Intermediate Algebra, Chapter 2, 2.1, Section 2.1, Problem 44

Solve the equation $2 [- (x - 1) + 4] = 5 + [-(6x - 7) + 9x]$, and check your solution. If applicable, tell whether the equation is an identity or contradiction.


$
\begin{equation}
\begin{aligned}

2 [- (x - 1) + 4] =& 5 + [-(6x - 7) + 9x]
&& \text{Given equation}
\\
2(-x + 1 + 4) =& 5 + (-6x + 7 + 9x)
&& \text{Distributive property}
\\
2(-x + 5) =& 5 + (3x + 7)
&& \text{Combine like terms}
\\
-2x + 10 =& 3x + 12
&& \text{Combine like terms}
\\
-2x - 3x =& 12 - 10
&& \text{Subtract $(3x + 5)$ from each side}
\\
-5x =& 2
&& \text{Combine like terms}
\\
\frac{-5x}{-5} =& \frac{2}{-5}
&& \text{Divide both sides by $-5$}
\\
x =& - \frac{2}{5}
&&


\end{aligned}
\end{equation}
$


Checking:


$
\begin{equation}
\begin{aligned}

2 \left[ - \left( - \frac{2}{5} - 1 \right) + 4 \right] =& 5 + \left[ - \left( 6 \left( - \frac{2}{5} \right) - 7 \right) + 9 \left( - \frac{2}{5} \right) \right]
&& \text{Substitute } x = - \frac{2}{5}
\\
\\
2 \left[ - \left( - \frac{7}{5} \right) + 4 \right] =& 5 + \left[ - \left( - \frac{47}{5} \right) + \left( - \frac{18}{5} \right) \right]
&& \text{Work inside the parentheses first}
\\
\\
2 \left( \frac{27}{5} \right) =& 5 + \frac{29}{5}
&& \text{Add inside parentheses}
\\
\\
\frac{54}{5} =& \frac{54}{5}
&& \text{True}

\end{aligned}
\end{equation}
$