# algebra

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

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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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