Results 1 to 13 of 13

Thread: help please with adding and subtracting rational expressions

  1. July 19th 2009, 07:40 PM #1
    icecreamfrk09
    icecreamfrk09 is offline
    Newbie
    Joined
    Jul 2009
    Posts
    16

    help please with adding and subtracting rational expressions

    please help me i dont remember how to do these anymore

    1. add (1/x+3) + (5/x-4)

    2. subtract (3/x) - (7/x-2)

    3. subract (3/x-3) - (4/x+5)
    Last edited by icecreamfrk09; July 19th 2009 at 07:56 PM.
    Follow Math Help Forum on Facebook and Google+
  2. July 19th 2009, 07:43 PM #2
    AlephZero
    AlephZero is offline
    Member
    Joined
    Jul 2009
    Posts
    152
    Quote Originally Posted by icecreamfrk09 View Post
    please help me i dont remember how to do these anymore

    1. add 1/x+3 + 5/x-4

    2. subtract 3/x - 7/x-2

    3. subract 3/x-3 - 4/x+5

    Do you want \frac{1}{x} + 3 + \frac{5}{x} -4, or do you want

    \frac{1}{x+3} + \frac{5}{x-4}?

    If you want the second one, you need to use parentheses!
    Follow Math Help Forum on Facebook and Google+
  3. July 19th 2009, 07:46 PM #3
    icecreamfrk09
    icecreamfrk09 is offline
    Newbie
    Joined
    Jul 2009
    Posts
    16
    the second one please and im sorry im new to this website
    Follow Math Help Forum on Facebook and Google+
  4. July 19th 2009, 07:55 PM #4
    AlephZero
    AlephZero is offline
    Member
    Joined
    Jul 2009
    Posts
    152
    Quote Originally Posted by icecreamfrk09 View Post
    the second one please and im sorry im new to this website
    Fair enough!

    What you need to do for these is to find a common denominator, which is most easily done by multiplying the two denominators. Here goes:

    \frac{1}{x+3}\cdot \frac{x-4}{x-4} + \frac{5}{x-4} \cdot \frac{x+3}{x+3} = \frac{x-4}{(x+3)(x-4)} + \frac{5(x+3)}{(x+3)(x-4)}

    = \frac{x-4+5x+15}{(x+3)(x-4)} = \frac{6x+11}{(x+3)(x-4)}
    Follow Math Help Forum on Facebook and Google+
  5. July 19th 2009, 08:03 PM #5
    icecreamfrk09
    icecreamfrk09 is offline
    Newbie
    Joined
    Jul 2009
    Posts
    16
    thank you so much but how to you subtract the rational expressions?
    Follow Math Help Forum on Facebook and Google+
  6. July 19th 2009, 08:11 PM #6
    AlephZero
    AlephZero is offline
    Member
    Joined
    Jul 2009
    Posts
    152
    Quote Originally Posted by icecreamfrk09 View Post
    thank you so much but how to you subtract the rational expressions?
    Well, by the same method. You find common denominators, and just subtract instead of adding. Here's how to start the second one:

    \frac{3}{x}\cdot\frac{x-2}{x-2} - \frac{7}{x-2}\cdot\frac{x}{x} = \frac{3(x-2)-7x}{x(x-2)}.

    Now simplify, and you should have your answer.

    Same method for #3.

    Let us know what you get, and we'll check your answers.
    Last edited by AlephZero; July 19th 2009 at 08:12 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+
  7. July 19th 2009, 08:17 PM #7
    AlephZero
    AlephZero is offline
    Member
    Joined
    Jul 2009
    Posts
    152
    Quote Originally Posted by icecreamfrk09 View Post
    please help me i dont remember how to do these anymore

    1. add (1/x+3) + (5/x-4)

    2. subtract (3/x) - (7/x-2)

    3. subract (3/x-3) - (4/x+5)
    By the way, and not to be a stick-in-the-mud, but you're still not expressing these in a mathematically correct way, which is important because it's hard for us to tell what you mean. The correct way to use the parentheses would be

    1. 1/(x+3) + 5/(x-4)
    2. 3/x - 7/(x-2)
    3. 3/(x-3) - 4/(x+5)

    In other words, you need to group all the numerators and denominators together, so we know what is being divided into what. I hope that makes sense.
    Follow Math Help Forum on Facebook and Google+
  8. July 19th 2009, 08:19 PM #8
    icecreamfrk09
    icecreamfrk09 is offline
    Newbie
    Joined
    Jul 2009
    Posts
    16
    okayy not sure if i did these right but i got (3-7x)/x for the second one and -1 for the third one is that right??
    Follow Math Help Forum on Facebook and Google+
  9. July 19th 2009, 08:26 PM #9
    AlephZero
    AlephZero is offline
    Member
    Joined
    Jul 2009
    Posts
    152
    Quote Originally Posted by icecreamfrk09 View Post
    okayy not sure if i did these right but i got (3-7x/x) for the second one and -1 for the third one is that right??
    Unfortunately, no. For the second one, I have shown you how to get to

    \frac{3(x-2)-7x}{x(x-2)}.

    What you need to do now is carry out the multiplication in the numerator and combine like terms. You won't have to do anything to the denominator.

    Give it a shot.
    Follow Math Help Forum on Facebook and Google+
  10. July 19th 2009, 08:41 PM #10
    icecreamfrk09
    icecreamfrk09 is offline
    Newbie
    Joined
    Jul 2009
    Posts
    16
    okayy i got (-4x-6)/x(x-2) for the second and (-x+27)/(x+5)(x-3) for the third but do you multiply the denominator for third one?
    Follow Math Help Forum on Facebook and Google+
  11. July 19th 2009, 08:43 PM #11
    yeongil
    yeongil is offline
    Super Member
    Joined
    May 2009
    Posts
    527
    No, you can leave the denominator in factored form. It's considered to in "simplified form" that way.'


    01
    Follow Math Help Forum on Facebook and Google+
  12. July 19th 2009, 08:47 PM #12
    icecreamfrk09
    icecreamfrk09 is offline
    Newbie
    Joined
    Jul 2009
    Posts
    16
    okayy so both those answers are right?
    Follow Math Help Forum on Facebook and Google+
  13. July 19th 2009, 08:55 PM #13
    AlephZero
    AlephZero is offline
    Member
    Joined
    Jul 2009
    Posts
    152
    Quote Originally Posted by icecreamfrk09 View Post
    okayy so both those answers are right?
    Yes, very good job!

    However the second one can be simplified further, depending upon what form your instructor wants:

    \frac{-4x-6}{x(x-2)} = \frac{-2(2x+3)}{x(x-2)}.
    Follow Math Help Forum on Facebook and Google+
« Averages | apps using equations with fractions »

Similar Math Help Forum Discussions

  1. Adding and Subtracting Rational Expressions
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 8th 2009, 05:12 AM
  2. adding and subtracting rational expressions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 9th 2008, 04:52 PM
  3. adding and subtracting rational expressions
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: November 2nd 2008, 08:39 AM
  4. Adding/ Subtracting rational expressions
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 3rd 2007, 03:12 PM
  5. Adding and Subtracting Rational Expressions
    Posted in the Algebra Forum
    Replies: 7
    Last Post: May 14th 2007, 09:06 PM

Search Tags


/mathhelpforum @mathhelpforum