6.111. $\left\{\begin{array}{l}2(\sqrt{x}+\sqrt{y})=3 \sqrt{x y} \\ x+y=5\end{array}\right.$

The system of equations is:
$\left\{\begin{array}{l}2(\sqrt{x}+\sqrt{y})=3 \sqrt{x y} \\ x+y=5\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \geq 0, \\ y \geq 0 .\end{array}\right.$

Rewrite the system of equations as $\left\{\begin{array}{l}2(\sqrt{x}+\sqrt{y})=3 \sqrt{x y}, \\ (\sqrt{x}+\sqrt{y})^{2}-\sqrt{x y}=5\end{array}\right.$ and introduce the substitution $\left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=u, \\ \sqrt{x y}=v,\end{array}\right.$ where $u \geq 0$ and $v \geq 0$. Then

$$
\left\{\begin{array} { l } 
{ 2 u = 3 v , } \\
{ u ^ { 2 } - 2 v = 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ v = \frac { 2 u } { 3 } , } \\
{ u ^ { 2 } - 2 v = 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ v = \frac { 2 u } { 3 } , } \\
{ u ^ { 2 } - 2 \cdot \frac { 2 u } { 3 } = 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=\frac{2 u}{3} \\
3 u^{2}-4 u-15=0
\end{array}\right.\right.\right.\right.
$$

From the second equation $u_{1}=-\frac{4}{3}, u_{2}=3 ; u_{1}=-\frac{4}{3}$ is not valid. Then $v=2$ and

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-330.jpg?height=142&width=720&top_left_y=93&top_left_x=62)

Using Vieta's theorem, we find: $\left\{\begin{array}{l}x_{1}=4, \\ y_{1}=1 ;\end{array} \quad\left\{\begin{array}{l}x_{2}=1, \\ y_{2}=4 .\end{array}\right.\right.$

Answer: $(4 ; 1)(1 ; 4)$.