Question

if 8x-9y=6xy and 10x+6y=19xy, the value of xy is? ​

Answer 1

Answer:

$xy = 1$

Step-by-step explanation:

GivenEquations

$8x - 9y = 6xy$

$10x + 6y = 19xy$

or

$\frac{8x - 9y}{xy} = 6$

$\frac{10x + 6y}{xy} = 19$

or

$\frac{8}{y} - \frac{9}{x} = 6........(1)$

$\frac{10}{y} + \frac{6}{x} = 19.........(2)$

Solution by elimination method

Multiplying (1) by 2 and (2) by 3 we get

$\frac{16}{y} - \frac{18}{x} = 12$

$\frac{30}{y} + \frac{18}{x} = 57$

Adding the equations gives

$\frac{30}{y} + \frac{16}{y} = 57 + 12$

$or \: \frac{46}{y} = 69$

$or \: y = \frac{46}{69}$

$or \: y = \frac{2}{3}$

Using this value of y in (1) we get

$\frac{8}{( \frac{2}{3} )} - \frac{9}{x} = 6$

$or \: \frac{8 \times 3}{2} - \frac{9}{x} = 6$

$or \: \frac{4 \times 3}{1} - 6 = \frac{9}{x}$

$or \: 12 - 6 = \frac{9}{x}$

$or \: \frac{9}{x} = 6$

$or \: 6x = 9$

$or \: x = \frac{3}{2}$

Hence

$xy = \frac{3}{2} \times \frac{2}{3} = 1$