Difference between revisions of "2018 AMC 10A Problems/Problem 12"
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+ | {{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #10]] and [[2018 AMC 10A Problems|2018 AMC 10A #12]]}} | ||
+ | |||
+ | == Problem == | ||
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> | ||
+ | |||
<cmath>\big||x|-|y|\big|=1</cmath> | <cmath>\big||x|-|y|\big|=1</cmath> | ||
− | |||
− | |||
− | |||
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− | + | <math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8 </math> | |
− | + | ||
+ | == 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. | 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. | ||
Line 25: | Line 26: | ||
dot((0,1)); | dot((0,1)); | ||
</asy> | </asy> | ||
− | Now, it becomes clear that there are <math>\boxed{\textbf{(C) } 3}</math> intersection points. | + | Now, it becomes clear that there are <math>\boxed{\textbf{(C) } 3}</math> intersection points. |
− | + | == Solution 2 == | |
<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. | ||
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<math>3-4y</math> will be positive so <math>3-4y = 1</math> <math>\rightarrow</math> <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> <math>\rightarrow</math> <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 <math>\boxed{\textbf{(C) } 3}</math>. | 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 3 == | |
Note that <math>||x| - |y||</math> can take on either of four values: <math>x + y</math>, <math>x - y</math>, <math>-x + y</math>, <math>-x -y</math>. | Note that <math>||x| - |y||</math> can take on either of four values: <math>x + y</math>, <math>x - y</math>, <math>-x + y</math>, <math>-x -y</math>. | ||
Solving the equations (by elimination, either adding the two equations or subtracting), | Solving the equations (by elimination, either adding the two equations or subtracting), | ||
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~minor edit, XxHalo711 | ~minor edit, XxHalo711 | ||
− | + | == Solution 4 == | |
Just as in solution <math>2</math>, we derive the equation <math>||3-3y|-|y||=1</math>. If we remove the absolute values, the equation collapses into four different possible values. <math>3-2y</math>, <math>3-4y</math>, <math>2y-3</math>, and <math>4y-3</math>, each equal to either <math>1</math> or <math>-1</math>. Remember that if <math>P-Q=a</math>, then <math>Q-P=-a</math>. Because we have already taken <math>1</math> and <math>-1</math> into account, we can eliminate one of the conjugates of each pair, namely <math>3-2y</math> and <math>2y-3</math>, and <math>3-4y</math> and <math>4y-3</math>. Find the values of <math>y</math> when <math>3-2y=1</math>, <math>3-2y=-1</math>, <math>3-4y=1</math> and <math>3-4y=-1</math>. We see that <math>3-2y=1</math> and <math>3-4y=-1</math> give us the same value for <math>y</math>, so the answer is <math>\boxed{\textbf{(C) } 3}</math> | Just as in solution <math>2</math>, we derive the equation <math>||3-3y|-|y||=1</math>. If we remove the absolute values, the equation collapses into four different possible values. <math>3-2y</math>, <math>3-4y</math>, <math>2y-3</math>, and <math>4y-3</math>, each equal to either <math>1</math> or <math>-1</math>. Remember that if <math>P-Q=a</math>, then <math>Q-P=-a</math>. Because we have already taken <math>1</math> and <math>-1</math> into account, we can eliminate one of the conjugates of each pair, namely <math>3-2y</math> and <math>2y-3</math>, and <math>3-4y</math> and <math>4y-3</math>. Find the values of <math>y</math> when <math>3-2y=1</math>, <math>3-2y=-1</math>, <math>3-4y=1</math> and <math>3-4y=-1</math>. We see that <math>3-2y=1</math> and <math>3-4y=-1</math> give us the same value for <math>y</math>, so the answer is <math>\boxed{\textbf{(C) } 3}</math> | ||
~Zeric Hang | ~Zeric Hang | ||
+ | |||
+ | == Solution 5 == | ||
+ | Just as in solution <math>2</math>, we derive the equation <math>x=3-3y</math>. Squaring both sides in the second equation gives <math>x^2+y^2-2|xy|=1</math>. Putting <math>x=3-3y</math> and doing a little calculation gives <math>10y^2-18y+9-2|3y-3y^2|=1</math>. From here we know that <math>3y-3y^2</math> is either positive or negative. | ||
+ | |||
+ | When positive, we get <math>2y^2-3y+1=0</math> and then, <math>y=1/2</math> or <math>y=1</math>. | ||
+ | When negative, we get <math>y^2-3y+2=0</math> and then, <math>y=2</math> or <math>y=1</math>. Clearly, there are <math>3</math> different pairs of values and that gives us <math>\boxed{\textbf{(C) } 3}</math> | ||
+ | |||
+ | ~OlutosinNGA | ||
+ | |||
+ | == Solution 6(Desperate) == | ||
+ | Since the absolute value is the square root of the square, we get that the first equation is quartic(degree <math>4</math>) and the other is linear. Subtract to get <math>\boxed{\textbf{(C) } 3}</math>. | ||
+ | |||
+ | ~integralarefun | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/o63EtwelFp0 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 01:56, 16 February 2021
- The following problem is from both the 2018 AMC 12A #10 and 2018 AMC 10A #12, so both problems redirect to this page.
Contents
Problem
How many ordered pairs of real numbers satisfy the following system of equations?
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.
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 .
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 . One of those equations overlap into so there's only 3 solutions.
~trumpeter, ccx09 ~minor edit, XxHalo711
Solution 4
Just as in solution , we derive the equation . If we remove the absolute values, the equation collapses into four different possible values. , , , and , each equal to either or . Remember that if , then . Because we have already taken and into account, we can eliminate one of the conjugates of each pair, namely and , and and . Find the values of when , , and . We see that and give us the same value for , so the answer is
~Zeric Hang
Solution 5
Just as in solution , we derive the equation . Squaring both sides in the second equation gives . Putting and doing a little calculation gives . From here we know that is either positive or negative.
When positive, we get and then, or . When negative, we get and then, or . Clearly, there are different pairs of values and that gives us
~OlutosinNGA
Solution 6(Desperate)
Since the absolute value is the square root of the square, we get that the first equation is quartic(degree ) and the other is linear. Subtract to get .
~integralarefun
Video Solution
~savannahsolver
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.