The number of the distinct solutions to the equation
$|x-|2x+1||=3$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$
We can create a tree of possibilities, progressively eliminating the absolute value signs by creating different cases.

So we have $4$ possible solutions: $-2, \frac{2}{3}, 2,$ and $-\frac{4}{3}$. Checking for extraneous solutions, we find that the only ones that work are $2$ and $-\frac{4}{3}$, so there are $2$ solutions, $\boxed{\text{C}}$.