Question

Solve the following pair of equations:

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

• A. $x=-1,y=\frac{\sqrt{3}+\sqrt{2}}{3}$
• B. $x=1,y=\frac{\sqrt{5}-\sqrt{2}}{3}$
• C. $x=1,y=\frac{\sqrt{7}-\sqrt{13}}{3}$
• D. $x=1,y=\frac{\sqrt{10}-\sqrt{5}}{3}$

Answer 1

The correct option is A $x=-1,y=\frac{\sqrt{3}+\sqrt{2}}{3}$

$\sqrt{2}x+3y=\sqrt{3}$ .....(i)
$\sqrt{3}x+3y=\sqrt{2}$ .....(ii)

Subtracting (ii) from (i)

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

$(\sqrt{2}-\sqrt{3})x=\sqrt{3}-\sqrt{2}$

$\Rightarrow x=-1$

On substituting $x=-1$ in (i). we get,

$\sqrt{2}(-1)+3y=\sqrt{3}$

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

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

$\therefore x=-1,y=\frac{\sqrt{3}+\sqrt{2}}{3}$