College Algebra, Chapter 1, 1.1, Section 1.1, Problem 48

The equation $\displaystyle \frac{1}{x+3} + \frac{5}{x^2 - 9} = \frac{2}{x-3}$ is either linear or equivalent to a linear equation. Solve the equation

$\begin{equation} \begin{aligned} \frac{1}{x+3} + \frac{5}{x^2 - 9} &= \frac{2}{x-3} && \text{Subtract both sides by } \frac{2}{x-3}\\ \\ \frac{1}{x+3} + \frac{5}{x^2- 9} - \frac{2}{x-3} &= \frac{2}{x-3} - \frac{2}{x-3} && \text{Simplify}\\ \\ \frac{1}{x+3} + \frac{5}{(x^2-9)} - \frac{2}{x-3} &= 0 && \text{Expand the denominator of 2nd term}\\ \\ \frac{1}{x+3} + \frac{5}{(x+3)(x-3)} - \frac{2}{x-3} &= 0 && \text{Get the LCD}\\ \\ \frac{x-3+5-2(x+3)}{(x+3)(x-3)} &= 0 && \text{Simplify}\\ \\ \frac{x-3+5-2x-6}{(x+3)(x-3)} &= 0 && \text{Combine like terms}\\ \\ \frac{-x-4}{(x+3)(x-3)} &= 0 && \text{Multiply both sides by } (x+3)(x-3)\\ \\ \cancel{(x+3)} \cancel{(x-3)} & \left[ \frac{-x-4}{\cancel{(x+3)}\cancel{(x-3)}} = 0 \right] (x+3)(x-3) && \text{Cancel out like terms}\\ \\ -x-4 &= 0 && \text{Add both sides by 4}\\ \\ -x - 4 + 4 &= 0 + 4 && \text{Simplify}\\ \\ -x &= 4 && \text{Multiply both sides by -1}\\ \\ -1 & [-x= 4] -1 && \text{Simplify}\\ \\ x &= -4 \end{aligned} \end{equation}$