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Difference between revisions of "2018 AMC 10A Problems/Problem 12"

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~Zeric Hang
 
~Zeric Hang
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===Solution 5===
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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.
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When positive, we get <math>2y^2-3y+1=0</math> and then, <math>y=1/2</math> or <math>y=1</math>.
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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>
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~OlutosinNGA
  
 
==See Also==
 
==See Also==

Revision as of 18:07, 23 January 2019

How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \[x+3y=3\] \[\big||x|-|y|\big|=1\] $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$

Solutions

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: [asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); [/asy] Now, it becomes clear that there are $\boxed{\textbf{(C) } 3}$ intersection points. (pinetree1) BOI

Solution 2

$x+3y=3$ can be rewritten to $x=3-3y$. Substituting $3-3y$ for $x$ in the second equation will give $||3-3y|-y|=1$. Splitting this question into casework for the ranges of $y$ will give us the total number of solutions.

$\textbf{Case 1:}$ $y>1$: $3-3y$ will be negative so $|3-3y| = 3y-3.$ $|3y-3-y| = |2y-3| = 1$ 

   Subcase 1: $y>\frac{3}{2}$

$2y-3$ is positive so $2y-3 = 1$ and $y = 2$ and $x = 3-3(2) = -3$ 

   Subcase 2: $1<y<\frac{3}{2}$

$2y-3$ is negative so $|2y-3| = 3-2y = 1$. $2y = 2$ and so there are no solutions ($y$ can't equal to $1$)

$\textbf{Case 2:}$ $y = 1$: It is fairly clear that $x = 0.$

$\textbf{Case 3:}$ $y<1$: $3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$ 

   Subcase 1: $y>\frac{4}{3}$

$3-4y$ will be negative so $4y-3 = 1$ $\rightarrow$ $4y = 4$. There are no solutions (again, $y$ can't equal to $1$) 

   Subcase 2: $y<\frac{4}{3}$

$3-4y$ will be positive so $3-4y = 1$ $\rightarrow$ $4y = 2$. $y = \frac{1}{2}$ and $x = \frac{3}{2}$. Thus, the solutions are: $(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)$, and the answer is $\boxed{\textbf{(C) } 3}$. $\text{\LaTeX}$ edit by pretzel, very minor $\text{\LaTeX}$ edits by Bryanli, very very minor $\text{\LaTeX}$ edit by ssb02

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: $(0, 1)$, $(-3,2)$, $(1.5, 0.5)$ so the answer is $\boxed{\textbf{(C) } 3}$. One of those equations overlap into $(0, 1)$ so there's only 3 solutions.

~trumpeter, ccx09 ~minor edit, XxHalo711

Solution 4

Just as in solution $2$, we derive the equation $||3-3y|-|y||=1$. If we remove the absolute values, the equation collapses into four different possible values. $3-2y$, $3-4y$, $2y-3$, and $4y-3$, each equal to either $1$ or $-1$. Remember that if $P-Q=a$, then $Q-P=-a$. Because we have already taken $1$ and $-1$ into account, we can eliminate one of the conjugates of each pair, namely $3-2y$ and $2y-3$, and $3-4y$ and $4y-3$. Find the values of $y$ when $3-2y=1$, $3-2y=-1$, $3-4y=1$ and $3-4y=-1$. We see that $3-2y=1$ and $3-4y=-1$ give us the same value for $y$, so the answer is $\boxed{\textbf{(C) } 3}$

~Zeric Hang

Solution 5

Just as in solution $2$, we derive the equation $x=3-3y$. Squaring both sides in the second equation gives $x^2+y^2-2|xy|=1$. Putting $x=3-3y$ and doing a little calculation gives $10y^2-18y+9-2|3y-3y^2|=1$. From here we know that $3y-3y^2$ is either positive or negative.

When positive, we get $2y^2-3y+1=0$ and then, $y=1/2$ or $y=1$. When negative, we get $y^2-3y+2=0$ and then, $y=2$ or $y=1$. Clearly, there are $3$ different pairs of values and that gives us $\boxed{\textbf{(C) } 3}$

~OlutosinNGA

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

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