How do you solve $(8(x-1))/(x^2-4)=4/(x-2)$ ?

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Answer 1 · 2016-09-06T02:48:23

$x=4$

Rewrite this as follows

$(8(x-1))/(x^2-4)=4/(x-2)$

$[8(x-1)]/[(x-2)(x+2)]-4/(x-2)=0$

$1/(x-2)*[(8(x-1))/(x+2)-4]=0$

$1/[(x-2)*(x+2)][8(x-1)-4(x+2)]=0$

$[8x-8-4x-8]/[(x-2)(x+2)]=0$

$(4(x-4))/[(x-2)(x+2)]=0$

From the last equation we get that $x=4$

Footnote

For the (initial) equation to hold must be $x!=2$ and $x!=-2$

How do you solve (8(x-1))/(x^2-4)=4/(x-2)(8(x-1))/(x^2-4)=4/(x-2)?