## Precalculus: Mathematics for Calculus, 7th Edition

$$|x+2|>1$$
The set of real numbers cover the regions: $(-\infty, -3) ∪ (-1, +\infty)$. From the graph we can see that it is an open interval; non-strict inequality. To find the actual inequality, involving absolute value, we can do the steps of solving inequality involving absolute value using reverse steps. So we have two inequality: $x-1$ Let's try to give them the form, from where we split normal absolute value inequality. In general, when solving $|x+a|1$ $-(x+2)>1$ $x+2>1$ Which is given by the inequality: $|x+2|>1$