# How do you solve 1/(x^2-5x)=(x+7)/x-1 and check for extraneous solutions?

Jan 17, 2017

$\left\{\frac{36}{7}\right\}$

#### Explanation:

Recall that ${x}^{2} - 5 x$ can be written as $x \left(x - 5\right)$.

$\frac{1}{x \left(x - 5\right)} = \frac{x + 7}{x} - 1$

Put on a common denominator:

$\frac{1}{x \left(x - 5\right)} = \frac{\left(x + 7\right) \left(x - 5\right)}{x \left(x - 5\right)} - \frac{x \left(x - 5\right)}{x \left(x - 5\right)}$

$1 = {x}^{2} + 7 x - 5 x - 35 - \left({x}^{2} - 5 x\right)$

$1 = {x}^{2} + 7 x - 5 x - 35 - {x}^{2} + 5 x$

$36 = 7 x$

$x = \frac{36}{7}$

This is not extraneous, as $x \ne 0 , 5$.

Hopefully this helps!