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 $\,1\,$’

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

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.

 Perform the indicated operation and write in simplest form:
 (MAX is 18; there are 18 different problem types.)