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Question
If x and y both are positive and (2x2 – 5y2) : xy = 1 : 3, find x : y.
Solution
(2x2 – 5y2) : xy = 1 : 3
`\implies (2x^2 - 5y^2)/(xy) = 1/3`
`\implies (2x)/y - (5y)/x = 1/3`
Put `x/y = a` we get
`\implies 2a - 5 1/a = 1/3`
`\implies` 3(2a2 – 5) = a
`\implies` 6a2 – a – 15 = 0
`\implies` 6a2 + 9a – 10a – 15 = 0
`\implies` 3a(2a + 3) – 5(2a + 3) = 0
`\implies` (2a + 3)(3a – 5) = 0
`\implies` (2a + 3) = 0 or (3a – 5) = 0
`\implies a = -3/2` or `a = 5/3`
`a = -3/2` is not acceptable, as x and y both are positive.
`a = 5/3`
`\implies x/y = 5/3`
`\implies` x : y = 5 : 3
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