Giải thích các bước giải:
$\dfrac{2x-4y}{3}=\dfrac{4z-3x}{2}=\dfrac{3y-2z}{4}=\dfrac{2x-4y+3y-2z}{3+4}=\dfrac{2x-y+z-3z}{7}=\dfrac{27+z}{7}$
$\to\begin{cases}\dfrac{4z-3x}{2}=\dfrac{27+z}{7}\\ \dfrac{3y-2z}{4}=\dfrac{27+z}{7}\end{cases}$
$\to\begin{cases}x=\dfrac{-54+26z}{21}\\ y=\dfrac{6(z+6)}{7}\end{cases}$
$\to 2\dfrac{-54+26z}{21}-\dfrac{6(z+6)}{7}+z=27$
$\to z=\dfrac{783}{55}$
$\to x,y$
