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

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

Given,

$\sqrt{2}x + \sqrt{3}y = 0$ ………(1)

$\sqrt{3}x - \sqrt{8}y = 0$ ……….(2)

Solving equation (1),

$\Rightarrow \sqrt{2}x + \sqrt{3}y = 0 \\[1em] \Rightarrow \sqrt{2}x = -\sqrt{3}y \\[1em] \Rightarrow x = -\dfrac{\sqrt{3}y}{\sqrt{2}} \text{ ……….(3)}$

Substituting above value of x in equation (2), we get :

$\Rightarrow \sqrt{3} \times -\dfrac{\sqrt{3}y}{\sqrt{2}} - \sqrt{8}y = 0 \\[1em] \Rightarrow -\dfrac{3y}{\sqrt{2}} - \sqrt{8}y = 0 \\[1em] \Rightarrow \dfrac{-3y - \sqrt{2}\sqrt{8}y}{\sqrt{2}} = 0 \\[1em] \Rightarrow \dfrac{-3y - \sqrt{16}y}{\sqrt{2}} = 0 \\[1em] \Rightarrow -3y - \sqrt{16}y = 0 \\[1em] \Rightarrow -3y - 4y = 0 \\[1em] \Rightarrow -7y = 0 \\[1em] \Rightarrow y = 0.$

Substituting value of y in equation (3), we get :

$\Rightarrow x = -\dfrac{\sqrt{3}}{\sqrt{2}} \times 0 \\[1em] \Rightarrow x = 0.$

Hence, x = 0 and y = 0.