2010/11 AHSMC Part 1 Problem 4

Note that $x$ must be odd, and $x = 9 - \frac{2y}{3}$. Since $y \ge 0$, $x \le 9$. Thus there are $5$ pairs $(x, y) = (9, 0)$, $(7,3)$, $(5,6)$, $(3,9)$ and $(1,12)$. The answer is (B).