Question

Solve the pair of equation $\sqrt{2}x+\sqrt{3}y=0,\sqrt{3}x-\sqrt{8}y=2$ by substitution method.

Answer 1

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

Put value of $x$ in $\sqrt{3}x+\sqrt{8}y=2$

$\Rightarrow \sqrt{3}\times \frac{-\sqrt{3}y}{\sqrt{2}}+\sqrt{8}y=2$

$\Rightarrow \frac{-3y}{\sqrt{2}}+2\sqrt{2}y=2$

$\Rightarrow \frac{-3y+4y}{\sqrt{2}}=2$

$\Rightarrow y=2\sqrt{2}$

Put $y=2\sqrt{2}$ in $x=\frac{-\sqrt{3}y}{\sqrt{2}}$

$\Rightarrow x=\frac{-\sqrt{3}\times 2\sqrt{2}}{\sqrt{2}}=-2\sqrt{3}$

$\therefore x=-2\sqrt{3},y=2\sqrt{2}$