Fraction with Ratio

Applying Addendo, the fraction should still be the same

\[\begin{align} \frac{y}{x-z} = \frac{x + y}{z} = \frac{x}{y} &= \color{Blue}{\frac{(y) + (x + y) + (x)}{(x-z) + (z) + (y)}} \\ \\ \frac{x}{y} &= \color{Blue}{\frac{2(x + y)}{x + y}} \\ \\ \frac{x}{y} &= \color{Blue}{\boxed{2}}

\end{align}\]

$\begin{cases} \dfrac y{x-z} = \dfrac xy \implies y^2 = x^2 - zx & \implies x^2 - y^2 = zx & ...(1) \\ \dfrac {x+y}z = \dfrac xy & \implies xy + y^2 = zx & ...(2) \end{cases}$

$(1)+(2): \quad x^2+xy = 2zx \implies \blue{x+y = 2z}$ for $x \ne 0$ . From $\dfrac xy = \dfrac {x+y}z = \dfrac {2z}z = \boxed 2$ .

Let $\dfrac{y}{x-z}=\dfrac{x+y}{z}=\dfrac{x}{y}=k$ .

Then $x=ky, y=kx-kz, x+y=kz\implies y=k^2y-ky-y\implies k^2-k-2=0\implies k=2$ (since $k$ has to be positive).

Therefore $\dfrac{x}{y}=\boxed 2$ .