How do you solve $(x-4) / (x+2) + 2 /( x-2) = 17 / (x^2-4)$ ?

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Answer 1 · 2017-04-03T14:54:40

$x=-1$ or $x=5$

Start from the left hand side:

$(x-4)/(x+2)+2/(x-2)=((x-4)(x-2)+2(x+2))/((x+2)(x-2))=(x^2-6x+8 +2x+4)/(x^2-4)=(x^2-4x+12)/(x^2-4)$

which is equal to the right hand side $17/(x^2-4)$ .

So we have
$x^2-4x+12=17$
$x^2-4x-5=0$
$(x+1)(x-5)=0$
$x=-1$ or $x=5$

How do you solve (x-4) / (x+2) + 2 /( x-2) = 17 / (x^2-4)(x-4) / (x+2) + 2 /( x-2) = 17 / (x^2-4)?