Difference between revisions of "2018 AMC 10A Problems/Problem 12"
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+ | ==Solution 4 (Similar to 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: <math>(0, 1)</math>, <math>(-2,3)</math>, <math>(1.5, 0.5)</math> so the answer is <math>C</math>. | ||
==See Also== | ==See Also== |
Revision as of 18:27, 9 February 2018
Contents
[hide]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
Solution 4 (Similar to 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.