Problem 5. Solve the system of equations

$$
\left\{\begin{array}{l}
5 x^{2}-14 x y+10 y^{2}=17 \\
4 x^{2}-10 x y+6 y^{2}=8
\end{array}\right.
$$
Answer: $(-1,-2),(11,7),(-11,-7),(1,2)$.

Solution. By adding and subtracting the two equations of the system, we obtain that the original system is equivalent to the following:

$$
\left\{\begin{aligned}
(3 x-4 y)^{2} & =25 \\
(x-2 y)^{2} & =9
\end{aligned}\right.
$$

From which we get 4 possible cases

$$
\left\{\begin{array} { r l } 
{ 3 x - 4 y } & { = 5 } \\
{ x - 2 y } & { = 3 }
\end{array} \quad \left\{\begin{array} { r l } 
{ 3 x - 4 y } & { = 5 } \\
{ x - 2 y } & { = - 3 }
\end{array} \quad \left\{\begin{array} { r l } 
{ 3 x - 4 y } & { = - 5 } \\
{ x - 2 y } & { = 3 }
\end{array} \quad \left\{\begin{array}{rl}
3 x-4 y & =-5 \\
x-2 y & =-3
\end{array}\right.\right.\right.\right.
$$

Solving each of these systems, we find 4 answers: $(-1,-2),(11,7),(-11,-7),(1,2)$.