Difference between revisions of "2018 AMC 10A Problems/Problem 12"
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Thus, the solutions are: <math>(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)</math>, and the answer is <math>3,</math> or <math>\boxed{\textbf{(C)}}</math> | Thus, the solutions are: <math>(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)</math>, and the answer is <math>3,</math> or <math>\boxed{\textbf{(C)}}</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 (do not use in the real test)== | ||
+ | List all of the cases out. | ||
+ | |||
+ | <math>x+3y=3</math> | ||
+ | |||
+ | $||x|-|y||=1 | ||
==See Also== | ==See Also== |
Revision as of 15:22, 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.
$||x|-|y||=1
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.