## Precalculus: Mathematics for Calculus, 7th Edition

$x=\dfrac{a-2b+6c-6}{6}$ or $x=\dfrac{a}{6}-\dfrac{b}{3}+c-1$
$a-2[b-3(c-x)]=6;$ for $x$ Eliminate the grouping symbols by perfoming the indicated operations: $a-2[b-3c+3x]=6$ $a-2b+6c-6x=6$ Solve for $x$: $a-2b+6c=6+6x$ $6+6x=a-2b+6c$ $6x=a-2b+6c-6$ $x=\dfrac{a-2b+6c-6}{6}$ or $x=\dfrac{a}{6}-\dfrac{b}{3}+c-1$