$x$, $y$, $z$ positive and $\frac{y}{x-z} = \frac{x + y}{z} = \frac{x}{y}$, find numerical value of $\frac{x}{y}$

Question

$x$, $y$, $z$ positive and $\frac{y}{x-z} = \frac{x + y}{z} = \frac{x}{y}$, find numerical value of $\frac{x}{y}$

The solution in the textbook seems straight forward. Because they are equal proportions we have:

$\frac{x}{y} = \frac{(y) + (x + y) + (x)}{(x - z) + (z) + (y)} = \frac{2(x + y)}{x + y} = 2$

When I solved the problem, I removed the $z$ by adding numerator and denominator of the first two:

$\frac{x}{y} = \frac{(y) + (x + y)}{(x - z) + (z)} = \frac{x + 2 y}{x} = 1 + 2 \frac{y}{x}$

So if we rename $w = \frac{x}{y}$ we have:

$w = 1 + 2 w^{-1}$

multiplying by $w$ and solving the quadratic $0 = w^2 - w - 2$, I get the solution $\frac{1 \pm \sqrt{5}}{2}$.

What is wrong with my method?

Edit: computed quadratic solution incorrectly