B3. Determine all pairs of positive integers $x$ and $y$ that are solutions to the equation:

$$
\frac{2}{x}+\frac{3}{y}=1
$$
B3. Both fractions must be less than 1, so it must hold that $x \geq 3, y \geq 4$. First, take $x=3$, then it must hold that $3 / y=1 / 3$ so $y=9$. For $y>9$ there are no more solutions, because the sum of the two fractions then becomes less than 1. Next, take $y=4$, then it must hold that $2 / x=1 / 4$ and thus $x=8$. For $x>8$ there are also no more solutions because then the sum of the two fractions again becomes less than 1. By trying values for $x$ between 3 and 8, you find two more solutions.

All solutions $(x, y)$: $(3,9),(4,6),(5,5)$ and $(8,4)$.