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Question
Solve the following pair of equations:
`(2xy)/(x + y) = 3/2, (xy)/(2x - y) = (-3)/10, x + y ≠ 0, 2x - y ≠ 0`
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Solution
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.
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