Beginning Algebra With Applications, Chapter 3, 3.2, Section 3.2, Problem 68

Solve $\displaystyle \frac{2}{3} = \frac{3}{4} - \frac{1}{2}y$ and check.


$
\begin{equation}
\begin{aligned}

\frac{2}{3} - \frac{3}{4} =& \frac{3}{4} - \frac{3}{4} - \frac{1}{2} y
&& \text{Subtract } \frac{3}{4}
\\
\\
\frac{-1}{12} =& \frac{-1}{2} y
&& \text{Simplify}
\\
\\
-2 \left( \frac{-1}{12} \right) =& \left( \frac{-1}{2} y \right) -2
&& \text{Multiply both sides by } -2
\\
\\
\frac{1}{6} =& y
&&

\end{aligned}
\end{equation}
$


Checking:


$
\begin{equation}
\begin{aligned}

\frac{2}{3} =& \frac{3}{4} - \frac{1}{2} \left( \frac{1}{6} \right)
&& \text{Substitute } y = \frac{1}{6}
\\
\\
\frac{2}{3} =& \frac{3}{4} - \frac{1}{12}
&& \text{Simplify}
\\
\\
\frac{2}{3} =& \frac{2}{3}
&&

\end{aligned}
\end{equation}
$

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