2018 AMC 10A Problem 12

The second equation says $|x|-|y|=1$ or $|x|-|y|=-1$, so it is enough to check the four linear possibilities $x=y\pm1$ and $x=-y\pm1$. Combining these with $x+3y=3$ gives $(x,y)=\left(\dfrac32,\dfrac12\right)$, $(0,1)$, $(0,1)$ again, and $(-3,2)$. These are three distinct ordered pairs, and each satisfies the original absolute-value equation. Thus, C is the correct answer.