Adding fractions

What the formula is doing

It rewrites both fractions over $bd$ and then adds the tops.

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

Once the bottom numbers match, the parts are the same size and the tops can be counted. The denominator never joins the addition, because it is a name for the part size, not a quantity.

Why you cannot add the denominators

$$\frac{2}{5} + \frac{1}{5} \ne \frac{3}{10}$$

Collecting more fifths does not make the fifths smaller. Two fifths plus one fifth is more than a half, while $\tfrac{3}{10}$ is less than a half — so adding the bottoms moves the answer the wrong way entirely.

The same reasoning as two metres plus one metre being three metres, not three square metres.

Use the LCD when it is smaller

The formula’s $bd$ always works but is not always the smallest choice.

$$\frac{1}{4} + \frac{1}{6}$$

The formula gives $\tfrac{6 + 4}{24} = \tfrac{10}{24} = \tfrac{5}{12}$. Using the least common denominator of $12$ gives $\tfrac{3}{12} + \tfrac{2}{12} = \tfrac{5}{12}$ directly, with no simplifying afterwards.

Worked examples

Work out $\tfrac{1}{2} + \tfrac{1}{3}$.

$$\frac{1 \times 3 + 2 \times 1}{2 \times 3} = \frac{5}{6}$$

Work out $\tfrac{3}{4} + \tfrac{2}{3}$.

$$\frac{9 + 8}{12} = \frac{17}{12} = 1\tfrac{5}{12}$$

Both fractions are over a half, so a total past $1$ is the expected size.