Multiplying and dividing algebraic fractions

Algebraic fractions are multiplied and divided in the same way as number fractions: \(\dfrac{a}{b}\times\dfrac{c}{d}=\dfrac{ac}{bd}\) and \(\dfrac{a}{b}\div\dfrac{c}{d}=\dfrac{a}{b}\times\dfrac{d}{c}\), where \(b\), \(c\) and \(d\) are non-zero.

Why dividing means multiplying by the reciprocal: \(\dfrac{c}{d}\times\dfrac{d}{c}=1\), so multiplying by \(\dfrac{d}{c}\) undoes multiplying by \(\dfrac{c}{d}\). With numbers, \(\dfrac23\div\dfrac49=\dfrac23\times\dfrac94=\dfrac32\), and indeed \(\dfrac32\times\dfrac49=\dfrac23\).

Method

1. For a division, change \(\div\) to \(\times\) and turn over only the fraction straight after the \(\div\) sign.
2. Write one fraction: the product of the numerators over the product of the denominators.
3. Cancel common factors before multiplying out, so the numbers stay small.
4. State the restrictions: the original denominators, and the numerator of any divisor, are non-zero.

Example

Simplify \(\dfrac{10a^2}{3b}\times\dfrac{9b^2}{25a}\).

\(\begin{aligned}\dfrac{10a^2}{3b}\times\dfrac{9b^2}{25a}&=\dfrac{10a^2\times9b^2}{3b\times25a} \\ &=\dfrac{90a^2b^2}{75ab} \\ &=\dfrac{6ab}{5},\qquad a\ne0,\ b\ne0\end{aligned}\)

Simplify \(\dfrac{6p^2}{5q}\div\dfrac{9pq}{10}\).

\(\begin{aligned}\dfrac{6p^2}{5q}\div\dfrac{9pq}{10}&=\dfrac{6p^2}{5q}\times\dfrac{10}{9pq} \\ &=\dfrac{60p^2}{45pq^2} \\ &=\dfrac{4p}{3q^2},\qquad p\ne0,\ q\ne0\end{aligned}\)

Check: turn over the second fraction only, never the first. In a chain such as \(\dfrac{a}{b}\div\dfrac{c}{d}\times\dfrac{e}{f}\), work from left to right, so only \(\dfrac{c}{d}\) is turned over.