College Algebra, Chapter 1, 1.1, Section 1.1, Problem 88

Solve the equation $\displaystyle a-2 [ b - 3 (c -x)] = 6$ for $x$

$
\begin{equation}
\begin{aligned}
a - 2 [ b - 3 (c - x) ] &= 6 && \text{Apply Distributive Property}\\
\\
a - 2 ( b- 3c - 3x ) &= 6 && \text{Apply Distributive Property}\\
\\
a - 2b + 6c + 6x &= 6 && \text{Add both sides by } (-a + 2b - 6c)\\
\\
a - 2b + 6c + 6x - a + 2b - 6c &= 6 - a + 2b - 6c && \text{Simplify}\\
\\
6x &= 6 - a + 2b-6c && \text{Divide both sides by 6}\\
\\
\frac{\cancel{6}x}{\cancel{6}} &= \frac{6-a+2b-6c}{6} && \text{Simplify}\\
\\
x &= \frac{6-a+2b-6c}{6}\\
\\
& \text{or}\\
\\
x &= 1 - \frac{a}{6} + \frac{b}{3} - c
\end{aligned}
\end{equation}
$