Question

Solve the following systems of equations:

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

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

Answer 1

The given system of equation is

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

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

`=> 1/y + 1/x = 2`    ...(i)

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

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

`=> 1/y - y/x = 6` .....(ii)

taking 1/y = v and `1/x= u` the above equations become

v + u = 2 ....(iii)

v - u = 6 ..........(iv)

Adding equation (iii) and equation (iv), we get

v + u + v - u = 2 + 6

=> 2v = 8

=> v = 8/2 = 4

Putting v = 4 in equation (iii), we get

4 + u = 2

=> u = 2 - 4 = 2

Hence ` x = 1/u = 1/(-2) = (-1)/2 and y = 1/v = 1/4`

So, the solution of the given system of equation  `x = (-1)/2, y = 1/4`