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

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Answer 1 · 2016-09-06T02:48:23

$x = 4$ $x=4$

Rewrite this as follows

$\frac{8 (x - 1)}{x^{2} - 4} = \frac{4}{x - 2}$ $(8(x-1))/(x^2-4)=4/(x-2)$

$\frac{8 (x - 1)}{(x - 2) (x + 2)} - \frac{4}{x - 2} = 0$ $[8(x-1)]/[(x-2)(x+2)]-4/(x-2)=0$

$\frac{1}{x - 2} \cdot [\frac{8 (x - 1)}{x + 2} - 4] = 0$ $1/(x-2)*[(8(x-1))/(x+2)-4]=0$

$\frac{1}{(x - 2) \cdot (x + 2)} [8 (x - 1) - 4 (x + 2)] = 0$ $1/[(x-2)*(x+2)][8(x-1)-4(x+2)]=0$

$\frac{8 x - 8 - 4 x - 8}{(x - 2) (x + 2)} = 0$ $[8x-8-4x-8]/[(x-2)(x+2)]=0$

$\frac{4 (x - 4)}{(x - 2) (x + 2)} = 0$ $(4(x-4))/[(x-2)(x+2)]=0$

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

Footnote

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

How do you solve (8(x-1))/(x^2-4)=4/(x-2)8(x−1)x2−4=4x−2(8(x-1))/(x^2-4)=4/(x-2)?