Difference between revisions of "2018 AMC 10A Problems/Problem 12"
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Now it's clear that there are <math>\boxed{3}</math> intersection points. (pinetree1) | Now it's clear that there are <math>\boxed{3}</math> intersection points. (pinetree1) | ||
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| + | x+3y=3 can be rewritten to x=3-3y. Substituting 3-3y for x in the second equation will give ||3-3y|-y|=1. Splitting this question into casework for the ranges of y will give us the total number of solutions. | ||
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| + | Case 1: y>1 | ||
| + | |||
| + | 3-3y will be negative so |3-3y| = 3y-3. | ||
| + | |3y-3-y| = |2y-3| = 1 | ||
| + | Subcase 1: y>3/2 | ||
| + | 2y-3 is positive so 2y-3 = 1 and y = 2 and x = 3-3(2) = -3 | ||
| + | Subcase 2: 1<y<3/2 | ||
| + | 2y-3 is negative so |2y-3| = 3-2y = 1. 2y = 2 and so there are no solutions (y can't equal to 1) | ||
| + | |||
| + | |||
| + | Case 2: y = 1 | ||
| + | x = 0 | ||
| + | |||
| + | |||
| + | Case 3: y<1 | ||
| + | |||
| + | 3-3y will be positive so |3-3y-y| = |3-4y| = 1 | ||
| + | Subcase 1: y>4/3 | ||
| + | 3-4y will be negative so 4y-3 = 1 --> 4y = 4. There are no solutions (again, y can't equal to 1) | ||
| + | Subcase 2: y<4/3 | ||
| + | 3-4y will be positive so 3-4y = 1 --> 4y = 2. y = 1/2 and x = 3/2 | ||
| + | |||
| + | NOTE: Please fix this up using latex I have no idea how | ||
| + | |||
| + | Solution by Danny Li JHS | ||
== See Also == | == See Also == | ||
Revision as of 17:57, 8 February 2018
Problem
How many ordered pairs of real numbers 
satisfy the following system of equations?
![\[x+3y=3\]](http://latex.artofproblemsolving.com/2/b/6/2b68f21d4456251f723b5b9780794891f520a3c0.png)
![\[\big||x|-|y|\big|=1\]](http://latex.artofproblemsolving.com/d/9/1/d91f752f8219148671d39d50f1e9d6a726c3ce35.png)
Solution
The graph looks something like this:
![[asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3,0)); dot((0,1)); [/asy]](http://latex.artofproblemsolving.com/0/f/9/0f9bc2d077559bb70dba8f9ded81e499a12acb2f.png)
Now it's clear that there are 
intersection points. (pinetree1)
x+3y=3 can be rewritten to x=3-3y. Substituting 3-3y for x in the second equation will give ||3-3y|-y|=1. Splitting this question into casework for the ranges of y will give us the total number of solutions.
Case 1: y>1
3-3y will be negative so |3-3y| = 3y-3. |3y-3-y| = |2y-3| = 1
Subcase 1: y>3/2
2y-3 is positive so 2y-3 = 1 and y = 2 and x = 3-3(2) = -3
Subcase 2: 1<y<3/2
2y-3 is negative so |2y-3| = 3-2y = 1. 2y = 2 and so there are no solutions (y can't equal to 1)
Case 2: y = 1
x = 0
Case 3: y<1
3-3y will be positive so |3-3y-y| = |3-4y| = 1
Subcase 1: y>4/3
3-4y will be negative so 4y-3 = 1 --> 4y = 4. There are no solutions (again, y can't equal to 1)
Subcase 2: y<4/3
3-4y will be positive so 3-4y = 1 --> 4y = 2. y = 1/2 and x = 3/2
NOTE: Please fix this up using latex I have no idea how
Solution by Danny Li JHS
See Also
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
| 2018 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
