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

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

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