Ta có: \(4x=3y\) hay \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\left(1\right)\)
\(4y=3z\) hay \(\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{12}=\dfrac{z}{16}\left(2\right)\)
Từ (1) và (2), suy ra:
\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\) \(\Rightarrow\dfrac{2x}{18}=\dfrac{y}{12}=\dfrac{z}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{18}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{2x+y-z}{18+12-16}=\dfrac{-14}{14}=-1\)
Do đó:
\(\dfrac{x}{9}=-1\Rightarrow x=9.\left(-1\right)=-9\)
\(\dfrac{y}{12}=-1\Rightarrow y=12.\left(-1\right)=-12\)
\(\dfrac{z}{16}=-1\Rightarrow z=16.\left(-1\right)=-16\)
Vậy x = -9 ; y = -12 ; z = -16
