9. Given that $x, y, z$ are integers from 1 to 9, and $x>y>z$, find the integer solutions $(x, y, z)$ for the equation $3(x-y)=4(y-z)$.
9. Answer: $(x, y, z)=(8,4,1),(9,5,2)$

Hint $\because(3,4)=1, \quad \therefore x-y(>0)$ is a multiple of 4. Similarly, $y-z(>0)$ is a multiple of 3.
$$
\begin{array}{l}
\text { Also } \because 9 \geqslant x>y>z \geqslant 1, \quad \therefore x-y=4, \\
y-z=6,3 .
\end{array}
$$