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:


$\begin{equation} \begin{aligned} \frac{1}{5} (-10) =& \frac{1}{2} (-10) + 3 && \text{Substitute } d = -10 \\ \\ \frac{-10}{5} =& \frac{-10}{2} + 3 && \text{Simplify} \\ \\ -2 =& -2 && \end{aligned} \end{equation}$