College Algebra, Chapter 1, 1.1, Section 1.1, Problem 32

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

$
\begin{equation}
\begin{aligned}
\frac{1}{t-1} + \frac{t}{3t-2} &= \frac{1}{2} && \text{Get the LCD of the left side}\\
\\
\frac{3t-2+t(t-1)}{(t-1)(3t-2)} &= \frac{1}{2} && \text{Simplify}\\
\\
\frac{3t-2+t^2-t}{(t-1)(3t-2)} &= \frac{1}{2} && \text{Combine like terms}\\
\\
\frac{t^2 + 2t - 2}{(t-1)(3t-2)} &= \frac{1}{2} && \text{Factor the numerator}\\
\\
\frac{\cancel{(t-1)}(t+2)}{\cancel{(t-1)}(3t-2)} &= \frac{1}{2} && \text{Cancel out like terms}\\
\\
\frac{t+2}{3t-2} &= \frac{1}{2} && \text{Multiply both sides by } 2(3t-2)\\
\\
2 (\cancel{3t-2}) & \left[ \frac{t+2}{\cancel{3t-2}} = \frac{1}{\cancel{2}} \right] \cancel{2} (3t - 2) && \text{Simplify}\\
\\
2(t+2) &= 3t -2 && \text{Apply Distributive property}\\
\\
2t +4 &= 3t - 2&& \text{Combine like terms}\\
\\
2t - 3t &= -4 -2 && \text{Simplify}\\
\\
-t &= -6 && \text{Multiply both sides by -1}\\
\\
-1 & [ -t = -6 ] -1 && \text{Simplify}\\
\\
t &= 6
\end{aligned}
\end{equation}
$