2021 AMC 10 A Fall

Complete problem set with solutions and individual problem pages

Problem 14 Hard

How many ordered pairs (x, y) of real numbers satisfy the following system of equations? \begin{array}{r} x^{2}+3 y=9 \\ (|x|+|y|-4)^{2}=1 \end{array}

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    5

  • E.

    7

Answer:D

Solution 1:

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 \Longrightarrow|x|+|y|-4=\pm 1 \Longrightarrow|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:

Solution 2:

We can manipulate the first equation to get y=-\frac{x^{2}}{3}+3. From the second equation, we have that |x|+|y|-4=1 or |x|+|y|-4=-1. We will consider each case separately. If |x|+|y|-4=1, then |x|+|y|=5. The graph of this is a square with vertices (5,0), (-5,0),(0,5) and (0,-5). The parabola from the first equation is downwards facing, and its vertex is inside this square; the parabola will clearly intersect the square twice. Therefore, this case gives us \underline{2} solutions. If |x|+|y|-4=-1, then |x|+|y|=3. The graph of this is a square with vertices (3,0),(-3,0),(0,3) and (0,-3). The vertex of the parabola from the first equation is on one of the corners of this square (in particular, (0,3) ). Also, at y=0, the parabola has x intercepts of \pm 3; the square passes through both of those points. If we continue to move down, the square narrows in, while the parabola continues to expand. Therefore, these are our only 3 intersection points in this case: (0,3),(3,0) and (-3,0). This case gives us \underline{3} solutions. Adding these two cases together, we get our final answer of (D) 5 .