# Difference between revisions of "1990 AHSME Problems/Problem 8"

## Problem

The number of real solutions of the equation $$|x-2|+|x-3|=1$$ is

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) more than } 3$

## Solution

For $2\le x\le 3$, the left-hand side is $(x-2)-(x-3)$ which is $1$ for all $x$ in the interval. $\fbox{E}$