Giải thích các bước giải:
Ta có :
$\dfrac{2x-y}{5}=\dfrac{3y-2z}{15}=\dfrac{(2x-y)-(3y-2z)}{5-15}=\dfrac{2x-y-3y+2z}{-10}$
$\to\dfrac{2x-y}{5}=\dfrac{3y-2z}{15}=\dfrac{2(x+z)-4y}{-10}$
$\to\dfrac{2x-y}{5}=\dfrac{3y-2z}{15}=\dfrac{2.2y-4y}{-10},x+z=2y$
$\to\dfrac{2x-y}{5}=\dfrac{3y-2z}{15}=0$
$\to 2x-y=0\to y=2x$
$3y-2z=0\to z=\dfrac{3}{2}y=3x$