3. For the numbers $x$ and $y, 0<x<y$, the equation $x^{2}+4 y^{2}=5 x y$ holds. Calculate the value of the expression $\frac{x+2 y}{x-2 y}$.
Solution. From the equalities $(x+2 y)^{2}=x^{2}+4 y^{2}+4 x y=9 x y \quad$ and $(x-2 y)^{2}=x^{2}+4 y^{2}-4 x y=x y$ we obtain that $\left(\frac{x+2 y}{x-2 y}\right)^{2}=9$, i.e., $\frac{x+2 y}{x-2 y}= \pm 3$. From $00$, and from $x<2 x<2 y$ it follows that $x-2 y<0$. Therefore, $\frac{x+2 y}{x-2 y}<0$, and hence $\frac{x+2 y}{x-2 y}=-3$.