6.229. $\left\{\begin{array}{l}x+\sqrt{y}-56=0 \\ \sqrt{x}+y-56=0\end{array}\right.$
Solution.

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

Subtracting the second equation of the system from the first, we get

$$
\begin{aligned}
& x+\sqrt{y}-\sqrt{x}-y=0, \quad(x-y)-(\sqrt{x}-\sqrt{y})=0 \\
& (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})-(\sqrt{x}-\sqrt{y})=0 \Leftrightarrow(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}-1)=0
\end{aligned}
$$

from which $\sqrt{x}-\sqrt{y}=0$ or $\sqrt{x}+\sqrt{y}-1=0$.

Then $\sqrt{y}=\sqrt{x}$ or $\sqrt{x}=1-\sqrt{y}$.

Substituting the value $\sqrt{y}=\sqrt{x}$ into the first equation of the system, we get $x+\sqrt{x}-56=0$. Solving this equation as a quadratic in terms of $\sqrt{x}$, we get $\sqrt{x}=-8, \sqrt{x}=7 ; \sqrt{x}=-8$ is not suitable. Then $x_{1}=49 ; y_{1}=49$.

Substituting the value $\sqrt{x}=1-\sqrt{y}$ into the second equation of the system, we get $y-\sqrt{y}-56=0$. Solving this equation as a quadratic in terms of $\sqrt{y}$, we get $\sqrt{y_{2}}=8$ or $y_{2}=64, \sqrt{y_{3}}=-7$ is not suitable. Then $\sqrt{x_{2}}=1-8=-7 ; \sqrt{x_{2}}=-7$ is not suitable.

Answer: $(49 ; 49)$.