Problem 5. Solve the system of equations

$$
\left\{\begin{array}{l}
26 x^{2}+42 x y+17 y^{2}=10 \\
10 x^{2}+18 x y+8 y^{2}=6
\end{array}\right.
$$
Answer: $(-1,2),(-11,14),(11,-14),(1,-2)$.

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

$$
\begin{cases}(6 x+5 y)^{2} & =16 \\ (4 x+3 y)^{2} & =4\end{cases}
$$

From which we get 4 possible cases

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

Solving each of these systems, we find 4 answers: $(-1,2),(-11,14),(11,-14),(1,-2)$.