$2 \cdot 100$ If $|x|+x+y=10$, and $x+|y|-y=12$, then $x+y$ equals
(A) -2 .
(B) 2 .
(C) $\frac{18}{5}$.
(D) $\frac{22}{3}$.
(E) 22 .
(39th American High School Mathematics Examination, 1988)
[Solution] Divide into cases $x>0, y>0 ; x>0, y<0$ and $x<0, y>0$. In this case, the system of equations becomes
$$
\begin{array}{l}
\quad\left\{\begin{array} { l } 
{ 2 x + y = 1 0 , } \\
{ x - 2 y = 1 2 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
x=\frac{32}{5}, \\
y=-\frac{14}{5} .
\end{array}\right.\right. \\
\therefore \quad x+y=\frac{18}{5} .
\end{array}
$$

Therefore, the answer is $(C)$.