Question

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

Answer 1

$\sqrt{2}\cdot x+\sqrt{3}y=0\Rightarrow \sqrt{2}x=-\sqrt{3}y$

$\sqrt{3x}-\sqrt{8y}=0\Rightarrow \sqrt{3x}=\sqrt{8y}$

$\Rightarrow 2x^2=3y^2$

$3x=8y\Rightarrow x=\frac{8}{3}y$

$=\Rightarrow 2\cdot {\left(\frac{8}{3}\right)}^2{y}^2=3y^2$

$2.64y^2=27y^2$

$(128-27)y^2=0\Rightarrow y=0\Rightarrow x=0$