Difference between revisions of "2018 AMC 10A Problems/Problem 12"
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dot((0,1)); | dot((0,1)); | ||
</asy> | </asy> | ||
− | Now, it becomes clear that there are <math>\boxed{3}</math> intersection points. (pinetree1) | + | Now, it becomes clear that there are <math>\boxed{\textbf{(C) } 3}</math> intersection points. (pinetree1) |
===Solution 2=== | ===Solution 2=== | ||
Line 45: | Line 45: | ||
Subcase 2: <math>y<\frac{4}{3}</math> | Subcase 2: <math>y<\frac{4}{3}</math> | ||
<math>3-4y</math> will be positive so <math>3-4y = 1</math> \rightarrow <math>4y = 2</math>. <math>y = \frac{1}{2}</math> and <math>x = \frac{3}{2}</math>. | <math>3-4y</math> will be positive so <math>3-4y = 1</math> \rightarrow <math>4y = 2</math>. <math>y = \frac{1}{2}</math> and <math>x = \frac{3}{2}</math>. | ||
− | Thus, the solutions are: <math>(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)</math>, and the answer is | + | Thus, the solutions are: <math>(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)</math>, and the answer is <math>\boxed{\textbf{(C) } 3}</math>. |
Solution by Danny Li JHS, <math>\text{\LaTeX}</math> edit by pretzel. | Solution by Danny Li JHS, <math>\text{\LaTeX}</math> edit by pretzel. | ||
− | ===Solution 3 | + | ===Solution 3=== |
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Note that ||x| - |y|| can take on either of four values: x + y, x - y, -x + y, -x -y. | Note that ||x| - |y|| can take on either of four values: x + y, x - y, -x + y, -x -y. | ||
Solving the equations (by elimination, either adding the two equations or subtracting), | Solving the equations (by elimination, either adding the two equations or subtracting), | ||
− | we obtain the three solutions: <math>(0, 1)</math>, <math>(-2,3)</math>, <math>(1.5, 0.5)</math> so the answer is <math>C</math>. | + | we obtain the three solutions: <math>(0, 1)</math>, <math>(-2,3)</math>, <math>(1.5, 0.5)</math> so the answer is <math>\boxed{\textbf{(C) } 3}</math>. |
==See Also== | ==See Also== |
Revision as of 21:06, 9 February 2018
How many ordered pairs of real numbers satisfy the following system of equations?
Solutions
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 . Solution by Danny Li JHS, edit by pretzel.
Solution 3
Note that ||x| - |y|| can take on either of four values: x + y, x - y, -x + y, -x -y. Solving the equations (by elimination, either adding the two equations or subtracting), we obtain the three solutions: , , so the answer is .
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.