# How do you solve 3/(x+4)-1/(x+3)=(x+9)/(x^2+7x+12) and check for extraneous solutions?

Dec 30, 2016

$x = 4$

#### Explanation:

Given:

$\frac{3}{x + 4} - \frac{1}{x + 3} = \frac{x + 9}{{x}^{2} + 7 x + 12}$

Multiply both sides by:

$\left(x + 4\right) \left(x + 3\right) = {x}^{2} + 7 x + 12$

to get:

$3 \left(x + 3\right) - 1 \left(x + 4\right) = x + 9$

Expanding, this becomes:

$3 x + 9 - x - 4 = x + 9$

Simplifying:

$2 x + 5 = x + 9$

Subtract $x + 5$ from both sides to get:

$x = 4$

This solution is valid since $\left(\textcolor{b l u e}{4} + 4\right) = 8 \ne 0$ and $\left(\textcolor{b l u e}{4} + 3\right) = 7 \ne 0$.