$3 \cdot 11$ If non-zero real numbers $x, y$ satisfy $|x|+y=3$, and $|x| y+x^{3}=0$. Then the integer closest to $x-y$ is
(A) -3 .
(B) -1 .
(C) 2 .
(D) 3 .
(E) 5 .
(45th American High School Mathematics Examination, 1994)
[Solution] If $x>0$, then $|x|=x$, the given equation becomes: $\left\{\begin{array}{l}x+y=3, \\ x y+x^{3}=0 .\end{array}\right.$ This leads to $x^{2}-x+3=0$ which has no real roots, hence it is not valid. If $x<0$, then $|x|=-x$, from the given equation we have $y-x=3$. Therefore, the correct choice is (D).