Difference between revisions of "2018 AMC 10A Problems/Problem 12"
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− | == Problem == | + | ==Problem== |
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How many ordered pairs of real numbers <math>(x,y)</math> satisfy the following system of equations? | How many ordered pairs of real numbers <math>(x,y)</math> satisfy the following system of equations? | ||
<cmath>x+3y=3</cmath> | <cmath>x+3y=3</cmath> | ||
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\textbf{(E) } 8 </math> | \textbf{(E) } 8 </math> | ||
− | == Solution == | + | ==Solution 1== |
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The graph looks something like this: | The graph looks something like this: | ||
<asy> | <asy> | ||
Line 26: | Line 23: | ||
dot((0,1)); | dot((0,1)); | ||
</asy> | </asy> | ||
+ | Now, it becomes clear that there are <math>\boxed{3}</math> intersection points. (pinetree1) | ||
− | + | ==Solution 2== | |
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− | == Solution 2 == | ||
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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 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. | ||
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<math>\textbf{Case 1:}</math> <math>y>1</math> | <math>\textbf{Case 1:}</math> <math>y>1</math> | ||
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<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> | ||
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<math>|3y-3-y| = |2y-3| = 1</math> | <math>|3y-3-y| = |2y-3| = 1</math> | ||
Subcase 1: <math>y>\frac{3}{2}</math> | Subcase 1: <math>y>\frac{3}{2}</math> | ||
Line 44: | Line 34: | ||
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>) | ||
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<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> | ||
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<math>\textbf{Case 3:}</math> <math>y<1</math> | <math>\textbf{Case 3:}</math> <math>y<1</math> | ||
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<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> | ||
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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>. | ||
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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> | ||
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Solution by Danny Li JHS, <math>\text{\LaTeX}</math> edit by pretzel. | Solution by Danny Li JHS, <math>\text{\LaTeX}</math> edit by pretzel. | ||
− | == See Also == | + | ==See Also== |
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{{AMC10 box|year=2018|ab=A|num-b=11|num-a=13}} | {{AMC10 box|year=2018|ab=A|num-b=11|num-a=13}} | ||
{{AMC12 box|year=2018|ab=A|num-b=9|num-a=11}} | {{AMC12 box|year=2018|ab=A|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:17, 9 February 2018
Contents
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.
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.