Question

Solve
$\frac{2}{x}+\frac{2}{3y}=\frac{1}{6},\frac{3}{x}+\frac{2}{y}=0$

Answer 1

$\frac{2}{x}+\frac{2}{3y}=\frac{1}{6}$ ; $\frac{3}{x}+\frac{2}{y}=0$

$6y+2x=\frac{3xy}{6}$ ; $3y+2x=0$

$2(3y+x)=\frac{xy}{2}$ ; $3y=-2x$

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

$\therefore 2\left(3\right(-\frac{2x}{3})+x)=\frac{x(-2x/3)}{2}$

$\Rightarrow -4x+2x=-\frac{x^2}{3}$

$-2x=-\frac{x^2}{3}$

$6x=x^2$

$x^2-6x=0$

$x(x-6)=0$

$\therefore x=0;x=6$

$\therefore y=0;y=-4$