Question

$\sqrt{2}x+\sqrt{3}y=0,\sqrt{3}x-\sqrt{8}y=0$,

Answer 1

$\sqrt{2}x+\sqrt{3y}=\mathrm{0......}\left(1\right)$

$\sqrt{3}x+\sqrt{8}y=\mathrm{0.......}\left(2\right)$

$by\left(1\right)$

$x=\frac{\sqrt{3y}}{\sqrt{2}}$

$putin\left(2\right)$

$\sqrt{3}\left(-\frac{\sqrt{3}y}{\sqrt{2}}\right)-\sqrt{8}y=0$

$-3y-\sqrt{16}y=0$

$-7y=0$

$y=0$

$x=0$