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

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