# 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<y<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)

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 = 1/2 and x = 3/2

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

Solution by Danny Li JHS