# How do you solve abs(x+2)-5>=8?

Solution: $x \le - 15 \mathmr{and} x \ge 11$. In interval notation: $\left(- \infty , - 15\right] \cup \left[11 , \infty\right)$
$| x + 2 | - 5 \ge 8 \mathmr{and} | x + 2 | \ge 8 + 5 \mathmr{and} x + 2 \ge 13 \mathmr{and} x \ge 11$ OR
$| x + 2 | - 5 \ge 8 \mathmr{and} | x + 2 | \ge 8 + 5 \mathmr{and} x + 2 \le - 13 \mathmr{and} x \le - 15$
Solution: $x \le - 15 \mathmr{and} x \ge 11$. In interval notation: $\left(- \infty , - 15\right] \cup \left[11 , \infty\right)$[Ans]