9. Suppose
$$
\frac{1}{x}=\frac{2}{y+z}=\frac{3}{z+x}=\frac{x^{2}-y-z}{x+y+z} .
$$

What is the value of $\frac{z-y}{x}$ ?
(A) 1 ;
(B) 2 ;
(C) 3 ;
(D) 4;
(E) 5 .
9. Ans: (B)
$$
\frac{1}{x}=\frac{2}{y+z} \text { and } \frac{1}{x}=\frac{3}{z+x}
$$
tells us
$$
\frac{y}{x}+\frac{z}{x}=2 \text { and } \frac{z}{x}+1=3 .
$$

Thus $y=0$ and $\frac{z}{x}=2$. Note we didn't use the last equality, but $x=-1 ; y=0$ and $z=-2$ satisfy all conditions.