Answer
$(-\displaystyle \infty,-\frac{75}{14})\cup(\frac{87}{14},\infty)$
Work Step by Step
Rewrite as
$|-2x+\displaystyle \frac{6}{7}|+\frac{3}{7} \gt 12\qquad $... $/\times 7$
$ 7|-2x+\displaystyle \frac{6}{7}|+3 \gt 84\qquad $... $/-3$
$|-14x+6| \gt 81$
... $|a-b|=|b-a|$
$|14x-6| \gt 81$
... $|u| \gt c $ is equivalent to ($ u\lt -c $) or ($ u\gt c $)
$ \begin{array}{lllll}
14x-6\lt-81 & /+6 & ...or... & 14x-6\gt 81 & /+6\\
14x\lt-75 & & & 14x \gt 87 & \\
x\lt-75/14 & & & x \gt 87/14 & \\
x\in(-\infty,-\frac{75}{14}) & & or & x\in(\frac{87}{14},\infty) & \\
& & & &
\end{array} $
Solution set: $(-\displaystyle \infty,-\frac{75}{14})\cup(\frac{87}{14},\infty)$
