1. Solve the equation in integers

$$
2 y^{2}-2 x y+x+9 y-2=0
$$
Answer: $(9 ; 1),(2 ; 0),(8 ; 2),(3 ;-1)$.

Solution: Express $x$ from this equation:

$$
x=\frac{2 y^{2}+9 y-2}{2 y-1}=y+5+\frac{3}{2 y-1} .
$$

Therefore, the number $2 y-1$ is a divisor of the number 3:

$$
\left[\begin{array} { c } 
{ 2 y - 1 = 1 } \\
{ 2 y - 1 = - 1 , } \\
{ 2 y - 1 = 3 } \\
{ 2 y - 1 = - 3 }
\end{array} \Rightarrow \left[\begin{array} { c } 
{ y = 1 } \\
{ y = 0 , } \\
{ y = 2 , } \\
{ y = - 1 }
\end{array} \Rightarrow \left[\begin{array}{l}
x=9 \\
x=2 \\
x=8 \\
x=3
\end{array}\right.\right.\right.
$$