AMC12 2018 A

The graph of the first equation is a line with $x$-intercept $(3,0)$ and $y$-intercept $(0,1)$. To draw the graph of the second equation, consider the equation quadrant by quadrant. In the first quadrant $x>0$ and $y>0$, and thus the second equation is equivalent to $\lvert x-y\rvert=1$, which in turn is equivalent to $y=x\pm1$. Its graph consists of the rays with endpoints $(0,1)$ and $(1,0)$, as shown. In the second quadrant $x<0$ and $y>0$. The corresponding graph is the reflection of the first quadrant graph across the $y$-axis. The rest of the graph can be sketched by further reflections of the first-quadrant graph across the coordinate axes, resulting in the figure shown. There are 3 intersection points: $(-3,2)$, $(0,1)$, and $\left(\frac{3}{2},\frac{1}{2}\right)$, as shown.