Results 1 to 7 of 7

Math Help - 1/x - 1/y = 1/2005

  1. #1
    Junior Member phgao's Avatar
    Joined
    May 2005
    Posts
    39

    1/x - 1/y = 1/2005

    Find all pairs of positive integers x and y that safisfy the equation:

    1/x - 1/y = 1/2005
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Apr 2005
    Posts
    103
    x, y

    1604, 8020

    1980, 158796

    2000, 802000

    2004, 4018020
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member phgao's Avatar
    Joined
    May 2005
    Posts
    39
    Thanks Paul for finding the solutions!
    Last edited by phgao; May 5th 2005 at 02:29 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member phgao's Avatar
    Joined
    May 2005
    Posts
    39
    Hi there, i was wondering how you rearrange my first equation:

    Namely: 1/x - 1/y = 1/2005

    Into: y = -2005 + (2005^2/2005-x)

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member phgao's Avatar
    Joined
    May 2005
    Posts
    39
    Ok solved. No need to post an answer.

    To solve add and subtract 2005^2 from the numerator... then it falls in place. Thanks anyway.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    SbD
    Guest
    how can you prove it tho? how do you work it out?


    x, y

    1604, 8020

    1980, 158796

    2000, 802000

    2004, 4018020

    ??
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Aug 2005
    Posts
    24
    Into: y = -2005 + (2005^2/2005-x)

    Why is it like that?

    Why isn't it y= -2005 + (2005^2/x-2005)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 8th 2010, 04:02 PM
  2. Need Help With Math B Regents June 2005
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: May 27th 2007, 06:05 PM
  3. 1/2005
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 8th 2005, 12:16 PM

Search Tags


/mathhelpforum @mathhelpforum