1. Reordenar términos:
Observamos que $\sqrt{2x+4y} = \sqrt{2(x+2y)} = \sqrt{2}\sqrt{x+2y}$ y $\sqrt{2x+2y} = \sqrt{2}\sqrt{x+y}$.
Sea $u = \sqrt{x+y}$ y $v = \sqrt{x+2y}$.

El sistema es:
$$ \begin{cases} u + \sqrt{2}v = \sqrt{2} + 4 & (1) \\ v - \sqrt{2}u = 2\sqrt{2} - 2 & (2) \end{cases} $$

2. Resolver el sistema para $u$ y $v$:
Multiplicamos (1) por $\sqrt{2}$:
$$\sqrt{2}u + 2v = 2 + 4\sqrt{2}$$
Sumamos con (2):
$$(\sqrt{2}u + 2v) + (v - \sqrt{2}u) = (2 + 4\sqrt{2}) + (2\sqrt{2} - 2)$$
$$3v = 6\sqrt{2} \implies v = 2\sqrt{2}$$

Sustituimos $v$ en (1):
$$u + \sqrt{2}(2\sqrt{2}) = \sqrt{2} + 4 \implies u + 4 = \sqrt{2} + 4 \implies u = \sqrt{2}$$

3. Volver a las variables originales:
$x + y = u^2 = (\sqrt{2})^2 = 2$
$x + 2y = v^2 = (2\sqrt{2})^2 = 8$

Restamos las ecuaciones:
$$(x + 2y) - (x + y) = 8 - 2$$
$$y = 6$$

Respuesta: (y = 6)