MULTIPLYING AND DIVIDING FRACTIONS WITH VARIABLES
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To multiply and divide fractions with variables:

EXAMPLE:

Multiply, and write your answer in simplest form:     [beautiful math coming... please be patient] $$ \frac{x^2-9}{5x^2+20x+15} \cdot \frac{x+1}{x+4} $$

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)
[beautiful math coming... please be patient] $\displaystyle \frac{-1-3}{5(-1+4)} = -\frac{4}{15}$ 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.

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