# How do you solve ((x-1) / (x^2-x-12)) - 4/(x^2-6x+8) = x/(x^2+x-6)?

Aug 7, 2016

Start by factoring the denominators to see what our denominator will have to be to put on equivalent bases.

#### Explanation:

$\frac{x - 1}{\left(x - 4\right) \left(x + 3\right)} - \frac{4}{\left(x - 4\right) \left(x - 2\right)} = \frac{x}{\left(x + 3\right) \left(x - 2\right)}$

Multiplying to put on a common denominator:

$\frac{\left(x - 1\right) \left(x - 2\right)}{\left(x - 4\right) \left(x + 3\right) \left(x - 2\right)} - \frac{4 \left(x + 3\right)}{\left(x - 4\right) \left(x - 2\right) \left(x + 3\right)} = \frac{x \left(x - 4\right)}{\left(x + 3\right) \left(x - 2\right) \left(x - 4\right)}$

We can now eliminate the denominators and solve as a regular equation.

${x}^{2} - x - 2 x + 2 - 4 x - 12 = {x}^{2} - 4 x$

${x}^{2} - {x}^{2} - 3 x - 10 = 0$

$- 3 x = 10$

$x = - \frac{10}{3}$

Checking in the original equation, we find that $x = - \frac{10}{3}$ satisfies. Hence, the solution set is $\left\{- \frac{10}{3}\right\}$.

Hopefully this helps!