Results 1 to 5 of 5

Math Help - solving rational equations

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    24

    solving rational equations

    I thought I got the whole concept but when I came across these problems, my answer came no where near that of the correct answer.

    1)  \frac{x}{x-3} + x = \frac{7x-18}{x-3}



    2)  \frac3{x} + \frac{x}{x+2} = - \frac{2}{x+2}


    I need to solve for "x". Help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    1)  \frac{x}{x-3} + x = \frac{7x-18}{x-3}
    The idea is to shed the denominators to get rid of the fractions. Do this by multiplying through by the LCD. In this case, x-3.

    Multiply through by x-3

    (x-3)\cdot\frac{x}{x-3}+(x-3)x=(x-3)\cdot\frac{7x-18}{x-3}

    Then we get x+x(x-3)=7x-18

    Now, it's easy to solve, huh?. Remember to check your solutions by plugging them back in. This one will give two solutions, but only one may be good.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2008
    Posts
    24
    Thanks.


    That's about how far I get and then I start running into trouble.
    After simplifying this:
    <br />
x+x(x-3)=7x-18

    I get this:
    x+x^2-3x=7x-18

    = x^2-2x=7x-18

    =  x^2-9x=-18

    Am I doing it wrong?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by Hypertension View Post
    Thanks.


    That's about how far I get and then I start running into trouble.
    After simplifying this:
    <br />
x+x(x-3)=7x-18

    I get this:
    x+x^2-3x=7x-18

    = x^2-2x=7x-18

    =  x^2-9x=-18

    Am I doing it wrong?
    No, that is correct but the method is incomplete. You must continue.

    Notice, you have  x^2-9x=-18 which is  x^2-9x+18=0.

    You can solve by factorising into 2 brackets and then equating each bracket to 0. Hence finding two values for x. If one of the solution give the denominator to equal 0 then reject this solution.

    EDIT: If are struggling to factorise, and when the quadratic equation is in the form ax^2 + bx + c =0 then you can use the quadratic formula which is x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}.
    Last edited by Simplicity; July 6th 2008 at 01:31 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2008
    Posts
    24
    Ok, I've got it now.

    x=6 or x=3


    But in this case x would equal 6 because when you plug in the 3 for x you get 3 equaling zero as the answer. I just didn't expect to have have to factor, since I haven't had a problem involving that yet.

    And for problem two, I got
    x=-3


    Thank you guys
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving Rational Equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 20th 2011, 04:19 AM
  2. Solving Rational equations.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 8th 2009, 05:28 PM
  3. I need help solving these rational equations.
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 27th 2009, 02:27 PM
  4. Solving rational equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 10th 2008, 03:26 AM
  5. Solving Rational Equations...help!!
    Posted in the Algebra Forum
    Replies: 14
    Last Post: January 7th 2007, 03:37 PM

Search Tags


/mathhelpforum @mathhelpforum