College Algebra, Chapter 1, 1.1, Section 1.1, Problem 46

The equation $\displaystyle \frac{1}{3-t} + \frac{4}{3+t} + \frac{16}{9-t^2} = 0 $ is either linear or equivalent to a linear equation. Solve the equation

$
\begin{equation}
\begin{aligned}
\frac{1}{3-t} + \frac{4}{3+t} + \frac{16}{9-t^2} &= 0 && \text{Expand the denominator of the 3rd term}\\
\\
\frac{1}{3-t} + \frac{4}{3+t} + \frac{16}{(3-t)(3+t)} &= 0 && \text{Get the LCD}\\
\\
\frac{3+t+4(3-t)+16}{(3-t)(3+t)} &= 0 && \text{Simplify}\\
\\
\frac{3+t+12-4t+16}{(3-t)(3+t)} &= 0 && \text{Combine like terms}\\
\\
\frac{31-3t}{(3-t)(3+t)} &= 0 && \text{Apply FOIL method in the denominator}\\
\\
\frac{31-3t}{0-t^2} &= 0 && \text{Multiply both sides by } (9-t^2)\\
\\
\cancel{(9-t^2)} & \left[ \frac{31-3t}{\cancel{9-t^2}} = 0 \right] (9-t^2) && \text{Cancel out like terms}\\
\\
31 - 3t &= 0 && \text{Solve for } t \\
\\
3t &= 31 && \text{Divide both sides by 3} \\
\\
\frac{\cancel{3}t}{\cancel{3}} &= \frac{31}{3} && \text{Simplify}\\
\\
t &= \frac{31}{3}
\end{aligned}
\end{equation}
$