Giải thích các bước giải:
Đặt
$\dfrac{2x-4y}{3}=\dfrac{4z-3x}{2}=\dfrac{3y-2z}{4}=k$
$\to \begin{cases}2x-4y=3k\\ 4z-3x=2k\\ 3y-2z=4k\end{cases}$
$\to x=2y+\dfrac{3k}{2},z=2k+3(2y+\dfrac{3k}{2})=\dfrac{13k}{2}+6y$
$\to 4(\dfrac{13k}{2}+6y)-3(2y+\dfrac{3k}{2})=2$
$\to 18y+\dfrac{43k}{2}=2$
$\to y=\dfrac{1}{9}-\dfrac{43k}{36}$
$\to x=2(\dfrac{1}{9}-\dfrac{43k}{36})+\dfrac{3k}{2}=\dfrac{2-8k}{9}$
$z=\dfrac{13k}{2}+6(\dfrac{1}{9}-\dfrac{43k}{36})=\dfrac{2-2k}{3}$
$\to 2x-y+z=2(\dfrac{2-8k}{9})-(\dfrac{1}{9}-\dfrac{43k}{36})+\dfrac{2-2k}{3}=27$
$\to k=\dfrac{-104}{5}$
$\to x=\dfrac{842}{45}, y=\dfrac{1123}{45}, z=\dfrac{218}{15}$
