Theo bài ra ta cs
\(3.x=y;5.y=4.z\)và \(6x+7y+8z=456\)
Ta lại cs
\(+,3x=y\Rightarrow\frac{x}{1}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{12}\)(1)
\(+,5y=4z\Rightarrow\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\)(2)
Từ (1) ; (2) \(\Rightarrow\frac{x}{4}=\frac{y}{12}=\frac{z}{15}\)
ADTC dãy tỉ số bằng nhau ta cs
\(\frac{x}{4}=\frac{y}{12}=\frac{z}{15}=\frac{6x+7y+8z}{6.4+7.12+8.15}=\frac{456}{228}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{4}=2\\\frac{y}{12}=2\\\frac{z}{15}=2\end{cases}\Rightarrow\hept{\begin{cases}x=2.4=8\\y=2.12=24\\z=2.15=30\end{cases}}}\)
