Difference between revisions of "2018 AMC 10A Problems/Problem 12"
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<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 <math>y</math> 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 <math>y</math> 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> | ||
<math>|3y-3-y| = |2y-3| = 1</math> | <math>|3y-3-y| = |2y-3| = 1</math> | ||
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<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> | ||
Subcase 1: <math>y>\frac{4}{3}</math> | Subcase 1: <math>y>\frac{4}{3}</math> |
Revision as of 20:03, 5 May 2018
How many ordered pairs of real numbers satisfy the following system of equations?
Solutions
Solution 1
We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
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 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 . There are no solutions (again, can't equal to )
Subcase 2:
will be positive so . and . Thus, the solutions are: , and the answer is . edit by pretzel, very minor edits by Bryanli, very very minor edit by ssb02
Solution 3
Note that can take on either of four values: , , , . Solving the equations (by elimination, either adding the two equations or subtracting), we obtain the three solutions: , , so the answer is
~trumpeter, ccx09
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 |
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