2018 AMC 10A Problems/Problem 12

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The following problem is from both the 2018 AMC 12A #10 and 2018 AMC 10A #12, so both problems redirect to this page.

Problem

How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations?

$$x+3y=3$$

$$\big||x|-|y|\big|=1$$

$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$

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.

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


$2y-3$ 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 $\boxed{\textbf{(C) } 3}$.

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

Solution 6(Desperate)

Since the absolute value is the square root of the square, we get that the first equation is quartic(degree $4$) and the other is linear. Subtract to get $\boxed{\textbf{(C) } 3}$.

~integralarefun

~savannahsolver