Results 1 to 8 of 8

Math Help - Simplifying/Reducing

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    3

    Simplifying/Reducing

    Could somebody please show me how to simplify this?

    ((1/4)x^(-3/4)*y^(3/4)/(3/4)x^(1/4)*y^(-1/4))

    That is, in words, one-fourth "x" raised to the negative three-fourths, "y" raised to the three-fourths, over/divided by three-fourths "x" raised to the one-fourth, "y" raised to the negative one-fourth.

    The answer is (y/3x), or in words, "y" over/divided by 3"x".

    I just need to see how to get to that and preferably the quickest way.

    Thank you so much for the help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Sep 2011
    From
    Oxford
    Posts
    17

    re: Simplifying/Reducing

    I've converted to Latex for now so that it's easier to read:

    \frac{\bigg(\frac{1}{4}x^{-\frac{3}{4}}y^{\frac{3}{4}}\bigg)}{\bigg(\frac{3}{  4}x^{\frac{1}{4}}y^{-\frac{1}{4}}\bigg)}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2011
    Posts
    3

    re: Simplifying/Reducing

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2011
    From
    Oxford
    Posts
    17

    re: Simplifying/Reducing

    Quote Originally Posted by mabentley View Post
    I've converted to Latex for now so that it's easier to read:

    \frac{\bigg(\frac{1}{4}x^{-\frac{3}{4}}y^{\frac{3}{4}}\bigg)}{\bigg(\frac{3}{  4}x^{\frac{1}{4}}y^{-\frac{1}{4}}\bigg)}
    When you divide with powers you subtract:

    \frac{1/4}{3/4}=\frac{4}{12}=\frac{1}{3}

    x^{-\frac{3}{4}}-x^\frac{1}{4}=x^\frac{4}{4}=x^{-1}=\frac{1}{x}

    y^\frac{3}{4}-y^{-\frac{1}{4}}=y^\frac{4}{4}=y^{1}=y

    Multiplying through:

    =\frac{y}{3x}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2011
    Posts
    3

    re: Simplifying/Reducing

    Thank you very much!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2011
    From
    Oxford
    Posts
    17

    re: Simplifying/Reducing

    You are welcome, feel free to hit the thanks button on the bottom left of your reply
    Follow Math Help Forum on Facebook and Google+

  7. #7
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1

    Re: Simplifying/Reducing

    Quote Originally Posted by mabentley View Post
    When you divide with powers you subtract:
    x^{-\frac{3}{4}}-x^\frac{1}{4}=x^\frac{4}{4}=x^{-1}=\frac{1}{x}

    y^\frac{3}{4}-y^{-\frac{1}{4}}=y^\frac{4}{4}=y^{1}=y
    That's not true. It's only the powers that are subtracted.

    . \dfrac{x^{-3/4}}{x^{1/4}} = x^{-3/4 - 1/4} = x^{-1}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Sep 2011
    From
    Oxford
    Posts
    17

    Re: Simplifying/Reducing

    Quote Originally Posted by e^(i*pi) View Post
    That's not true. It's only the powers that are subtracted.

    . \dfrac{x^{-3/4}}{x^{1/4}} = x^{-3/4 - 1/4} = x^{-1}
    Sorry the x^{\frac{4}{4}} should have read x^{-\frac{4}{4}}. I can see that my wording could also have been clearer. Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. reducing quadratic
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 10th 2011, 10:00 PM
  2. [SOLVED] Reducing the modulo
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: March 17th 2011, 12:19 PM
  3. Reducing the polynomial
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 9th 2009, 04:30 AM
  4. Reducing the set
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: October 7th 2009, 12:02 PM
  5. Reducing Roots
    Posted in the Algebra Forum
    Replies: 5
    Last Post: June 6th 2009, 01:54 PM

Search Tags


/mathhelpforum @mathhelpforum