Question

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

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

• A. $x=0$ and $y=1$
• B. $x=1$ and $y=0$
• C. $x=1$ and $y=1$
• D. None of these

Answer 1

The correct option is D None of these

Given Equations are
$\sqrt{2}x+\sqrt{3}y=0$ .....(1)
$\sqrt{3}x-\sqrt{8}y=0$ .......(2)
From (1), we get
$y=\frac{\sqrt{2}x}{\sqrt{3}}$ ......(3)
Substituting $y=\frac{\sqrt{2}x}{\sqrt{3}}$ in (2), we get
$\sqrt{3}x-\sqrt{8}\left(\frac{\sqrt{2}x}{\sqrt{3}}\right)=0$
$\Rightarrow \sqrt{3}x+\frac{4x}{\sqrt{3}}=0$
$\Rightarrow 3x+4x=0$
$\Rightarrow 7x=0$
$\Rightarrow x=0$
Substituting $x=0$ in (3), we get

$y=\frac{-\sqrt{2}\times 0}{\sqrt{3}}=0$
Hence, the solution is $x=0$ and $y=0$.