$\frac{x}{4} = \frac{y}{x} = \frac{7}{y}$ Could someone please explain to me what's wrong here?
$\cfrac{x}{4} = \cfrac{y}{x} = \cfrac{7}{y}$
So $$\cfrac{x^2}{4} = y \ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \ y  = \sqrt{7x}$$
Hence $$\cfrac{x^2}{4} =\sqrt{7x}$$
With a solution of $x = 2.57$
But $$\cfrac{2.57^2}{4} \neq \sqrt{7 (2.57)}$$

$x\neq 2.57$. You want to solve for $x^2/4 = \sqrt{7x}$, which corresponds to solving the polynomial equation $x^4/16 - 7x = 0$, which has a real and nontrivial zero at $\sqrt[3]{112}$. Hence, $x = \sqrt[3]{112}$ and $y = (112)^{2/3} / 4$. You can check that this in fact, is a solution to your system.
