Results 1 to 4 of 4

Math Help - Algebra/Simplifcation

  1. #1
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167

    Algebra/Simplifcation

    Hi I was wondering how one can get from

     2x = \lambda(2x-2) to  x = \frac{\lambda}{\lambda - 1}

    and

     2y = \lambda(2y-4) to y = \frac{2\lambda}{\lambda - 1}

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2007
    Posts
    144
    Hi FalconPUNCH!,

    We have 2x=\lambda(2x-2) . Divide throughout by 2 hence x=\lambda x-\lambda . Take the x and the \lambda over therefore \lambda = x(\lambda-1) and thus dividing throughout by (\lambda -1) gives the required results.

    The seconds is similar. Hope this helps.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by FalconPUNCH! View Post
    Hi I was wondering how one can get from

     2x = \lambda(2x-2) to  x = \frac{\lambda}{\lambda - 1}

    and

     2y = \lambda(2y-4) to y = \frac{2\lambda}{\lambda - 1}

    thanks
     2x = \lambda(2x-2)

    2x=2x\lambda-2\lambda

    x-\lambda x=-\lambda

    x(1-\lambda)=-\lambda

    x=\frac{-\lambda}{1-\lambda}

    x=\frac{\lambda}{\lambda-1}

    The other one is done in a similar manner.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167
    Thank you guys
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 4th 2011, 08:39 AM
  2. Replies: 2
    Last Post: December 6th 2010, 03:03 PM
  3. algebra
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 30th 2010, 05:29 PM
  4. Algebra or Algebra 2 Equation Help Please?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 12th 2010, 11:22 AM
  5. Replies: 0
    Last Post: April 23rd 2010, 11:37 PM

Search Tags


/mathhelpforum @mathhelpforum