SOLUTION: find two positive integers that differ by 8 and whose reciprocal differ by 1/6. I would really appreciate it if yall could help me with this question!!! Thank you soo much for

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Question 860872: find two positive integers that differ by 8 and whose reciprocal differ by 1/6.
I would really appreciate it if yall could help me with this question!!! Thank you soo much for taking your time to help me too:) god bless

Found 3 solutions by mananth, stanbon, josmiceli:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
let one number be x
other number is x+8
1/x - 1/(x+8) = 1/6
LCD = 6x(x+8) multiply the terms by the LCD
6(x+8)-6x=x(x+8)
6x+48-6x=x^2+8x
x^2+8x-48=0
x^2+8x-48=0

x^2+12x-4x+48=0
x(x+12)-4(x+12)=0
(x+12)(x-4)=0
x= 4
x+8=12
4, 12

Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
find two positive integers that differ by 8 and whose reciprocal differ by 1/6.
-------
Equations:
x - y = 8
(1/x) - (1/y) = 1/6
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Modify:
x = y + 8
---------
Substitute:
(1/(y+8)) - 1/y = 1/6
-------
Multiply thru by 6y(y+8)
6y - 6(y+8) = y(y+8)
-------
y^2 + 8y + 48 = 0
This quadratic has no integer roots.
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Cheers,
Stan H.

Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website!
Let the integers be and
Assume that is the larger
(1)
(2)
Note that 1 / ( small number) is greater than 1 / ( big number )
-----------------------
(2)
Multiply both sides by
(2)
(2)
(2)
and
(1)
By substitution:
(2)
(2)
(2)
(2)
(2)
( choose the positive root )
and
(1)
(1)
(1)
The integers are 4 and 12
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check:
(2)
(2)
(2)
(2)
OK