Question

Solve the following equations:
$x^2+y^2-3=3xy$,
$2x^2-6+y^2=0$.

Answer 1

$x^2+y^2-3=3xy$

$x^2+y^2-3xy=3$

$2x^2-6+y^2=02x^2+y^2=6$ .......(i)

Put $y=vx$

$x^2+v^2{x}^2-3v{x}^2=3$ ........(ii)

$2x^2+v^2{x}^2=6$ .........(iii)

Dividing (ii) by (iii), we have

$\frac{x^2+v^2{x}^2-3v{x}^2}{2x^2+v^2{x}^2}=\frac{3}{6}$

$\Rightarrow \frac{1+v^2-3v}{2+v^2}=\frac{1}{2}$

$\Rightarrow 2+2v^2-6v=2+v^2$

$\Rightarrow v^2-6v=0$

$\Rightarrow v(v-6)=0$

$\Rightarrow v=0,6\Rightarrow y=0,6x$

Substituting $y$ in (i), we have
$2x^2+y^2=6$

(i) Put $y=0$

Therefore, $2x^2=6$

$\Rightarrow x=\pm \sqrt{3}$

Thus $y=0$
(ii) Put $y=6x$

Therefore, $2x^2+(6x{)}^2=6$

$\Rightarrow 38x^2=6$

$\Rightarrow x^2=\frac{3}{19}$

$\Rightarrow x=\pm \sqrt{\frac{3}{19}}$

Thus $y=6(\pm \sqrt{\frac{3}{19}})=\pm 6\sqrt{\frac{3}{19}}$