Solve $LaTeX: \displaystyle \frac{x}{x + 4} + \frac{1}{x + 2}=\frac{2}{x^{2} + 6 x + 8}$ .

Factoring the denominator on the right hand side gives $LaTeX: \displaystyle \left(x + 2\right) \left(x + 4\right)$ . This gives the LCD as $LaTeX: \displaystyle \left(x + 2\right) \left(x + 4\right)$ . Multiplying by the LCD gives $LaTeX: \displaystyle x \left(x + 2\right) + x + 4 = 2$ . Getting zero on one side gives $LaTeX: \displaystyle x^{2} + 3 x + 2=0$ . Factoring gives $LaTeX: \displaystyle \left(x + 1\right) \left(x + 2\right)=0$ . The two possible solutions are $LaTeX: \displaystyle x = -1$ and $LaTeX: \displaystyle x = -2$ . Checking the possible solutions gives:
Since $LaTeX: \displaystyle -1$ is not zero of the denominator it is a solution. Since $LaTeX: \displaystyle -2$ is zero of the denominator it is not in the domain and must be rejected as a solution.