# How do you solve abs(x+a)>=a?

Solution: $x \ge 0 \mathmr{and} x \le - 2 a$ , In interval notation: $\left(- \infty , - 2 a\right] \cup \left[0 , \infty\right)$
$| x + a | \ge a \mathmr{and} x + a \ge a \mathmr{and} x \ge a - a \mathmr{and} x \ge 0$ OR
$| x + a | \ge a \mathmr{and} x + a \le - a \mathmr{and} x \le - a - a \mathmr{and} x \le - 2 a$
Solution: $x \ge 0 \mathmr{and} x \le - 2 a$ , In interval notation: $\left(- \infty , - 2 a\right] \cup \left[0 , \infty\right)$ [Ans]