Solve the following system of equations:
$\frac{2}{x}\ +\ \frac{3}{y}\ =\ \frac{9}{xy}$
$\frac{4}{x}\ +\ \frac{9}{y}\ =\ \frac{21}{xy}$

Given:

The given system of equations is:

$\frac{2}{x}\ +\ \frac{3}{y}\ =\ \frac{9}{xy}$

$\frac{4}{x}\ +\ \frac{9}{y}\ =\ \frac{21}{xy}$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations can be written as,

$\frac{2}{x}+\frac{3}{y}=\frac{9}{xy}$

$\frac{2y+3x}{xy}=\frac{9}{xy}$

$3x+2y=9$---(i)

$\frac{4}{x}+\frac{9}{y}=\frac{21}{xy}$

$\frac{4y+9x}{xy}=\frac{21}{xy}$

$9x+4y=21$

$9x=21-4y$

$x=\frac{21-4y}{9}$---(ii)

Substituting $x=\frac{21-4y}{9}$ in equation (i), we get,

$3(\frac{21-4y}{9})+2y=9$

$\frac{21-4y}{3}+2y=9$

Multiplying  both sides of the equation by $3$, we get,

$3(\frac{21-4y}{3})+3(2y)=3(9)$

$21-4y+6y=27$

$2y=27-21$

$2y=6$

$y=\frac{6}{2}$

$y=3$

Using $y=3$ in equation (i), we get,

$3x+2(3)=9$

$3x+6=9$

$3x=9-6$

$3x=3$

$x=\frac{3}{3}$

$x=1$

Therefore, the solution of the given system of equations is $x=1$ and $y=3$.