Question

Solve the simultaneous equation:
$\frac{2}{x}+\frac{2}{3y}=\frac{1}{6}$; $\frac{3}{x}+\frac{2}{y}=0$

Answer 1

Given equations

$\frac{2}{x}+\frac{2}{y}\times \frac{1}{3}=\frac{1}{6}.....\left(1\right)$

and $\frac{3}{x}+\frac{2}{y}=0$

$\Rightarrow \frac{2}{y}=-\frac{3}{x}....\left(2\right)$

Putting (2) in (1) , we get

$\Rightarrow \frac{2}{x}+\frac{-3}{x}\times \frac{1}{3}=\frac{1}{6}$

$\Rightarrow \frac{2}{x}-\frac{1}{x}=\frac{1}{6}$

$\Rightarrow \frac{1}{x}=\frac{1}{6}$

$\Rightarrow x=6$

Putting value of x in (2)

$\Rightarrow \frac{2}{y}=-\frac{3}{6}$

$\Rightarrow \frac{2}{y}=-\frac{1}{2}$

$\Rightarrow y=-4$