Question

The solution of $\sqrt{2}x+\sqrt{3}y=0$ and $\sqrt{3}x+\sqrt{8}y=0$ is

• A. $(1,2)$
• B. $(\sqrt{2},\sqrt{8})$
• C. $(0,0)$
• D. $(\sqrt{3},\sqrt{2})$

Answer 1

The correct option is C $(0,0)$

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

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

subtracting both

$\left(\sqrt{6}x+3y\right)-\left(\sqrt{6}x+4y\right)=0-0$

$-y=0\Rightarrow y=0$

From (1) $\sqrt{2}x+\sqrt{3}y=0$

$\Rightarrow \sqrt{2}x+0=0$

$\Rightarrow x=0$

So, point of intersection is $(0,0)$