Answer
$x=\dfrac{a-2b+6c-6}{6}$ or $x=\dfrac{a}{6}-\dfrac{b}{3}+c-1$
Work Step by Step
$a-2[b-3(c-x)]=6;$ 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 $x=\dfrac{a}{6}-\dfrac{b}{3}+c-1$
