Question

Solve: $6x+3y=7xy\mathrm{and}3x+9y=11xy$

Answer 1

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

For equation (i), we have:

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

For equation (ii), we have:

$\frac{3x+9y}{xy}=11$
$\Rightarrow \frac{3x}{xy}+\frac{9y}{xy}=11\Rightarrow \frac{3}{y}+\frac{9}{x}=11$ ....(iv)
On substituting $\frac{1}{y}=v$ and $\frac{1}{x}=u$ in (iii) and (iv), we get:
6v + 3u = 7 ....(v)
3v + 9u = 11 ....(vi)
On multiplying (v) by 3, we get:
18v + 9u = 21 ....(vii)
On subtracting (vi) 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 (iii), 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 $y=\frac{3}{2}$.