Question

solve the following system of equation

Answer 1

$\frac{2}{x} + \frac{3}{y} = \frac{9}{xy} \\ \\ => \frac{2y + 3x}{xy} = \frac{9}{xy} \\ \\ => 2y + 3x = 9 \: \: \: \: \: \: \: - - - - - 1equation\\ \\ \\ and \: \: \: \frac{4}{x} + \frac{9}{y} = \frac{21}{xy} \\ \\ => \frac{4y + 9x}{xy} = \frac{21}{xy} \\ \\ => 4y + 9x = 21 \: \: \: \: \: - - - - 2equation$

Hence,

2y + 3x = 9 ----: ( 1 ) equation
4y + 9x = 21 ----: ( 2 ) equation

Multiply ( 1 ) equation by 4 and ( 2 ) equation by ( 2 )

8y + 12x = 36 ---: ( 3 ) equation
8y + 18x = 42 ---: ( 4 ) equation

Subtract ( 4 ) equation from ( 3 ) equation

8y + 12x = 36
8y + 18x = 42
-_(-)____(-)__
- 6x = - 6
__________

x = 1

Putting the value of x in 1equation,

2y + 3x = 9

=> 2y + 3( 1 ) = 9

=> 2y + 3 = 9

=> 2y = 6

=> y = 3

Hence, x = 1 and y = 3