AMC10 2021 10A, Problem 14

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}

Solution: 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 $\fbox{5}$.

Answer: 5