6.237. $\left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=3 \\ \sqrt{x+5}+\sqrt{y+3}=5\end{array}\right.$
## Solution.

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

Rewrite the system of equations as $\left\{\begin{array}{l}\sqrt{x}=3-\sqrt{y} \\ \sqrt{x+5}=5-\sqrt{y+3}\end{array} \Rightarrow\right.$

$$
\begin{aligned}
& \Rightarrow\left\{\begin{array} { l } 
{ x = 9 - 6 \sqrt { y } + y , } \\
{ x + 5 = 2 5 - 1 0 \sqrt { y + 3 } + y + 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=9-6 \sqrt{y}+y \\
x=23-10 \sqrt{y+3}+y
\end{array} \Rightarrow\right.\right. \\
& \Rightarrow x=9-6 \sqrt{y}+y, 9-6 \sqrt{y}+y=23-10 \sqrt{y+3}+y \Leftrightarrow 5 \sqrt{y+3}=7+3 \sqrt{y} \Rightarrow \\
& \Rightarrow 25 y+75=49+42 \sqrt{y}+9 y, 8 y+13=21 \sqrt{y} \Rightarrow 64 y^{2}-233 y+169=0,
\end{aligned}
$$

from which $y_{1}=1, y_{2}=\frac{169}{64}$. Then $x_{1}=4, x_{2}=\frac{121}{64}$.

Answer: $(4 ; 1),\left(\frac{121}{64} ; \frac{169}{64}\right)$.