6.233. $\left\{\begin{array}{l}\sqrt{2 x+y+1}-\sqrt{x+y}=1, \\ 2 x+2 y=4\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}2 x+y+1 \geq 0 \\ x+y \geq 0\end{array}\right.$

Squaring the first equation, we have

$$
\begin{aligned}
& 2 x+y+1-2 \sqrt{(2 x+y+1)(x+y)}+x+y=1 \Leftrightarrow \\
& \Leftrightarrow 3 x+2 y-2 \sqrt{(2 x+y+1)(x+y)}=0
\end{aligned}
$$

Since $3 x+2 y=4$, from the second equation of the system we get

$$
4-2 \sqrt{(2 x+y+1)(x+y)}=0 \Leftrightarrow \sqrt{(2 x+y+1)(x+y)}=2
$$

After squaring, we have $(2 x+y+1)(x+y)=4$.

$$
\begin{aligned}
& y=\frac{4-3 x}{2} \Rightarrow\left(2 x+\frac{4-3 x}{2}+1\right)\left(x+\frac{4-3 x}{2}\right)=4 \Leftrightarrow x^{2}+2 x-8=0, x_{1}=-4, \\
& x_{2}=2 . \text { Then } y_{1}=8, y_{2}=-1
\end{aligned}
$$

By verification, we find that $x_{1}$ and $y_{1}$ are extraneous solutions. Answer: $(2, -1)$.