2018 AMC 12A Problem 10

Attempt Problem 10 of the 2018 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12A solutions, or check the answer key.

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10.

How many ordered pairs of real numbers

(x, y)

satisfy the following system of equations?

x + 3y = 3

\big|\,|x| - |y|\,\big| = 1

1

2

3

4

8

The equation

\big|\,|x| - |y|\,\big| = 1

means

|x| - |y| = \pm 1,

which unfolds into

x = \pm y \pm 1

Solve

x + 3y = 3

together with each of

x = y + 1,

x = y - 1,

x = -y + 1,

x = -y - 1,

then discard repeats

The second equation gives

|x| - |y| = \pm 1,

equivalently

x = \pm y \pm 1.

Substituting into

x + 3y = 3:

x = y + 1,

then

(x, y) = \left(\tfrac32, \tfrac12\right).

x = y - 1,

then

(x, y) = (0, 1).

x = -y + 1,

then again

(x, y) = (0, 1).

x = -y - 1,

then

(x, y) = (-3, 2).

The distinct solutions are

(-3, 2),

(0, 1),

and

\left(\tfrac32, \tfrac12\right),

all of which check, so there are

3.

Thus, the correct answer is C.

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