Results 1 to 7 of 7

Math Help - How does x/(x-1) - x/(x+1) equal 1/(x+1) + 1/(x-1) ??

  1. #1
    Junior Member
    Joined
    May 2011
    Posts
    26

    How does x/(x-1) - x/(x+1) equal 1/(x+1) + 1/(x-1) ??

    I'm looking at my notes and one of my professors steps goes from:

    \frac{x}{x-1}-\frac{x}{x+1}

    to

    \frac{1}{x-1}+\frac{1}{x+1}

    What happened to the numerator? I know this is not a mistake because when I enter the first line into my ti-89 it spits out the second line.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2009
    Posts
    177
    First, notice that \frac{x}{x-1}=\frac{x-1}{x-1}+\frac{1}{x-1}=1+\frac{1}{x-1} and that \frac{x}{x+1}=\frac{x+1}{x+1}-\frac{1}{x+1}=1-\frac{1}{x+1}

    What do you get if you put that into the given equation?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2011
    Posts
    26
    Thanks a lot. That explains it.

    Although it seems like such an unusual (and unnecessary) step to take... I guess I just have to try to remember this if I see something like it again. Does this technique have a name?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Carbon View Post
    Thanks a lot. That explains it.

    Although it seems like such an unusual (and unnecessary) step to take... I guess I just have to try to remember this if I see something like it again. Does this technique have a name?
    I doubt if there's a name to it. Note that his (rather clever) way of doing the problem can be done in a more direct, if somewhat longer, manner. Add the two fractions, then use a partial fraction decomposition. It takes no more than five lines.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2011
    Posts
    26
    Quote Originally Posted by topsquark View Post
    I doubt if there's a name to it. Note that his (rather clever) way of doing the problem can be done in a more direct, if somewhat longer, manner. Add the two fractions, then use a partial fraction decomposition. It takes no more than five lines.

    -Dan
    Can you show this? Because when I do it, I don't get 1 in the numerator.

    Adding the fractions, you get 2x/[(x-1)(x+1)]
    Using partial fraction I go to 2x=A(x+1) + B(x-1)
    Solving for A and B, I end up getting x/(x-1) -x/(x+1)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Carbon View Post
    Can you show this? Because when I do it, I don't get 1 in the numerator.

    Adding the fractions, you get 2x/[(x-1)(x+1)]
    Using partial fraction I go to 2x=A(x+1) + B(x-1)
    Solving for A and B, I end up getting x/(x-1) -x/(x+1)
    \frac{2x}{(x + 1)(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}

    2x = A(x + 1) + B(x - 1)

    2x = (A + B)x + (A - B)

    So we know that
    2 = A + B
    0 = A - B

    I get A = B = 1 and the theorem follows.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Nov 2009
    Posts
    177
    x^1: 2 = A + B
    x^0: 0 = A - B

    That gives me A = B = 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Equal power sets -> Equal sets?
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: July 5th 2012, 10:23 AM
  2. Another cant get this to equal this :)
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 26th 2010, 10:26 AM
  3. less than or equal!!!
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: February 9th 2010, 12:32 PM
  4. Replies: 2
    Last Post: March 23rd 2009, 08:11 AM
  5. Never equal
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 24th 2008, 08:19 AM

Search Tags


/mathhelpforum @mathhelpforum