Intermediate Algebra, Chapter 2, 2.1, Section 2.1, Problem 30

Solve the equation $6x - 3(5x + 2) = 4(1 - x)$, and check your solution. If applicable, tell whether the equation is an identity or contradiction.


$
\begin{equation}
\begin{aligned}

6x - 3(5x + 2) =& 4(1 - x)
&& \text{Given equation}
\\
6x - 15x - 6 =& 4 - 4x
&& \text{Distributive property}
\\
-9x - 6 =& 4 - 4x
&& \text{Combine like terms}
\\
-9x + 4x =& 4+6
&& \text{Add $(4x + 6)$ from each side}
\\
-5x =& 10
&& \text{Combine like terms}
\\
\frac{-5x}{-5} =& \frac{10}{-5}
&& \text{Divide both sides by $-5$}
\\
x =& -2
&&

\end{aligned}
\end{equation}
$


Checking:


$
\begin{equation}
\begin{aligned}

6(-2) - 3 [5(-2) + 2] =& 4 [1 - (-2)]
&& \text{Substitute } x = -2
\\
6(-2) - 3 (-10 + 2) =& 4 [1 - (-2)]
&& \text{Multiply inside parentheses first}
\\
6(-2) - 3 (-8) =& 4(3)
&& \text{Work inside parentheses first}
\\
-12 + 24 =& 12
&& \text{Multiply}
\\
12 =& 12
&& \text{True}

\end{aligned}
\end{equation}
$