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}
$