How do you solve #x/(x+2) - 2/(x-2) = (x^2+4)/(x^2-4)#?

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How do you solve #x/(x+2) - 2/(x-2) = (x^2+4)/(x^2-4)#?

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Answer 1 · 2016-02-27T03:59:26

There are no real solutions

Explanation:

First we must exclude the values that nullifies the denominators
which are #x=2# and #x=-2#
It is

#x/(x+2) - 2/(x-2) = (x^2+4)/(x^2-4)=> (x*(x-2)-2*(x+2))/(x^2-4)=(x^2+4)/(x^2-4)=> (x^2-4x-4)/(x^2-4)=(x^2+4)/(x^2-4)=> 1/(x^2-4)*[x^2-4x-4-x^2-4]=0=> 1/(x^2-4)*(-4x-8)=0=> -4*(x+2)/[(x-2)*(x+2)]=0=> -4/(x-2)=0#

From the last equation is apparent that there no real solutions

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How do you solve #x/(x+2) - 2/(x-2) = (x^2+4)/(x^2-4)#?

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