# 1984 AHSME Problems/Problem 20

## Problem

The number of the distinct solutions to the equation $|x-|2x+1||=3$ is $\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$

## Solution

We can create a tree of possibilities, progressively eliminating the absolute value signs by creating different cases. $[asy] unitsize(2cm); draw((0,0)--(2,-2)); draw((0,0)--(-2,-2)); label("|x-|2x+1||=3",(0,0),N); label("x-|2x+1|=3",(-2,-2),S); label("x-|2x+1|=-3",(2,-2),S); label("|2x+1|=x-3",(-2,-2.25),S); label("|2x+1|=x+3",(2,-2.25),S); draw((2,-2.5)--(3,-3.5)); draw((2,-2.5)--(1,-3.5)); draw((-2,-2.5)--(-3,-3.5)); draw((-2,-2.5)--(-1,-3.5)); label("2x+1=x-3",(-3,-3.5),S); label("x=-2",(-3,-3.75),S); label("2x+1=-x+3",(-1,-3.5),S); label("x=\frac{2}{3}",(-1,-3.75),S); label("2x+1=x+3",(1,-3.5),S); label("x=2",(1,-3.75),S); label("2x+1=-x-3",(3,-3.5),S); label("x=-\frac{4}{3}",(3,-3.75),S); [/asy]$

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, $\boxed{\text{C}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 