How do you solve |5x - 8| > 12?

$x < - \frac{4}{5} \mathmr{and} x > 4$ In interval notation x is expressed as $\left(- \infty , \frac{4}{5}\right) \bigcup \left(4 , \infty\right)$
$| 5 x - 8 | > 12 \therefore 5 x - 8 > 12 \mathmr{and} 5 x - 8 < - 12 \therefore 5 x > 20 \therefore x > 4 \mathmr{and} 5 x < - 4 \therefore x < - \frac{4}{5}$ In interval notation x is expressed as $\left(- \infty , \frac{4}{5}\right) \bigcup \left(4 , \infty\right)$[Ans]