Question

Solve:
$\frac{3}{x}+\frac{6}{y}=7;\frac{9}{x}+\frac{3}{y}=11(x\ne 0,y\ne 0)$

Answer 1

$\frac{3}{x}+\frac{6}{y}=7$ ....$\left(1\right)$

$\frac{9}{x}+\frac{3}{y}=11$ ....$\left(2\right)$

Let $u=\frac{1}{x},v=\frac{1}{y}$ the $\left(1\right)$ and $\left(2\right)$ becomes,

$3u+6v=7$ .....$\left(3\right)$

$9u+3v=11$ .....$\left(4\right)$

$3\times$eqn$\left(4\right)-9\times$eqn$\left(3\right)$ gives

$27u+9v-27u-54v=33-63$

$\Rightarrow -45v=-30$

$\Rightarrow v=\frac{30}{45}=\frac{2}{3}$

Substitute the value of $v=\frac{2}{3}$ in $\left(3\right)$ we get

$3u+6v=7\Rightarrow 3u=7-6\times \frac{2}{3}=7-4=3$

$\therefore u=\frac{3}{3}=1$

$\therefore u=\frac{1}{x}=1\Rightarrow x=1$

$\therefore v=\frac{1}{y}=\frac{2}{3}\Rightarrow y=\frac{3}{2}$

Hence $x=1,y=\frac{3}{2}$