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

## 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

The graph looks something like this: $[asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3,0)); dot((0,1)); [/asy]$

Now it's clear that there are $\boxed{3}$ intersection points. (pinetree1)

## 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.

Case 1: $y>1$

$3-3y$ will be negative so $|3-3y| = 3y-3$

$|3y-3-y| = |2y-3| = 1$

   Subcase 1: $y>3/2$


$2y-3$ 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 $1$)

Case 2: $y = 1 x = 0$

Case 3: $y<1$

$3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$

   Subcase 1: $y>4/3$


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

   Subcase 2: y<4/3


$3-4y$ will be positive so $3-4y = 1$ --> $4y = 2$. $y = \frac{1}{2}$ and $x = \frac{3}{2}$ Solutions: $(-3,2) (0,1) (\frac{3}{2},\frac{1}{2})$ $\boxed{\textbf{(C)} \ 3}$

NOTE: Please fix this up using latex I have no idea how

Solution by Danny Li JHS