ta có:\(5x-3y=4y\Rightarrow5x=7y\Leftrightarrow x=\dfrac{7y}{5}\)(1)
mà \(4y=3z+10x\Rightarrow4y=3z+14y\)
\(\Leftrightarrow-10y=3z\Leftrightarrow z=\dfrac{-10y}{3}\) (2)
thay (1), (2) vào 3x+2y+z=989, ta co:
\(\dfrac{21y}{5}+2y-\dfrac{10y}{3}=989\Leftrightarrow\dfrac{43y}{15}=989\)
\(\Leftrightarrow y=345\)
thay y=345 vào (1), (2) ta dc: \(\left\{{}\begin{matrix}x=\dfrac{7\times345}{5}=483\\z=\dfrac{-10\times345}{3}=-1150\end{matrix}\right.\)
vậy \(\left\{{}\begin{matrix}x=483\\y=345\\z=-1150\end{matrix}\right.\)
Đúng 0
Bình luận (3)
