The easiest solution is to break this into linear equations, based on the absolute values. There are three intervals we must consider: $x \le -\frac{1}{2}$ In this interval, the equation becomes $-(2x+1) - (x-1)=2-x$, or $-3x=2-x$. Thus, $\boxed{x=-1}$ is our solution here. $-\frac{1}{2} \le x \le 1$ In this interval, the equation becomes $(2x+1) - (x-1) = x+2 = 2-x$. Thus $\boxed{x=0}$ is another solution. $1 \le x \le 2$ ($x \le 2$ because the LHS is positive) $(2x+1) + (x-1) = 3x = 2-x$. This solves to $x=\frac{1}{2}$, but that's not in the interval, and thus is not a solution Our final answer is $\boxed{x=-1, 0}$