I took a test a couple of days ago and was stuck on this question... I tried to figure it out but could not... can you help me...
~Question~
The sum of the reciprocal of a number and the reciprocal of 6 less
than the number is 7 times the reciprocal of the original number.
Find the original number.
Thank you all so much for your help!
jonathon.shine@rogers.com
Favorite Answer
let the number be n
reciprocal of n = 1/n
6 less than n = n-6
therefore we have:
(1/n) + [1/(n-6)] = 7/n
6/n = 1/(n-6)
6(n-6) = n
6n-36 = n
5n = 36
n = 36/5
modulo_function
Well, let's see.
Let n be the number
then
the number plus its reciprocal is 6 less than 7 times its reciprocal:( is that what the words mean?)
n+1/n = 7*(1/n)-6
mult by n:
n^2+1 = 7 -6n
n^2+6n-6=0
Doesn't seem like I translated the statement correctly. Is that a verbatem wording?
raj
1/x+1/(x-6)=7/x
6/x=1/(x-6)
cross multiplying
6(x-6)=x
6x-36=x
5x=36
x=36/5
check 5/36+5/6=5+30/36=35/36
checks
bob h
1/x + 1/(x-6) = 7*(1/x)
1/x + 1/(x-6) = 7/x
(x-6)/(x(x-6)) + x/(x(x-6)) = 7/x
(2x-6) / (x^2-6x) = 7/x
2x-6 = (7x^2 -42x) /x
2x-6 = 7x - 42
5x = 36
x = 36/5