2A. For the positive real numbers $x$ and $y$, the equality

$$
x^{2}+y^{2}=6 x y
$$

is satisfied. Calculate the value of the expression $\frac{x+y}{x-y}$.
Solution. If we add $2xy$ to both sides of the equation $x^{2}+y^{2}=6xy$, we get

$$
x^{2}+2xy+y^{2}=8xy \text{, i.e., }(x+y)^{2}=8xy
$$

If we add $-2xy$ to both sides of the same equation, we get

$$
x^{2}-2xy+y^{2}=4xy, \text{ i.e., }(x-y)^{2}=4xy
$$

According to this,

$$
\left(\frac{x+y}{x-y}\right)^{2}=\frac{(x+y)^{2}}{(x-y)^{2}}=\frac{8xy}{4xy}=2
$$

Finally, $\frac{x+y}{x-y}= \pm \sqrt{2}$.