$\dfrac{x}{y}=\dfrac{y}{x-z}=\dfrac{x+y}{z}\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} \textbackslash{} \dfrac{x}{y}=\dfrac{x+y+(x+y)}{y+\left( x-z\right)+z}=\dfrac{2x+2y}{x+y}=\dfrac{2\left( x+y\right)}{x+y}=2.$ 
If~$\dfrac{y}{x-z}=\dfrac{x+y}{z}=\dfrac{x}{y}$~for three positive numbers x, y and z, all different, then what is the value of~$\dfrac{x}{y}$. 
Options:
a) $$3$$ 
b) $$5$$ 
c) $$1$$ 
d) $$2$$ 
e) $$6$$ 