9. Let $x \neq y$, and the sequences $x, a_{1}, a_{2}, a_{3}, y$ and $b_{1}, x, b_{2}, b_{3}, y, b_{1}$ are both arithmetic sequences, then $\frac{b_{4}-b_{3}}{a_{2}-a_{1}}=$ $\qquad$ .
$9.2 \frac{2}{3}$.
Brief solution: Let the common differences of the two sequences be $d, d'$, then $y-x=4d=3d'$. Therefore, $\frac{d}{d'}=\frac{3}{4}$. Thus, $\frac{b_{4}-b_{3}}{a_{2}-a_{1}}$ $=\frac{2d'}{d}=2 \times \frac{4}{3}=2 \frac{2}{3}$.