get $\dfrac{3}{12}$ . You are multiplying by $\dfrac{3}{3} = 1$ , so the value of the fraction never changes — only its name.

Any common denominator works, not just the least one — you could always use the product of the two denominators. The LCD just keeps the numbers smaller and the final simplifying easier.

Simplify, and handle mixed numbers

Always check the final answer for common factors. $\dfrac{4}{8}$ should be reported as $\dfrac{1}{2}$ . If the top and bottom share no factor other than $1$ , the fraction is in simplest form.

For mixed numbers, the reliable path is to convert to improper fractions first: $2\dfrac{1}{3} = \dfrac{7}{3}$ . Then add or subtract as usual and convert back at the end. This avoids the borrowing headaches that come from working with the whole-number parts separately.

Worked examples

Example 1: same denominator

Add $\dfrac{3}{8} + \dfrac{1}{8}$ .

The denominators match, so add the numerators

\dfrac{3 + 1}{8} = \dfrac{4}{8}

Simplify — top and bottom share a factor of

4

\dfrac{1}{2}

Answer: $\dfrac{1}{2}$

Example 2: different denominators

Add $\dfrac{1}{4} + \dfrac{2}{3}$ .

Find the LCD of

4

and

3

12

Rewrite each fraction over

12

\dfrac{3}{12} + \dfrac{8}{12}

Add the numerators, keep the denominator

\dfrac{11}{12}

Answer: $\dfrac{11}{12}$

Example 3: subtracting with unlike denominators

Subtract $\dfrac{5}{6} - \dfrac{1}{4}$ .

Find the LCD of

6

and

4

12

Rewrite each fraction over

12

\dfrac{10}{12} - \dfrac{3}{12}

Subtract the numerators, keep the denominator

\dfrac{7}{12}

Answer: $\dfrac{7}{12}$

Example 4: mixed numbers

Add $2\dfrac{1}{3} + 1\dfrac{1}{2}$ .

Convert both to improper fractions

\dfrac{7}{3} + \dfrac{3}{2}

Rewrite over the LCD,

6

\dfrac{14}{6} + \dfrac{9}{6}

Add the numerators

\dfrac{23}{6}

Convert back to a mixed number

3\dfrac{5}{6}

Answer: $3\dfrac{5}{6}$

Try one yourself

Common questions

Why do I need a common denominator to add fractions?

Because you can only count pieces together when they are the same size. A fourth plus a third is not two of anything until you rename both as twelfths — then $\dfrac{3}{12} + \dfrac{4}{12}$ is clearly $7$ twelfths.

Does rewriting a fraction over a new denominator change its value?

No. Multiplying the top and bottom by the same number is multiplying by $1$ , so $\dfrac{1}{4}$ and $\dfrac{3}{12}$ are the same amount. Only the size of the pieces changes, and the count changes to match.

What if I used a common denominator that is not the least one?

Your answer will still be correct — it will just need more simplifying at the end. Using $4 \cdot 3 = 12$ versus some larger shared multiple changes the size of the numbers you carry, not the final value.

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