What are all values of $x$ in $\mathbb{R}$ that satisfy $4 < |x+2| + |x-1| < 5$? I am having some problems getting started with this problem, as I never had to deal with an inequality that was between two values with absolute values. Any help is appreciated. The problem is find all values of $x$ in $\mathbb{R}$ that satisfy $4 < |x+2| + |x-1| < 5$. I keep trying to find cases with $x < -2$ or $x \geq 1$, but that is not getting me anywhere.

If $4 < |x+2| + |x-1| < 5$ then $\ \pm (x+2) \pm (x-1) < 5$. Note that for an opposite choice of sign the inequality is never satisfied since $(x+2) - (x-1)=3$. So the inequality implies 
$$ 4 < (x+2) + (x-1) < 5$$
or 
$$4 < -(x+2) -(x-1) < 5$$
The inequality $ 4 < (x+2) + (x-1) < 5$ is equivalent to $4 < 2x+1<5 \iff 1.5 < x < 2$ and on this interval both $x+2$, $x-1$ are $>0$ so we have part of the solution.
The inequality $ 4 < -(x+2) -(x-1) < 5$ is equivalent to $4 < -2x-1<5 \iff -3 < x < -2.5$ and on this interval both $x+2$, $x-1$ are $<0$ so we have another part of the solution. 
Therefore, the set of solutions is $(-3,-2.5) \cup (1.5,2)$
