\(2x=3y\Leftrightarrow\frac{x}{21}=\frac{y}{14}\)
\(7z=5x\Leftrightarrow\frac{x}{21}=\frac{z}{35}\)
\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{35}=\frac{3x-7y+5z}{63-98+175}=\frac{80}{140}=\frac{4}{7}\)
-Vậy: \(\frac{x}{21}=\frac{4}{7}=>x=\frac{21.4}{7}=12\)
\(\frac{y}{14}=\frac{4}{7}=>y=\frac{14.4}{7}=8\)
\(\frac{z}{35}=\frac{4}{7}=>z=\frac{35.4}{7}=20\)