2018 AMC 12A Problem 10

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.