Solusi

Persamaan (1)
$x-2\sqrt{xy}+y=4xy$
$(\sqrt{x}-\sqrt{y})^2=4xy$

Persamaan (2)
$x+2\sqrt{xy}+y=8xy$
$(\sqrt{x}+\sqrt{y})^2=8xy$

Substitusi (1) ke (2):
$(\sqrt{x}+\sqrt{y})^2=2(\sqrt{x}-\sqrt{y})^2$

Ada dua kemungkinan:

Kemungkinan 1: $\sqrt{x} \ge \sqrt{y}$

$\sqrt{x}+\sqrt{y}=\sqrt{2}.(\sqrt{x}-\sqrt{y})$

$(\sqrt{2}+1).\sqrt{y}=(\sqrt{2}-1).\sqrt{x}$

$\large{\frac{\sqrt{x}}{\sqrt{y}}}=\large{\frac{\sqrt{2}+1}{\sqrt{2}-1}}$

$\large{\frac{\sqrt{x}}{\sqrt{y}}}=\large{\frac{\sqrt{2}+1}{\sqrt{2}-1}}.\large{\frac{\sqrt{2}+1}{\sqrt{2}+1}}$

$\large{\frac{\sqrt{x}}{\sqrt{y}}}=(\sqrt{2}+1)^2=3+2\sqrt{2}$

$\large{\frac{x}{y}}=(3+2\sqrt{2})^2$

Kemungkinan 2: $\sqrt{x} \lt \sqrt{y}$

$\sqrt{x}+\sqrt{y}=\sqrt{2}.(\sqrt{y}-\sqrt{x})$

$(\sqrt{2}+1).\sqrt{x}=(\sqrt{2}-1).\sqrt{y}$

$\large{\frac{\sqrt{x}}{\sqrt{y}}}=\large{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$

$\large{\frac{\sqrt{x}}{\sqrt{y}}}=\large{\frac{\sqrt{2}-1}{\sqrt{2}+1}}.\large{\frac{\sqrt{2}-1}{\sqrt{2}-1}}$

$\large{\frac{\sqrt{x}}{\sqrt{y}}}=(\sqrt{2}-1)^2=3-2\sqrt{2}$

$\large{\frac{x}{y}}=(3-2\sqrt{2})^2$