Difference between revisions of "2018 AMC 10A Problems/Problem 12"
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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. | + | <math>x+3y=3</math> can be rewritten to <math>x=3-3y</math>. Substituting <math>3-3y</math> for <math>x</math> in the second equation will give <math>||3-3y|-y|=1</math>. Splitting this question into casework for the ranges of y will give us the total number of solutions. |
− | Case 1: y>1 | + | Case 1: <math>y>1</math> |
− | 3-3y will be negative so |3-3y| = 3y-3 | + | <math>3-3y</math> will be negative so <math>|3-3y| = 3y-3</math> |
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+ | <math>|3y-3-y| = |2y-3| = 1</math> | ||
+ | Subcase 1: <math>y>3/2</math> | ||
+ | <math>2y-3</math> is positive so <math>2y-3 = 1</math> and <math>y = 2</math> and <math>x = 3-3(2) = -3</math> | ||
+ | Subcase 2: <math>1<y<3/2</math> | ||
+ | <math>2y-3</math> is negative so <math>|2y-3| = 3-2y = 1</math>. <math>2y = 2</math> and so there are no solutions (<math>y</math> can't equal to <math>1</math>) | ||
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+ | Case 2: <math>y = 1 | ||
+ | x = 0</math> | ||
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− | 3-3y will be positive so |3-3y-y| = |3-4y| = 1 | + | Case 3: <math>y<1</math> |
− | 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) | + | <math>3-3y</math> will be positive so <math>|3-3y-y| = |3-4y| = 1</math> |
+ | Subcase 1: <math>y>4/3</math> | ||
+ | <math>3-4y</math> will be negative so <math>4y-3 = 1</math> --> <math>4y = 4</math>. There are no solutions (again, <math>y</math> can't equal to <math>1</math>) | ||
Subcase 2: y<4/3 | Subcase 2: y<4/3 | ||
− | 3-4y will be positive so 3-4y = 1 --> 4y = 2. y = 1/ | + | <math>3-4y</math> will be positive so <math>3-4y = 1</math> --> <math>4y = 2</math>. <math>y = \frac{1}{2}</math> and <math>x = \frac{3}{2}</math> |
+ | Solutions: <math>(-3,2) (0,1) (\frac{3}{2},\frac{1}{2})</math> | ||
+ | <math>\boxed{\textbf{(C)} \ 3}</math> | ||
NOTE: Please fix this up using latex I have no idea how | NOTE: Please fix this up using latex I have no idea how |
Revision as of 19:14, 8 February 2018
Contents
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)
Solution 2
can be rewritten to . Substituting for in the second equation will give . Splitting this question into casework for the ranges of y will give us the total number of solutions.
Case 1:
will be negative so
Subcase 1:
is positive so and and
Subcase 2:
is negative so . and so there are no solutions ( can't equal to )
Case 2:
Case 3:
will be positive so
Subcase 1:
will be negative so --> . There are no solutions (again, can't equal to )
Subcase 2: y<4/3
will be positive so --> . and Solutions:
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.