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

$x = 4$

#### Explanation:

Rewrite this as follows

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

$\frac{8 \left(x - 1\right)}{\left(x - 2\right) \left(x + 2\right)} - \frac{4}{x - 2} = 0$

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

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

$\frac{8 x - 8 - 4 x - 8}{\left(x - 2\right) \left(x + 2\right)} = 0$

$\frac{4 \left(x - 4\right)}{\left(x - 2\right) \left(x + 2\right)} = 0$

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

Footnote

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