\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\) và \(x+2y+4z=-93\)
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{x+2y+4z}{3+2.4+4.5}=\dfrac{-93}{31}=-3\)
\(\dfrac{x}{3}=-3\Rightarrow x=-3.3=-9\)
\(\dfrac{y}{4}=-3\Rightarrow y=-3.4=-12\)
\(\dfrac{z}{5}=-3\Rightarrow z=-3.5=-15\)
Ta có : \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)và x+2y+4z=-93
Như vậy ta suy ra được nhờ váo tính chất của dãy tỉ số bằng nhau :
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{4z}{20}=\dfrac{x+2y+4z}{3+8+20}=\dfrac{-93}{31}=-3\)\(\dfrac{x}{3}=-3\Rightarrow x=3\left(-3\right)\Rightarrow x=-9\)
\(\dfrac{2y}{8}=-3\Rightarrow2y=8\left(-3\right)\Rightarrow2y=-24\Rightarrow y=\dfrac{-24}{2}\Rightarrow y=-12\)
\(\dfrac{4z}{20}=-3\Rightarrow4z=20\left(-3\right)\Rightarrow4z=-60\Rightarrow z=\dfrac{-60}{4}\Rightarrow z=-15\)