# How do you solve 4/(x^2-x-2)+40/(x^2-2x-3)=(x+44)/(x^2-x-2) and check for extraneous solutions?

Oct 3, 2016

Factor.

$\frac{4}{\left(x - 2\right) \left(x + 1\right)} + \frac{40}{\left(x - 3\right) \left(x + 1\right)} = \frac{x + 44}{\left(x - 2\right) \left(x + 1\right)}$

Place on a common denominator.

$\frac{4 \left(x - 3\right)}{\left(x - 2\right) \left(x + 1\right) \left(x - 3\right)} + \frac{40 \left(x - 2\right)}{\left(x - 3\right) \left(x + 1\right) \left(x - 2\right)} = \frac{\left(x + 44\right) \left(x - 3\right)}{\left(x - 2\right) \left(x - 3\right) \left(x + 1\right)}$

$4 x - 12 + 40 x - 80 = {x}^{2} + 44 x - 3 x - 132$

$0 = {x}^{2} - 3 x - 40$

$0 = \left(x - 8\right) \left(x + 5\right)$

$x = 8 \mathmr{and} - 5$

Neither of these solutions are extraneous, because neither render the equation undefined.

Hopefully this helps!