Multiplying and Dividing Fractions with Variables

To multiply and divide fractions with variables:

factor all numerators and denominators completely

use the rules for multiplying and dividing fractions:

$$ \frac{A}{B}\cdot\frac{C}{D} = \frac{AC}{BD} $$	(to multiply fractions, multiply ‘across’)
$$ \frac{A}{B}\div\frac{C}{D} = \frac{A}{B}\cdot\frac{D}{C} = \frac{AD}{BC} $$	(to divide by a fraction, instead multiply by its reciprocal)

cancel any common factors; that is, get rid of any extra ‘factors of $\,1\,$’
leave your final answer in factored form

EXAMPLE:

Multiply, and write your answer in simplest form: $$ \frac{x^2-9}{5x^2+20x+15} \cdot \frac{x+1}{x+4} $$

SOLUTION:

$\displaystyle \frac{x^2-9}{5x^2+20x+15} \cdot \frac{x+1}{x+4} $	$=$	$\displaystyle \frac{(x-3)(x+3)}{5(x^2+4x+3)}\cdot\frac{x+1}{x+4}$	factor: difference of squares (numerator), common factor (denominator)
	$=$	$\displaystyle \frac{(x-3)(x+3)}{5(x+3)(x+1)}\cdot\frac{x+1}{x+4} $	factor the trinomial in the denominator
	$=$	$\displaystyle \frac{\hphantom{5}(x+1)(x+3)(x-3)}{5(x+1)(x+3)(x+4)} $	multiply, re-order
	$=$	$\displaystyle \frac{(x-3)}{5(x+4)} $	cancel the two extra factors of $\,1\,$

It is interesting to compare the original expression (before simplification), and the simplified expression (after cancellation).
Although they are equal for almost all values of $\,x\,$, they do differ a bit, because of the cancellation:

VALUES OF $\,x\,$	ORIGINAL EXPRESSION: $\displaystyle \frac{x^2-9}{5x^2+20x+15} \cdot \frac{x+1}{x+4} $ in factored form: $\displaystyle \frac{\hphantom{5}(x+1)(x+3)(x-3)}{5(x+1)(x+3)(x+4)}$	SIMPLIFIED EXPRESSION: $\displaystyle \frac{(x-3)}{5(x+4)}$	COMPARISON
$x = -4$	not defined (division by zero)	not defined (division by zero)	behave the same: both are not defined
$x = -1$	not defined (division by zero)	$\displaystyle \frac{-1-3}{5(-1+4)} = -\frac{4}{15}$	the presence of $\,\frac{x+1}{x+1}\,$ causes a puncture point at $\,x = -1\,$; see the first graph below
$x = -3$	not defined (division by zero)	$\displaystyle \frac{-3-3}{5(-3+4)} = -\frac{6}{5}$	the presence of $\,\frac{x+3}{x+3}\,$ causes a puncture point at $\,x = -3\,$; see the first graph below
all other values of $\,x\,$	both defined; values are equal		behave the same: values are equal

GRAPH OF: $\displaystyle \frac{\hphantom{5}(x+1)(x+3)(x-3)}{5(x+1)(x+3)(x+4)}$	GRAPH OF: $\displaystyle \frac{(x-3)}{5(x+4)}$

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Adding and Subtracting Fractions With Variables

For more advanced students, a graph is displayed.
For example, the expression $\,\frac{x+1}{x+2}\cdot\frac{x+3}{x+1}\,$ is optionally accompanied by the graph of $\,y = \frac{x+1}{x+2}\cdot\frac{x+3}{x+1}\,$.
A puncture point occurs at $\,x = -1\,$, due to the presence of $\,\frac{x+1}{x+1}\,$.
The graph of the simplified expression would not have this puncture point.
Horizontal/vertical asymptote(s) are shown in light grey.
Note: A puncture point may occasionally occur outside the viewing window;
use the arrows in the lower-right graph corner to navigate left/up/down/right.
Click the “show/hide graph” button if you prefer not to see the graph.