Difference between revisions of "2018 AMC 10A Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
− | <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. | + | <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. |
+ | |||
<math>\textbf{Case 1:}</math> <math>y>1</math> | <math>\textbf{Case 1:}</math> <math>y>1</math> | ||
<math>3-3y</math> will be negative so <math>|3-3y| = 3y-3.</math> | <math>3-3y</math> will be negative so <math>|3-3y| = 3y-3.</math> | ||
Line 34: | Line 35: | ||
Subcase 2: <math>1<y<\frac{3}{2}</math> | Subcase 2: <math>1<y<\frac{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>) | <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>) | ||
+ | |||
<math>\textbf{Case 2:}</math> <math>y = 1</math> | <math>\textbf{Case 2:}</math> <math>y = 1</math> | ||
It is fairly clear that <math>x = 0.</math> | It is fairly clear that <math>x = 0.</math> | ||
+ | |||
<math>\textbf{Case 3:}</math> <math>y<1</math> | <math>\textbf{Case 3:}</math> <math>y<1</math> | ||
<math>3-3y</math> will be positive so <math>|3-3y-y| = |3-4y| = 1</math> | <math>3-3y</math> will be positive so <math>|3-3y-y| = |3-4y| = 1</math> |
Revision as of 16:08, 9 February 2018
Problem
How many ordered pairs of real numbers satisfy the following system of equations?
Solution 1
The graph looks something like this: Now, it becomes 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.
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 )
It is fairly clear that
will be positive so
Subcase 1:
will be negative so \rightarrow . There are no solutions (again, can't equal to )
Subcase 2:
will be positive so \rightarrow . and . Thus, the solutions are: , and the answer is or Solution by Danny Li JHS, edit by pretzel.
Solution 3 (do not use in the real test)
List all of the cases out.
, ,
We can see that there is 3 solutions, so the answer is
-Baolan
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.