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. $\fbox{E}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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