Answer 1

The given system of equations is

`(x + y)/(xy) = 2`

`=> x/(xy) + y/(xy) = 2`

`=> 1/y + 1/x = 2`

`=> 1/x + 1/y = 2` ............(i)

And

`(x - y)/(xy) = 6`

`=> x/(xy) - y/(xy) = 6`

`=> 1/y - 1/x = 6`

`=> 1/x - 1/y = 6`   ......(ii)

Taking `u = 1/x and v = 1/y` we get

`u + v = 2 => u + v - 2 = 0` ....(iii)

And u - v = -6 => u - v+ 6 = 0 ........(iv)

Here

`a_1 = 1, b_1 = 1,c_1 = -2`

`a_2 = 1, b_2 = -1, c_2 = 6`

By cross multiplication

`=> u/(1xx6-(-2)xx(-1)) = v/(1xx6-(-2)xx1) = 1/(1xx(-1) -1 xx1)`

`=> u/(6-2) = (-v)/(6+2) = 1/(-1-1)`

`=> u/4 = (-v)/8 = 1/(-2)`

Now, `u/4 = 1/(-2)`

`=> u = 4/(-2) = -2`

And `(-v)/8 = 1/(-2)`

`=> -v  = 8/(-2) = -4`

`=> -v = -4`

=> v = 4

Now `x = 1/u = (-1)/2, y = 1/v = 1/4`

Hence `x = (-1)/2, y = 1/4 `is the solution of the given system of equations.