17. If $|x|+x+y=10, x+|y|-y=12$, then $x+y=$
(A) -2 .
(B) $2 \cdot$ (C) $\frac{18}{5}$
(D) $\frac{22}{3}$
(E) 22 .
17
17. Given the equations $|x|+x+y=10$,(1) $x+|y|-y=12$. (2)

If $x \leqslant 0$, then from (1) we get $y=10$, and from (2) we get $x=12$, which is a contradiction. Therefore, $x>0$. In this case, equation (1) becomes
$$
2 x+y=10 .
$$

If $y \geqslant 0$, from (2) we get $x=12$, and from (3) we get $y=-14$, which is a contradiction. Therefore, $y<0$. In this case, equation (2) becomes
$$
x-2 y=12 .
$$

From (3) and (4), we solve to get $x=\frac{32}{5}, y=-\frac{14}{5}$. Thus, $x+y=\frac{18}{5}$. Therefore, the correct answer is (C).