# 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 31, 2016

$\left\{2\right\}$.

#### Explanation:

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

Put on a common denominator.

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

Cancel the denominators and solve the resulting quadratic.

${x}^{2} + 3 x + 4 x + 12 + {x}^{2} - 14 x + 2 x - 28 = {x}^{2} + x - 8 x - 8$

${x}^{2} + 2 x - 8 = 0$

$\left(x + 4\right) \left(x - 2\right) = 0$

$x = - 4 \mathmr{and} 2$

However, $x = - 4$ is extraneous since it is one of the restrictions on the original equation (it renders the denominator $0$). The solution set is therefore $\left\{2\right\}$.

Hopefully this helps!