Question 86680
Given:
.
{{{(x/(x+4))-(36/(x^2-x-20))}}}
.
Subtract
.
Notice first that {{{x^2-x-20}}} factors into {{{(x +4)*(x -5)}}}. Substitute the factored
form into the problem, and the problem becomes:
.
{{{(x/(x+4))-(36/((x+4)*(x-5))))}}}
.
Next, create a common denominator in the two terms by multiplying the first term by
{{{(x-5)/(x-5)}}} which is equivalent to multiplying by 1 because the numerator of this 
multiplier is the same as the denominator.  This multiplication makes the problem 
become:
.
{{{(x/(x+4))*((x-5)/(x-5))-(36/((x+4)*(x-5))))}}}
.
and this simplifies to:
.
{{{((x*(x-5))/((x+4)*(x-5)))-(36/((x+4)*(x-5))))}}}
.
The numerator of the first term multiplies out to {{{x^2-5x}}}. Substituting this into
the polynomial makes it become:
.
{{{((x^2 - 5x)/((x+4)*(x-5)))-(36/((x+4)*(x-5))))}}}
.
Now since the two terms have the same denominator, they can be combined by just subtracting
the two numerators and putting that result over the common denominator.  This results
in:
.
{{{((x^2 - 5x -36)/((x+4)*(x-5))))}}}
.
Now notice that the numerator {{{x^2 - 5x - 36}}} factors into {{{(x - 9)*(x + 4)}}}.
Substitute this for the numerator and the result is:
.
{{{(((x-9)*(x+4)))/((x+4)*(x-5))))}}}
.
Cancel the like terms in the numerator and denominator:
.
{{{(((x-9)*(cross(x+4))))/((cross(x+4))*(x-5))))}}}
.
and this reduces to the form:
.
{{{(x-9)/(x-5)}}}
.
This is the answer you are looking for. Hope this helps you to understand the problem
and how to work it through to an answer.
.