College Algebra, Chapter 1, 1.1, Section 1.1, Problem 26

The equation $\displaystyle \frac{2}{3} y + \frac{1}{2} (y - 3) = \frac{y+1}{4}$ is either linear or equivalent to a linear equation. Solve the equation

$\begin{equation} \begin{aligned} \frac{2}{3} y + \frac{1}{2} (y - 3) &= \frac{y+1}{4} && \text{Apply Distributive proeperty}\\ \\ \frac{2y}{3} + \frac{y-3}{2} &= \frac{y+1}{4} && \text{Get the LCD of the left side}\\ \\ \frac{4y + 3y -9}{6} &= \frac{y+1}{4} && \text{Simplify}\\ \\ \frac{7y-9}{6} &= \frac{y+1}{4} && \text{Group}\\ \\ \frac{7y}{6} - \frac{9}{6} &= \frac{y}{4} + \frac{1}{4} && \text{Combine like terms}\\ \\ \frac{7y}{6} - \frac{y}{4} &= \frac{9}{6} + \frac{1}{4} && \text{Get the LCD of both sides}\\ \\ \frac{14y-3y}{12} &= \frac{18+3}{12} && \text{Simplify}\\ \\ \frac{11y}{12} &= \frac{21}{12} && \text{Multiply both sides by 12}\\ \\ \cancel{12} & \left[\frac{11y}{\cancel{12}} = \frac{21}{\cancel{12}} \right] \cancel{12} && \text{Simplify}\\ \\ 11y &= 21 && \text{Divide both sides by 11}\\ \\ \frac{\cancel{11}y}{\cancel{11}} &= \frac{21}{11} && \text{Simplify}\\ \\ y &= \frac{21}{11} \end{aligned} \end{equation}$