algebra

Find two possible numbers that differ by 8 and whose reciprocals differ by 1/6.

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asked by Keith
  1. x-y = 8 or y = x-8

    1/x - 1/y = 1/6
    multiply by 6xy ....
    6y - 6x = xy
    6y - xy = 6x
    y(6-x) = 6x
    y = 6x/(6-x)

    then 6x/(6-x) = x-8
    6x = 6x - 48 - x^2 + 8x
    x^2 - 8x + 48 = 0
    Solving we get two imaginary numbers , namely

    4 ± √-32

    so one pair is
    4 + √-32 and 4+√-32 - 8 or -4 + √-32

    check: (4+√-32) - (-4+√-32) = 8
    1/ (4+√-32) - 1/(-4+√-32)
    = [(-4+√-32) - (4+√-32)]/(-32 - 16)
    =-8/-48
    = 1/6

    Ok, I found two possible numbers and verified that they work
    4+√-32 and -4+√-32

    There is a second such pair. Can you find it?

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    posted by Reiny

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