I'm looking to solve the following system of equations and find the value of $\sqrt3xz + yw$:
$\begin{cases} 3x^2 + y^2 - 3xy = 3 + 2\sqrt 2 \\ y^2 + z^2 - yz = 9 + 6\sqrt 2 \\ z^2 + w^2 + \sqrt 3zw = 3 + 2\sqrt 2 \\ w^2 + 3x^2 + \sqrt 3wx = 9 + 6\sqrt 2 \end{cases}$
I'd greatly appreciate any assistance or insights from the community. Thank you!
We begun the process of setting matching equations as I suggested in a comment. If we continue, we eliminate the squares and end up with an almost linear eqution. We still have terms to eliminate but we can solve for these terms, substitute into earlier equations, and repeat the process. Can you take it from here?
$\begin{cases} 3x^2 + y^2 - 3xy = 3 + 2\sqrt 2 \\ y^2 + z^2 - yz = 9 + 6\sqrt 2 \\ z^2 + w^2 + \sqrt 3zw = 3 + 2\sqrt 2 \\ w^2 + 3x^2 + \sqrt 3wx = 9 + 6\sqrt 2 \end{cases}$ $\implies$ $\begin{cases} 3x^2 + y^2 - 3xy - 3 - 2\sqrt 2=0 \\ y^2 + z^2 - yz - 9 - 6\sqrt 2 =0 \\ z^2 + w^2 + \sqrt 3zw - 3 - 2\sqrt 2=0 \\ w^2 + 3x^2 + \sqrt 3wx - 9 - 6\sqrt 2=0 \end{cases}$
Match odd equations with even equations \begin{align*} \implies \begin{cases} 3x^2 + y^2 - 3xy - 3 - 2\sqrt 2&= y^2 + z^2 - yz - 9 - 6\sqrt 2\\ z^2 + w^2 + \sqrt 3zw - 3 - 2\sqrt 2&= w^2 + 3x^2 + \sqrt 3wx - 9 - 6\sqrt 2 \end{cases} \end{align*} Eliminate $\,y^2,\, w^2\,$ by subtraction from both sides \begin{align*} \implies \begin{cases} 3x^2 - 3xy &= z^2 - yz - 6 - 8\sqrt 2\\ z^2 + \sqrt 3zw &=3x^2 + \sqrt 3wx - 6 - 4\sqrt 2 \end{cases} \end{align*} \begin{align*} \implies \begin{cases} 3x^2 - 3xy - z^2 + yz + 6 + 8\sqrt 2&=0\\ z^2 + \sqrt 3zw - 3x^2 - \sqrt 3wx + 6 + 4\sqrt 2&=0 \end{cases} \end{align*} Eliminate $\,z^2,\, 3x^2\,$ by adding equations $$ \implies - 3xy+ \sqrt 3zw - \sqrt 3wx + yz + 12 + 12\sqrt 2=0 $$