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We have $\frac{x+y}{z}=\frac{x}{y}⟹xy+y^2=xz$ and $\frac{y}{x-z}=\frac{x}{y}⟹y^2=x^2-xz⟹x^2-y^2=xz$. Equating the two expressions for $xz$ gives $xy+y^2=x^2-y^2⟹x^2-xy-2y^2=0⟹(x+y)(x-2y)=0$, so as $x+y$ cannot be $0$ for positive $x$ and $y$, we must have $x-2y=0⟹x=2y⟹\frac{x}{y}=2$. $\boxed{E}$ # Solution 2 We cross multiply the first and third fractions and the second and third fractions, respectively, for $$x(x-z)=y^2$$ $$y(x+y)=xz$$ Notice how the first equation can be expanded and rearranged to contain an $(x+y)$ term. $$x^2-xz=y^2$$ $$x^2-y^2=xz$$ $$(x+y)(x-y)=xz$$ We can divide this by the second equation to get $$\frac{(x+y)(x-y)}{y(x+y)}=\frac{xz}{xz}$$ $$\frac{x-y}{y}=1$$ $$\frac{x}{y}-1=1$$ $$\frac{x}{y}=2\to \boxed{2}$$