Theo bài ra ta cs
\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\left(1\right)\)
\(4y=5z\Rightarrow\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{15}=\frac{z}{12}\left(2\right)\)
Từ (1) ; (2) => \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)
ADTC dãy tỉ số bằng nhau ta cs
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{-x-y+z}{-10-15+12}=-\frac{52}{-13}=4\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=4\\\frac{y}{15}=4\\\frac{z}{12}=4\end{cases}\Rightarrow\hept{\begin{cases}x=40\\y=60\\z=48\end{cases}}}\)
