Question

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

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

Solve the following pair of linear equations by the substitution method.

Answer 1

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

Substituting value of ${}^{\prime }{x}^{\prime }$ in equation $\left(i\right)$, we get
$\sqrt{2}\left(\frac{\sqrt{8}y}{\sqrt{3}}\right)+\sqrt{3}y=0$

$\Rightarrow \frac{\sqrt{16}y}{\sqrt{3}}+\sqrt{3}y=0$

$\Rightarrow \frac{4y+3y}{\sqrt{3}}=0$

$\Rightarrow 7y=0$
$\Rightarrow y=0$

Substituting value of y in equation (ii), we get
$x=\frac{\sqrt{8}\left(0\right)}{\sqrt{3}}=0$
So $x=0,y=0$