Question

Solve the pair equations

i)$\sqrt{2}x+\sqrt{3}y=0,\sqrt{3}y-\sqrt{8}y=0$

ii)$3x-5y+1=0;x-y+1=0$

iii)$0.2x+0.3y=13,0.4x+0.5y=2.3$

by substitution method $(P.S)$

Answer 1

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

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

$\sqrt{3}\left(\frac{-\sqrt{3y}}{\sqrt{2}}\right)-\sqrt{8}y=0$

$\Rightarrow -3y-4y=0$

$y=0,x=0$

(ii) $3x-5y+1=0\Rightarrow x=\frac{5y-1}{3}$

$x-y+1=0$

$\frac{5y-1}{3}-y+1=0\Rightarrow 5y-1-3y+3=0.$

$\Rightarrow 2y=-2\Rightarrow y=-1$

$\therefore y=-1$

$x-y+1=0\Rightarrow x+1+1=0$

$y=-1,x=-2$

(iii) $0.2x+0.3y=1.3,0.4x+0.5y=2.3$

$x=\frac{1.3-0.3y}{02}$

$0.4\left(\frac{1.3-0.3y}{0.2}\right)+0.5y=2.3$

$\Rightarrow 2.6-0.6y+0.5y=2.3$

$-0.1y=-0.3\Rightarrow y=3$

$x=\frac{1.3-0.9}{0.2}=2$

$\therefore x=2,y=3$