Intermediate Algebra, Chapter 2, 2.1, Section 2.1, Problem 36

Solve the equation $4[2x - (3-x) + 5] = -(2 + 7x)$, and check your solution. If applicable, tell whether the equation is an identity or contradiction.


$
\begin{equation}
\begin{aligned}

4[2x - (3-x) + 5] =& -(2 + 7x)
&& \text{Given equation}
\\
4[2x - 3 + x + 5] =& -2-7x
&& \text{Distributive property}
\\
4 [3x+2] =& -2-7x
&& \text{Combine like terms}
\\
12x + 8 =& -2 - 7x
&& \text{Distributive property}
\\
12x + 7x =& -2-8
&& \text{Add $(7x-8)$ from each side}
\\
19x =& -10
&& \text{Combine like terms}
\\
\frac{19x}{19} =& \frac{-10}{19}
&& \text{Divide both sides by $19$}
\\
x =& \frac{-10}{19}
&&

\end{aligned}
\end{equation}
$


Checking:


$
\begin{equation}
\begin{aligned}

4 \left[ 2 \left( \frac{-10}{19} \right) - \left( 3 - \left( \frac{-10}{19} \right) \right) + 5 \right] =& - \left( 2 + 7 \left( \frac{-10}{19} \right) \right)
&& \text{Substitute } x = \frac{-10}{19}
\\
\\
4 \left[ 2 \left( \frac{-10}{19} \right) - \left( \frac{67}{19} \right) + 5 \right] =& - \left( 2 - \frac{70}{19} \right)
&& \text{Work inside parentheses first}
\\
\\
4 \left[ \frac{-20}{19} - \frac{67}{19} + 5 \right] =& - \left( - \frac{32}{19} \right)
&& \text{Work inside parentheses first}
\\
\\
4 \left( \frac{8}{19} \right) =& \frac{32}{19}
&& \text{Simplify}
\\
\\
\frac{32}{19} =& \frac{32}{19}
&& \text{True}

\end{aligned}
\end{equation}
$