950. Solve the inequality in natural numbers

$$
x y < x + 2 y
$$
$\triangle$ Here, the constraint of the type $x \leqslant y$ on the unknowns $x$ and $y$ (similar to what we applied when solving problem 946) should not be used, since the inequality is not symmetric with respect to $x$ and $y$.

Gather all terms of the inequality on the left side and add 2 to both sides of the resulting inequality:

$$
x y - x - 2 y < 0, \quad x y - x - 2 y + 2 < 2, \quad (y - 1)(x - 2) < 2
$$

Now let's consider several cases.

If $y - 1 = 0$, i.e., $y = 1$, then $x$ is any natural number.

If $x = 1$ or $x = 2$, then $y$ is any natural number.

In addition, the case $y - 1 = 1, x - 2 = 1$, i.e., $x = 3, y = 2$ also works.

Answer: $(x; 1) (x \in \mathbb{N}); (1; y) (y \in \mathbb{N}, y \neq 1); (2; y) (y \in \mathbb{N}, y \neq 1); (3; 2)$.