If $\frac{a}{b}=\frac{c}{d}=\frac{e}{z}=\frac{2}{3}$, then what is the value of the expression $\frac{2a-6c-3e}{2b-6d-3z}$, while $bdz(2b-6c-3z)\ne0$
I attempted to solve the question as follows:
$\frac{2a-6c-3e}{2b-6d-3z}=\frac{2a-6c-2z}{3a-6d-3z}=\frac{2(a-2c-z)-2c}{3(a-2d-z)}$
This is as far as I got. My intuition tells me that the expression is equal to $\frac{2}{3}$ (something which I get after trying a few different values of $a,b,c,d,e,z$), but I don't know how to prove it. Could you please explain to me how to solve this question?