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

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