How do you combine #(x^2)/(x^2 - 49) - x/(x-7)#?

2 Answers
Apr 16, 2018

#-(7x)/((x-7)(x+7))#

#=>-(7x)/(x^2-49)#

Explanation:

First, you have to realize that the denominator of the first fraction is a difference between two squares:

If you have
#(a^2-b^2)#

You can rewrite it as
#(a-b)(a+b)#

In this situation
#a=x#
#b=7#

So we can rewrite the first fraction as
#x^2/((x-7)(x+7))#

Now we have
#x^2/((x-7)(x+7))-x/(x-7)#

So in order to combine them, we need an LCD. In order to do so, we can multiply the second fraction by #(x+7)/(x+7)#. This works because it is the same thing as multiplying it by #1#, which wouldn't change it.

Now we have

#x^2/((x-7)(x+7))-((x+7)/(x+7))x/(x-7)#

#=>x^2/((x-7)(x+7))-(x(x+7))/((x-7)(x+7))#

Now distribute the #x# in the numerator of the second fraction

#x^2/((x-7)(x+7))-(x^2+7x)/((x-7)(x+7))#

Now subtract the fractions

#(x^2-x^2-7x)/((x-7)(x+7))#

#=>-(7x)/((x-7)(x+7))#

Answer

#-(7x)/((x-7)(x+7))#

or

#-(7x)/(x^2-49)#

Apr 16, 2018

#-(7x)/((x-7)(x+7))#
or
#-(7x)/(x^2-49)#

Explanation:

#(x^2)/(x^2-49)-(x)/(x-7)#
Notice that #x^2-49# is a difference of squares (#x^2-7^2#), so it can be factored to #(x+7)(x-7)# Inputting this,
#(x^2)/((x+7)(x-7))-(x)/(x-7)#
Now, multiply #(x)/(x-7)# by #(x+7)/(x+7)# to get common denominators:
#(x^2)/((x+7)(x-7))-((x)(x+7))/((x-7)(x+7))#
Since the denominators are now common, you can subtract the numerators and you get
#((x^2)-((x)(x+7)))/((x-7)(x+7))#. This simplifies to
#-(7x)/((x-7)(x+7))#
or
#-(7x)/(x^2-49)#