# How do you solve 3/(x+1)-1/(x+4)=(x+12)/(x^2+5x+4) and check for extraneous solutions?

Aug 29, 2016

Solution is $x = 1$

#### Explanation:

$\frac{3}{x + 1} - \frac{1}{x + 4} = \frac{x + 12}{{x}^{2} + 5 x + 4}$

$\Leftrightarrow \frac{3 \left(x + 4\right) - 1 \left(x + 1\right)}{{x}^{2} + 5 x + 4} = \frac{x + 12}{{x}^{2} + 5 x + 4}$ or

$\frac{3 x + 12 - x - 1}{{x}^{2} + 5 x + 4} = \frac{x + 12}{{x}^{2} + 5 x + 4}$ or

$\frac{2 x + 11}{{x}^{2} + 5 x + 4} = \frac{x + 12}{{x}^{2} + 5 x + 4}$

Now assuming ${x}^{2} + 5 x + 4 \ne 0$ (i.e.$x$ is neither equal to $- 1$ nor equal to $- 4$), and multiplying each side by ${x}^{2} + 5 x + 4$, we get

$2 x + 11 = x + 12$ or

$x = 1$

As $x$ is neither equal to $- 1$ nor equal to $- 4$, there is no extraneous solution.