1. The number of solutions to the system of equations $\left\{\begin{array}{l}|x|+y=12, \\ x+|y|=6\end{array}\right.$ is ( ) .
(A) 1
(B) 2
(C) 3
(D) 4
-、1.A.
If $x \geqslant 0$, then $\left\{\begin{array}{l}x+y=12, \\ x+|y|=6 \text {. }\end{array}\right.$
Thus, $|y|-y=6$, which is obviously impossible.
If $x<0$, then $\left\{\begin{array}{l}-x+y=12, \\ x+|y|=6 \text {. }\end{array}\right.$ Thus, $|y|+y=18$.
Solving this gives $y=9$, and subsequently, $x=-3$.
Therefore, the original system of equations has only 1 solution: $x=-3, y=9$.