793. Solve the equation in integers

$$
2 x^{2}-2 x y+9 x+y=2
$$
$\triangle$ Express from the equation the unknown that enters it only in the first degree - in this case $y:$

$$
2 x^{2}+9 x-2=2 x y-y, \quad y=\frac{2 x^{2}+9 x-2}{2 x-1}
$$

For the last fraction, let's separate the integer part - either using the rule of polynomial division by "corner" or by breaking down the numerator of the fraction into addends, all of which, except for the free term, are divisible by $2 x-1$. We get:

$$
y=x+5+\frac{3}{2 x-1}
$$

Therefore, the difference $2 x-1$ can only take the values $-3, -1, 1,$ and $3$. It remains to check these four cases.

Answer: $(1 ; 9),(2 ; 8),(0 ; 2),(-1 ; 3)$.