2021 Fall AMC 10A Problem 14

How many ordered pairs

(x,y)

of real numbers satisfy the following system of equations?

\begin{aligned} x^2+3y&=9, \\ (|x|+|y|-4)^2&=1. \end{aligned}

Show solution

Solution

Put

t=|x|.

The second equation gives

t+|y|=3

5,

while the first gives

y=3-\frac{t^2}{3}.

y\ge0,

then

0\le t\le3.

For

t+y=3,

substitution gives

t-\frac{t^2}{3}=0,

t=0

3.

These yield

x=0,3,-3,

for three points. For

t+y=5,

substitution gives

t^2-3t+6=0,

which has no real root. If

y<0,

then

t>3.

The equation

t-y=3

gives

t^2+3t-18=0,

whose only nonnegative root is the excluded boundary value

t=3.

The equation

t-y=5

gives

t^2+3t-24=0,

with exactly one positive root

t=\frac{-3+\sqrt{105}}2>3.

It yields two points, one for each sign of

x.

The total number of ordered pairs is

3+2=5.

Thus, D is the correct answer.

Tagged: absolute value · system of equations · casework