1984 AHSME Problems/Problem 20

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

See Also

1984 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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