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Question
Divide 15 into two parts such that the sum of their reciprocals is `3/10`.
Solution
Let the two parts be x and 15 – x.
`1/x + 1/(15 - x) = 3/10`
`(15 - x + x)/(x(15 - x)) = 3/10`
`15/(15x - x^2) = 3/10`
`cancel(15)^5/(15x - x^2) xx 10/cancel(3)_1 = 0`
`5/(15x - x^2) xx 10 = 0`
`50 = 15x - x^2`
`x^2 - 15x + 50 = 0`
`x^2 - 10x - 5x + 50 = 0`
(x – 10)(x – 5) = 0
∴ x – 10 = 0
∴ x = 10
∴ x – 5 = 0
∴ x = 5
Thus the required two parts are 5 and 10.
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