Beginning Algebra With Applications, Chapter 3, 3.2, Section 3.2, Problem 128

Solve $\displaystyle \frac{1}{5} d = \frac{1}{2}d + 3$ and check.


$
\begin{equation}
\begin{aligned}

\frac{1}{5} d =& \frac{1}{2}d + 3
&& \text{Given equation}
\\
\\
\frac{1}{5}d - \frac{1}{2}d =& 3
&& \text{Subtract } \frac{1}{2}d
\\
\\
\frac{-3}{10} d =& 3
&& \text{Simplify}
\\
\\
\frac{-10}{3} \left( \frac{-3}{10} d \right) =& 3 \left( \frac{-10}{3} \right)
&& \text{Multiply both sides by } \frac{-10}{3}
\\
\\
d =& -10
&&

\end{aligned}
\end{equation}
$


Checking:


$
\begin{equation}
\begin{aligned}

\frac{1}{5} (-10) =& \frac{1}{2} (-10) + 3
&& \text{Substitute } d = -10
\\
\\
\frac{-10}{5} =& \frac{-10}{2} + 3
&& \text{Simplify}
\\
\\
-2 =& -2
&&

\end{aligned}
\end{equation}
$