<s>[INST] solve for $x$ and $y$ in the following equation $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$ Solve for $x$ and $y$ in the following equations: $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$. I made $y^2$ the subject of the formula in eqn 2.
This gives $y^2 = 6 -2x^2$.
I substitute this into the first eqn.
This gives $x^2+3xy -3$. There's where am stuck.
 Can someone pull me out.
 [/INST] If 
$$
  \begin{cases}
    2x^2+y^2=6\\
x^2+y^2=3+3xy \\ 
  \end{cases}
$$
then by setting $x^2=a$, $y^2=b$
$$
  \begin{cases}
     2a+b=6 \\
    a+b=3+3xy \\
  \end{cases}
$$
Hence, by Cramer's rule (of linear algebra)
$$
\begin{cases}
a=x^2=3-3xy & (1)\\
b=y^2=6xy & (2)
\end{cases}
$$
Hence, by (2) $y=0$ or $y=6x$. Plug it into (1) yields 
$$
x^2=3-3x\cdot0\qquad\text{or}\qquad
x^2=3-3x\cdot6x
$$
Hence $(x,y)=(\pm\sqrt{3},0)$ or $(x,y)=(\pm\frac{\sqrt{57}}{19},\pm\frac{6\sqrt{57}}{19})$. 
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