Ta có: \(\hept{\begin{cases}3x=4y;2y=5z\\2x-3y+z=8\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{4}=\frac{y}{3};\frac{y}{5}=\frac{z}{2}\\2x-3y+z=8\end{cases}}}\) \(\Rightarrow\frac{x}{20}=\frac{y}{15}=\frac{z}{6}\Rightarrow\frac{2x-3y+z}{40-45+6}=\frac{8}{1}=8\)
Vậy : \(x=8.20=160;y=8.15=120;z=8.6=48\)