6.231. $\left\{\begin{array}{l}\sqrt{x+y}+\sqrt{2 x+y+2}=7, \\ 3 x+2 y=23 .\end{array}\right.$
Solution.

Domain of definition: $\left\{\begin{array}{l}x+y \geq 0, \\ 2 x+y+2 \geq 0 .\end{array}\right.$

Squaring the first equation of the system, we get

$$
\begin{aligned}
& x+y+2 \sqrt{(x+y)(2 x+y+2)}+2 x+y+2=49 \Leftrightarrow \\
& \Leftrightarrow 3 x+2 y+2 \sqrt{(x+y)(2 x+y+2)}=47 \Leftrightarrow \\
& \Leftrightarrow 23+2 \sqrt{(x+y)(2 x+y+2)}=47, \sqrt{(x+y)(2 x+y+2)}=12
\end{aligned}
$$

Squaring, we have $(x+y)(2 x+y+2)=144$. The system of equations takes the form $\left\{\begin{array}{l}(x+y)(2 x+y+2)=144, \\ 3 x+2 y=23\end{array} \Rightarrow y=\frac{23-3 x}{2}\right.$ and $\left(x+\frac{23-3 x}{2}\right)\left(2 x+\frac{23-3 x}{2}+2\right)=144 \Leftrightarrow x^{2}+4 x-45=0$, from which $x_{1}=-9, x_{2}=5$. Then $y_{1}=25, y_{2}=4$.

Answer: $(-9 ; 25),(5 ; 4)$.