$8 \quad$ Given that $x$ and $y$ are real numbers satisfying the following equations:
$$
x+x y+y=2+3 \sqrt{2} \text { and } x^{2}+y^{2}=6 \text {, }
$$
find the value of $|x+y+1|$.
(A) $1+\sqrt{3}$
(B) $2-\sqrt{3}$
(C) $2+\sqrt{3}$
(D) $3-\sqrt{2}$
(E) $3+\sqrt{2}$
8 Answer: (E)
$$
\begin{array}{l}
x^{2}+y^{2}=6 \Rightarrow(x+y)^{2}=2 x y+6 \text {. So } 2 x+\left\{(x+y)^{2}-6\right\}+2 y=2(2+3 \sqrt{2}) \\
\Rightarrow(x+y)^{2}+2(x+y)+1=11+6 \sqrt{2} \Rightarrow(x+y+1)^{2}=(3+\sqrt{2})^{2} .
\end{array}
$$