# Multiplication and Division of Fractions

The following rules for multiplication and division of fractions can be derived from the basic algebraic properties of real numbers and the definition of a quotient.

(1) Rule for Multiplication of Fractions

If $p$, $q$, $r$and $s$ are real numbers, $q \ne 0$, and $s \ne 0$, then
$\frac{p}{q} \cdot \frac{r}{s} = \frac{{pr}}{{qs}}$

For instance, $\frac{2}{3} \cdot \frac{5}{7} = \frac{{2 \cdot 5}}{{3 \cdot 7}} = \frac{{10}}{{21}}$

(2) Rule for Division of Fractions

If $p$, $q$, $r$and $s$ are real numbers, $q \ne 0$, $r \ne 0$and $s \ne 0$, then
$\frac{p}{q} \div \frac{r}{s} = \frac{{p/q}}{{r/s}} = \frac{p}{q} \cdot \frac{s}{r} = \frac{{ps}}{{qr}}$

For instance, $\frac{2}{3} \div \frac{5}{7} = \frac{2}{3} \cdot \frac{7}{5} = \frac{{14}}{{15}}$   and   $\frac{{3/4}}{{4/5}} = \frac{3}{4} \cdot \frac{5}{4} = \frac{{15}}{{16}}$

Of course, the same rules apply to multiplying or dividing fractional expressions in general. Before multiplying fractions, it is good idea to factor the numerators and denominators completely to reveal any factors that can be canceled.

Example:

Perform the indicated operation and simplify the result.

(a) $\frac{{x + 4}}{y} \cdot \frac{{{y^3}}}{{5x + 20}}$
(b) $\frac{{{x^3} – 1}}{{5{x^2} – 26x + 5}} \cdot \frac{{5{x^2} + 9x – 2}}{{{x^2} + x – 2}}$
(c) $\frac{{{t^2} – 49}}{{{t^2} – 5t – 14}} \div \frac{{2{t^2} + 15t + 7}}{{2{t^2} – 13t – 7}}$

Solution:

(a)
$\frac{{x + 4}}{y} \cdot \frac{{{y^3}}}{{5x + 20}} = \frac{{x + 4}}{y} \cdot \frac{{y \cdot {y^2}}}{{5\left( {x + 4} \right)}} = \frac{{\left( {x – 4} \right)y \cdot {y^2}}}{{y\left( 5 \right)\left( {x + 4} \right)}} = \frac{{{y^2}}}{5}$

(b) $\frac{{{x^3} – 1}}{{5{x^2} – 26x + 5}} \cdot \frac{{5{x^2} + 9x – 2}}{{{x^2} + x – 2}} = \frac{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {5x – 1} \right)\left( {x – 5} \right)}} \cdot \frac{{\left( {5x – 1} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x – 1} \right)}}$
$= \frac{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)\left( {5x – 1} \right)\left( {x + 2} \right)}}{{\left( {5x – 1} \right)\left( {x – 5} \right)\left( {x + 2} \right)\left( {x – 1} \right)}}$
$= \frac{{\left( {{x^2} + x + 1} \right)}}{{\left( {x – 5} \right)}}$

(c)
$\frac{{{t^2} – 49}}{{{t^2} – 5t – 14}} \div \frac{{2{t^2} + 15t + 7}}{{2{t^2} – 13t – 7}} = \frac{{{t^2} – 49}}{{{t^2} – 5t – 14}} \cdot \frac{{2{t^2} – 13t – 7}}{{2{t^2} + 15t + 7}}$
$= \frac{{\left( {t – 7} \right)\left( {t + 7} \right)}}{{\left( {t + 2} \right)\left( {t – 7} \right)}} \cdot \frac{{\left( {2t + 1} \right)\left( {t – 7} \right)}}{{\left( {2t + 1} \right)\left( {t + 7} \right)}}$
$= \frac{{\left( {t – 7} \right)\left( {t + 7} \right)\left( {2t + 1} \right)\left( {t – 7} \right)}}{{\left( {t + 2} \right)\left( {t – 7} \right)\left( {2t + 1} \right)\left( {t + 7} \right)}}$
$= \frac{{t – 7}}{{t + 2}}$