Question

$$\begin{gathered} 3x+9y=11xy, \\ 6x+3y=7xy\left(x\ne 0,y\ne 0\right). \end{gathered}$$

Answer 1

The given equations are:
3x + 9y = 11xy .......(i)
6x + 3y = 7xy .........(ii)

From equation (i), we have:
$\frac{3x+9y}{xy}=11$
$\Rightarrow \frac{3x}{xy}+\frac{9y}{xy}=11\Rightarrow \frac{3}{y}+\frac{9}{x}=11$ .........(iii)

From equation (ii), we have:

$\frac{6x+3y}{xy}=7$
$\Rightarrow \frac{6x}{xy}+\frac{3y}{xy}=7\Rightarrow \frac{6}{y}+\frac{3}{x}=7$ ........(iv)

On substituting $\frac{1}{y}=v$ and $\frac{1}{x}=u$ , we get:
3v + 9u = 11 ...........(v)
6v + 3u = 7 ...........(vi)

On multiplying (vi) by 3, we get:
18v + 9u = 21 ..............(vii)

On subtracting (v) from (vii), we get:
15v = 10 ⇒ v = $\frac{10}{15}=\frac{2}{3}$
⇒ $\frac{1}{y}=\frac{2}{3}\Rightarrow y=\frac{3}{2}$
On substituting $y=\frac{3}{2}$ in (iv), we get:
$\frac{6}{\left({\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}+\frac{3}{x}=7$
$\Rightarrow 4+\frac{3}{x}=7\Rightarrow \frac{3}{x}=3\Rightarrow 3x=3$
$\Rightarrow x=1$
Hence, the required solution is x = 1 and $y=\frac{3}{2}$.