Results 1 to 3 of 3

Math Help - Algebraic Simplification

  1. #1
    Newbie
    Joined
    Jan 2008
    From
    Australia
    Posts
    11

    Algebraic Simplification

    Please help me simplify these questions:

    a - b - (a^2 / a+b)


    (y^3 + 1) / (y^2 - 1) divided by (y^3 - y^2 + y) / (y^2 - 2y + 1)


    And perhaps solve for 'y':

    y^2 = 3y


    2y^2 = y


    y / y + 2 + 2 / y^2 - 4 = 1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Ogsta101 View Post
    (y^3 + 1) / (y^2 - 1) divided by (y^3 - y^2 + y) / (y^2 - 2y + 1)
    \frac{\frac{y^3 + 1}{y^2 - 1}}{\frac{y^3 - y^2 + y}{y^2 - 2y + 1}}

    First, let's get rid of those pesky fractions: Multiply the numerator and denominator by (y^2 - 1)(y^2 - 2y + 1)

    = \frac{\frac{y^3 + 1}{y^2 - 1}}{\frac{y^3 - y^2 + y}{y^2 - 2y + 1}} \cdot \frac{(y^2 - 1)(y^2 - 2y + 1)}{(y^2 - 1)(y^2 - 2y + 1)}
    (Pop quiz: Note that we only need to multiply by (y - 1)^2(y + 1), why?)

    = \frac{(y^3 + 1)(y^2 - 2y + 1)}{(y^3 - y^2 + y)(y^2 - 1)}

    Now let's do some factoring:
    = \frac{[(y + 1)(y^2 - y + 1)][(y - 1)^2]}{[y(y^2 - y + 1)][(y + 1)(y - 1)]}

    Now let's do some(!) canceling:
    = y - 1;~y \neq -1, 1

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Ogsta101 View Post
    y^2 = 3y
    y^2 = 3y

    y^2 - 3y = 0

    y(y - 3) = 0

    Thus
    y = 0
    or
    y - 3 = 0 \implies y = 3

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. having a problem with algebraic simplification
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 20th 2011, 04:07 AM
  2. Replies: 1
    Last Post: August 25th 2011, 06:09 AM
  3. Algebraic Simplification
    Posted in the Algebra Forum
    Replies: 6
    Last Post: August 7th 2011, 08:04 PM
  4. Algebraic Simplification
    Posted in the Calculus Forum
    Replies: 20
    Last Post: August 5th 2011, 07:06 AM
  5. Algebraic Simplification
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 8th 2008, 01:02 PM

Search Tags


/mathhelpforum @mathhelpforum