Tìm x, y, z biết $\frac{6\text{x}-3\text{z}}{5}=\frac{4y-6x}{7}=\frac{3\text{z}-4y}{9}$  và 2x + 3y – 5z = –21.

Áp dụng tính chất dãy tỉ số bằng nhau có:

$\frac{6\text{x}-3\text{z}}{5}=\frac{4y-6x}{7}=\frac{3\text{z}-4y}{9}=\frac{6\text{x}-3\text{z}+4y-6\text{x}+3\text{z}-4y}{5+7+9}=0$

$\iff \left\{\begin{array}{l}\frac{6\text{x}-3z}{5}=0\\ \frac{4y-6x}{7}=0\\ \frac{3\text{z}-4y}{9}=0\end{array}\right.\iff \left\{\begin{array}{l}6\text{x}-3z=0\\ 4y-6x=0\\ 3\text{z}-4y=0\end{array}\right.\iff \left\{\begin{array}{l}6\text{x}=3\text{z}\\ 4y=6\text{x}\\ 3\text{z}=4y\end{array}\right.\iff 6\text{x}=4y=3\text{z}$

$\iff \frac{x}{2}=\frac{y}{3}=\frac{z}{4}\iff \frac{2x}{4}=\frac{3y}{9}=\frac{5z}{20}$

Áp dụng tính chất dãy tỉ số bằng nhau có:

$\frac{2x}{4}=\frac{3y}{9}=\frac{5z}{20}=\frac{2\text{x}+3y-5\text{z}}{4+9-20}=\frac{-21}{-7}=3$

$\iff \left\{\begin{array}{l}\frac{2x}{4}=3\\ \frac{3y}{9}=3\\ \frac{5z}{20}=3\end{array}\right.\iff \left\{\begin{array}{l}2\text{x}=12\\ 3y=27\\ 5\text{z}=60\end{array}\right.\iff \left\{\begin{array}{l}x=6\\ y=9\\ z=12\end{array}\right.$

Vậy x = 6, y = 9, z = 12.