Question

Solve the following pair of equations:

`(2xy)/(x + y) = 3/2, (xy)/(2x - y) = (-3)/10,  x + y ≠ 0, 2x - y ≠ 0`

Answer 1

Given pair of equations is

`(2xy)/(x + y) = 3/2`, where x + y ≠ 0

⇒ `(x + y)/(2xy) = 2/3`

⇒ `x/(xy) + y/(xy) = 4/3`

⇒ `1/y + 1/x = 4/3`   .....(i)

And `(xy)/(2x - y) = (-3)/10`, where 2x – y ≠ 0

⇒ `(2x - y)/(xy) = (-10)/3`

⇒ `(2x)/(xy) - y/(xy) = (-10)/3` 

⇒ `2/y - 1/x = (-10)/3`  .....(ii)

Now, put `1/x` = u and `1/y` = v, then the pair of equations becomes

v + u = `4/3`   .....(iii)

And 2v – u = `(-10)/3`  .....(iv)

On adding both equations, we get

3v = `4/3 - 10/3`

= `(-6)/3`

⇒ 3v = – 2

⇒ v = `(-2)/3`

Now, put the value of v in equation (iii), we get

`(-2)/3 + u = 4/3`

⇒ u = `4/3 + 2/3`

= `6/3`

= 2

∴ x = `1/u = 1/2`

And y = `1/v`

= `1/((-2)/3)`

= `(-3)/2`

Hence, the required values of x and y are `1/2` and `(-3)/2`, respectively.