Easy or not?

Here is my solution. Let's mark each equation as (1) and (2). Since $x\neq0$ so:

$(1)\Leftrightarrow 2(\frac{y}{x})^2-1=\frac{1}{x^2}$ and $(2)\Leftrightarrow 2-(\frac{y}{x})^3=2.\frac{y}{x}.\frac{1}{x^2}-\frac{1}{x^2}$ .

Subtitute like this:

$a=\frac{y}{x}$ and $b=\frac{1}{x^2}$ .

Do that and we have a new system:

$2a^2-1=b$ $(3)$ and $2-a^3=2ab-b$ $(4)$

From $(3)$ and $(4)$ $\Rightarrow a=1$ which means:

$\frac{y}{x}=1\Leftrightarrow y=x$ .

Combine it with $(1)$ and we have the following result:

$x=y=1$ or $x=y=-1$ .

Solution ends.

let y=kx and substitute into the to equations. divide the second by the first equation. There we go

that's pretty easy!! as you can place each of the options in given equation! :P