9.2. Non-zero real numbers $x, y$ satisfy the equation $2 x^{2}+2 y^{2}=5 x y$. Find all possible values of the expression $\frac{x+y}{x-y}$.
Answer: 3 and -3.

Solution. Add $4xy$ to both sides of the equality in the condition, we get $2x^2 + 4xy + 2y^2 = 2(x + y)^2 = 9xy$, from which $x + y = \pm \frac{3}{\sqrt{2}} \sqrt{xy}$. Similarly, subtract $4xy$ from both sides of the equality in the condition, we get $2x^2 - 4xy + 2y^2 = 2(x - y)^2 = xy$, from which $x - y = \pm \frac{1}{\sqrt{2}} \sqrt{xy}$. The signs in the two cases do not have to be consistent. Therefore, the ratio $\frac{x + y}{x - y}$ can take one of two values, 3 and -3. Let's check if they are actually achieved. Solve the system $\left\{\begin{array}{l}x + y = \frac{3}{\sqrt{2}} \\ x - y = \frac{1}{\sqrt{2}}\end{array}\right.$, we find $x = \sqrt{2}, y = \frac{1}{\sqrt{2}}$, for them $2x^2 + 2y^2 = 4 + 1 = 5 = 5 \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 5xy$ - the numbers satisfy the condition of the problem. Therefore, the number 3 is the answer to the problem. Similarly, solve the system $\left\{\begin{array}{l}x + y = \frac{3}{\sqrt{2}}, \\ x - y = -\frac{1}{\sqrt{2}} .\end{array}\right.$, we find $x = \frac{1}{\sqrt{2}}, y = \sqrt{2}$, for them $2x^2 + 2y^2 = 1 + 4 = 5 = 5 \cdot \frac{1}{\sqrt{2}} \cdot \sqrt{2} = 5xy$ - the numbers satisfy the condition of the problem. Therefore, the number -3 is also the answer to the problem.

Grading criteria. Answers and examples of $x, y$ when they are achieved: 2 points. Proof that there are no other answers: 5 points.