Question

How many ordered pairs of real numbers $(x, y)$ satisfy the following system of equations?

$x + 3y = 3$ $\big|\,|x| - |y|\,\big| = 1$

$1$

$2$

$3$

$4$

$8$

Answer: C

Concepts:absolute value system of equations casework

Difficulty rating: 1560

Solution:

The second equation gives $|x| - |y| = \pm 1,$ equivalently $x = \pm y \pm 1.$ Substituting into $x + 3y = 3:$

If $x = y + 1,$ then $(x, y) = \left(\tfrac32, \tfrac12\right).$ If $x = y - 1,$ then $(x, y) = (0, 1).$ If $x = -y + 1,$ then again $(x, y) = (0, 1).$ If $x = -y - 1,$ 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.

2018 AMC 12A Problem 10

Answer: C

Difficulty rating: 1560

Solution:

Problem 10 in Other Years