Beginning Algebra With Applications, Chapter 3, 3.2, Section 3.2, Problem 164

Solve $5+3 [1+2 (2x-3)] = 6(x+5)$ and check.


$
\begin{equation}
\begin{aligned}

5+3 [1+2 (2x-3)] =& 6(x+5)
&& \text{Given equation}
\\
\\
5+3 (1+4x - 6) =& 6x + 30
&& \text{Apply Distributive Property}
\\
\\
5+3+12x - 18 =& 6x + 30
&& \text{Apply Distributive Property}
\\
\\
5+3-18-30 =& 6x - 12x
&& \text{Subtract $12x$ and subtract } 30
\\
\\
-40 =& -6x
&& \text{Simplify}
\\
\\
\frac{-40}{-6} =& \frac{\cancel{-6}x}{\cancel{-6}}
&& \text{Divide by } -6
\\
\\
\frac{20}{3} =& x
&&


\end{aligned}
\end{equation}
$


Checking:


$
\begin{equation}
\begin{aligned}

5 + 3 \left[ 1 + 2 \left( 2 \left( \frac{20}{3} \right) - 3 \right)\right] =& 6 \left( \frac{20}{3} + 5 \right)
&& \text{Substitute } x = \frac{20}{3}
\\
5 + 3 \left( \frac{65}{3} \right) =& 6 \left( \frac{35}{3} \right)
&& \text{Simplify}
\\
70 =& 70
&&

\end{aligned}
\end{equation}
$