1984 AHSME Problems/Problem 20
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Problem
The number of the distinct solutions to the equation
is
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]](http://latex.artofproblemsolving.com/c/f/a/cfaa7a2656b5dd04faf01404307808955a55054d.png)
So we have 
possible solutions: 
and 
. Checking for extraneous solutions, we find that the only ones that work are 
and , so there are solutions, 
.
See Also
| 1984 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
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