# How do you solve (x+3)/(x^2-6x-16)+(x-14)/(x^2-4x-32)=(x+1)/(x^2+6x+8) and check for extraneous solutions?

Aug 12, 2016

$x = 2$ (extraneous solution $x = - 4$)

#### Explanation:

First factor all of the denominators and multiply through by the least common multiple:

${x}^{2} - 6 x - 16 = \left(x + 2\right) \left(x - 8\right)$

${x}^{2} - 4 x - 32 = \left(x - 8\right) \left(x + 4\right)$

${x}^{2} + 6 x + 8 = \left(x + 2\right) \left(x + 4\right)$

So the least common multiple of the denominators is:

$\left(x + 2\right) \left(x + 4\right) \left(x - 8\right)$

Multiplying the original equation by this we get:

$\left(x + 3\right) \left(x + 4\right) + \left(x - 14\right) \left(x + 2\right) = \left(x + 1\right) \left(x - 8\right)$

Which expands out to:

${x}^{2} + 7 x + 12 + {x}^{2} - 12 x - 28 = {x}^{2} - 7 x - 8$

Subtract the right hand side from the left and simplify to get:

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

Hence zeros $x = - 4$ and $x = 2$

The value $x = - 4$ is extraneous, since it makes two of the denominators of the original equation zero and division by $0$ is undefined.

That leaves one solution $x = 2$