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}\\
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\cancel{12} & \left[\frac{11y}{\cancel{12}} = \frac{21}{\cancel{12}} \right] \cancel{12} && \text{Simplify}\\
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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}
$