Question

$$\begin{gathered} \frac{2}{x}+\frac{3}{y}=\frac{9}{xy}, \\ \frac{4}{x}+\frac{9}{y}=\frac{21}{xy}\left(x\ne 0,y\ne 0\right). \end{gathered}$$

Answer 1

The given equations are:
$\frac{2}{x}+\frac{3}{y}=\frac{9}{xy}$ ............(i)

$\frac{4}{x}+\frac{9}{y}=\frac{21}{xy}$............(ii)
On multiplying (i) and (ii) by xy, we get:
$\frac{2xy}{x}+\frac{3xy}{y}=9$
⇒ 2y + 3x = 9 .............(iii)
and $\frac{4xy}{x}+\frac{9xy}{y}=21$
⇒ 4y + 9x = 21 ...........(iv)
On multiplying (iii) by 3, we get:
6y + 9x = 27 .............(v)
On subtracting (iv) from (v), we get:
2y = 6 ⇒ y = 3
On substituting y = 3 in (iii), we get:
$2\times 3+3x=9$
⇒ $6+3x=9\Rightarrow 3x=\left(9-6\right)=3$
$\Rightarrow x=\frac{3}{3}=1$
Hence, the required solution is x = 1 and y = 3.