$\frac{x}{4} = \frac{y}{x} = \frac{7}{y}$ 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 $$\cfrac{2.57^2}{4} \neq \sqrt{7 (2.57)}$$

Hint $ $ Multiplyimg $\,\overbrace{ \dfrac{\color{#c00}x}{4} = \dfrac{\color{#0a0}y}{\color{#c00}x} = \dfrac{7}{\color{#0a0}y}}^{\Large a}\,\Rightarrow\,a^{\large 3} = \dfrac{7}4$
hence  $\, x = 4a,\ \ y = xa = 4a^2,\ \ 7 = ya = 4a^3\  \ \checkmark$
Remark $ $ This telescoping product view works generally
$\,\overbrace{ \dfrac{\color{#c00}{x_1}}{x_0} = \dfrac{\color{#0a0}{x_2}}{\color{#c00}{x_1}} = \dfrac{x_3}{\color{#0a0}{x_2}}= \,\cdots\, = \dfrac{x_n}{x_{n-1}} }^{\Large a}\ \Rightarrow\ a^{\large n} = \dfrac{x_n}{x_0}$
so $\, x_1 = a x_0,\ x_2 = a x_1 = a^2 x_0,\,\ldots,\, x_n = a^n x_0$
See here for many more examples of multiplicative telescopy.
