# How do you solve |x - 7| >10?

$x > 17 \mathmr{and} x < - 3$ In interval notation: $\left(- \infty , - 3\right) \cup \left(17 , \infty\right)$
$| x - 7 | > 10 \therefore$ $x - 7 > 10 \mathmr{and} x - 7 < - 10 \therefore x > 17 \mathmr{and} x < - 3$
In interval notation the solution is $\left(- \infty , - 3\right) \cup \left(17 , \infty\right)$[Ans]