# How do you solve (x)/(x-2) – (8)/(x+3) = 10/(x^2+x-6)?

Aug 29, 2016

$x = 3 , x \ne - 3 , 2$

#### Explanation:

Start by factoring everything to see what the denominators are and what we can cancel out.

$\implies \frac{x}{x - 2} - \frac{8}{x + 3} = \frac{10}{\left(x + 3\right) \left(x - 2\right)}$

The common denominator will be $\left(x + 3\right) \left(x - 2\right)$.

$\implies \frac{x \left(x + 3\right)}{\left(x + 3\right) \left(x - 2\right)} - \frac{8 \left(x - 2\right)}{\left(x + 3\right) \left(x - 2\right)} = \frac{10}{\left(x + 3\right) \left(x - 2\right)}$

We can now cancel the denominators and solve as a regular quadratic.

$\implies {x}^{2} + 3 x - 8 x + 16 = 10$

$\implies {x}^{2} - 5 x + 6 = 0$

$\implies \left(x - 3\right) \left(x - 2\right) = 0$

$\implies x = 3 \mathmr{and} 2$

However, the $x = 2$ is extraneous, since it is a restriction on the variable (it makes the denominator equal to $0$ and therefore undefined).

Checking in the original equation, $x = 3$ works.

Hopefully this helps!