# 2021 Fall AMC 10A Problems/Problem 14

## Problem

How many ordered pairs $(x,y)$ of real numbers satisfy the following system of equations? \begin{align*} x^2+3y&=9 \\ (|x|+|y|-4)^2 &= 1 \end{align*} $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7$

## Solution 1 (Graphing)

The second equation is $(|x|+|y| - 4)^2 = 1$. We know that the graph of $|x| + |y|$ is a very simple diamond shape, so let's see if we can reduce this equation to that form: $$(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5\}.$$ We now have two separate graphs for this equation and one graph for the first equation, so let's put it on the coordinate plane: $[asy] Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 1.0,0.5)); yaxis(-8,8,Ticks(f, 1.0,0.5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return -x-3; } draw(graph(f,0,-3)); real f(real x) { return 5-x; } draw(graph(f,0,5)); real f(real x) { return 5+x; } draw(graph(f,0,-5)); real f(real x) { return x-5; } draw(graph(f,0,5)); real f(real x) { return -x-5; } draw(graph(f,0,-5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return (-x^2)/3+3; } draw(graph(f,-5,5)); [/asy]$ We see from the graph that there are $5$ intersections, so the answer is $\boxed{\textbf{(D) } 5}$.

~KingRavi

## Video Solution

~Education, the Study of Everything

~savannahsolver