If $\dfrac{2x-3y}{3z+y}=\dfrac{z-y}{z-x}=\dfrac{x+3z}{2y-3x}$, prove that each of these ratios is equal to $\dfrac{x}{y}$; hence show that either $x=y$, or $z=x+y$
$\Rightarrow \dfrac{2x-3y}{3z+y}=\dfrac{z-y}{z-x}=\dfrac{x+3z}{2y-3x}=k$
$$2x-3y=(3z+y)k\tag{1}$$ $$z-y=(z-x)k\tag{2}$$ $$x+3z=(2y-3x)k\tag{3}$$
Then $3 \cdot (2)-(3)-(1)$ to get $\dfrac{x}{y}=k$, proving all the ratios are equal to it. My question is how do I show $x=y$ or $z=x+y$? I could for example just plug in $z=x+y$ into $(2)$ to get $\dfrac{x}{y}=k$, but I'm not sure I'm meant to just literally plug it back in. I think I'm suppose to prove it somehow, but I don't know how. Thanks for the help.