Beginning Algebra With Applications, Chapter 3, 3.2, Section 3.2, Problem 202

Solve $\displaystyle \frac{1}{5} (25-10 b) + 4 = \frac{1}{3} (9b-15)-6$. If the equation has no solution, write "No solution."



$
\begin{equation}
\begin{aligned}

\frac{1}{5} (25-10b) + 4 =& \frac{1}{3} (9b-15)-6
&& \text{Given equation}
\\
\\
\frac{25}{5}- \frac{10b}{5}+ 4 =& \frac{9b}{3} - \frac{15}{3} - 6
&& \text{Apply Distributive Property}
\\
\\
5-2b + 4 =& 3b-5-6
&& \text{Simplify}
\\
\\
-2b-3b =& -5-5-6-4
&& \text{Combine like terms}
\\
\\
-5b =& -20
&& \text{Simplify}
\\
\\
\frac{\cancel{-5}b}{\cancel{-5}} =& \frac{-20}{-5}
&& \text{Divide by } -5
\\
\\
b =& 4
&&

\end{aligned}
\end{equation}
$