5. Find all pairs of positive numbers $(x, y)$ that satisfy the system of equations $\left\{\begin{array}{l}2 x-\sqrt{x y}-4 \sqrt{\frac{x}{y}}+2=0 \\ 2 x^{2}+x^{2} y^{4}=18 y^{2} .\end{array}\right.$
Answer: $(2 ; 2),\left(\frac{\sqrt[4]{286}}{4} ; \sqrt[4]{286}\right)$.

Solution. Let $\sqrt{\frac{x}{y}}=u, \sqrt{x y}=v \quad$ (with $u>0, \quad v>0$ ). Then $u v=\sqrt{\frac{x}{y}} \cdot \sqrt{x y}=\sqrt{x^{2}}=|x|=x$, $\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{x}{y}}=\sqrt{y^{2}}=|y|=y$, since $x$ and $y$ are positive by condition. The system takes the form

$$
\left\{\begin{array} { l } 
{ 2 u v - v - 4 u + 2 = 0 , } \\
{ 2 u ^ { 2 } v ^ { 2 } + \frac { v ^ { 6 } } { u ^ { 2 } } = \frac { 1 8 v ^ { 2 } } { u ^ { 2 } } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(v-2)(2 u-1)=0 \\
v^{4}+2 u^{4}=18
\end{array}\right.\right.
$$

From the first equation, it follows that $v=2$ or $u=\frac{1}{2}$.

If $v=2$, then $16+2 u^{4}=18$, from which $u=1$; then $y=\frac{v}{u}=2, x=u v=2$.

If $u=\frac{1}{2}$, then $v^{4}+\frac{1}{8}=18$, from which $v=\frac{\sqrt[4]{286}}{2}$; then $y=\frac{v}{u}=\sqrt[4]{286}, x=u v=\frac{\sqrt[4]{286}}{4}$.