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

Homework: $x^{2}+2 y+1 \geq 0$.

Let $\sqrt{x^{2}+2 y+1}=t$, where $t \geq 0$; then $x^{2}+2 y=t^{2}-1 \Leftrightarrow t^{2}-1+t=1$, or $t^{2}+t-2=0$, from which $t_{1}=-2, t_{2}=1 ; t_{1}=-2<0$ is not suitable. Then $\sqrt{x^{2}+2 y+1}=1, x^{2}+2 y+1=1, x^{2}+2 y=0$. We have $\left\{\begin{array}{l}x^{2}+2 y=0, \\ 2 x+y=2\end{array} \Rightarrow\right.$ $y=2-2 x, x^{2}+2(2-2 x)=0 \Leftrightarrow x^{2}-4 x+4=0,(x-2)^{2}=0$, from which $x=2$. Then $y=-2$.

Answer: $(2 ;-2)$.