Question

How do you solve `|x+2|>2`?

Answer 1

x > 0 and x < `-4`. The graduation x on the real line, under this condition, is outside the closed interval `[-4, 0]`.

Explanation:

`|x+2|>2` is equivalent to the pair of inequalities

`x+2 > 2 and -(x+2) > 2`.

So, x > 0 and (rearranging for +x) `x < -4`.