Difference between revisions of "2018 AMC 10A Problems/Problem 12"
(→Solution) |
|||
Line 28: | Line 28: | ||
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) | ||
+ | |||
+ | 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 == | == See Also == |
Revision as of 18:57, 8 February 2018
Problem
How many ordered pairs of real numbers satisfy the following system of equations?
Solution
The graph looks something like this:
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.