Results 1 to 3 of 3

Math Help - Literal fraction

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    22

    Literal fraction

    [Math]\frac{10}{r-3}+\frac{4}{3-r}=6[/tex]

    [MATh]\frac{10(3-r)}{(r-3)(3-r)}+\frac{4(r-3)}{(r-3)(3-r)}=\frac{6(r-3)(3-r)}{(r-3)(3-r)}[/tex]

    Is it correct to this point? The latest fraction is 0. Then 30 -10r +4r -12 = 0, r= 3
    My book says r=4 so at some point I must have made a mistake.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    Quote Originally Posted by strunz View Post
    [Math]\frac{10}{r-3}+\frac{4}{3-r}=6[/tex]
    Note that 3 - r = -r + 3 = -(r - 3). So the common denominator is just r - 3:

    . . . . . \frac{10}{r\, -\, 3}\, -\, \frac{4}{r\, -\, 3}\, =\, 6

    A useful method for solving rational equations is to multiply through by the common denominator. In this case, you would end up with:

    . . . . . 10\, -\, 4\, =\, 6(r\, -\, 3)

    . . . . . 1\, =\, r\, -\, 3

    Solve the linear equation. Remember to check the value of "r" in the original equation, to make sure there are no "division by zero" problems.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    \frac{10}{r-3}+\frac{4}{3-r}=\frac{10}{r-3}-\frac{4}{r-3}=\frac{6}{r-3}

    The equation becomes \frac{6}{r-3}=6\Rightarrow r-3=1\Rightarrow r=4
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Literal Equations Calculator
    Posted in the Math Software Forum
    Replies: 1
    Last Post: September 22nd 2011, 07:19 PM
  2. Literal Equation help
    Posted in the Algebra Forum
    Replies: 5
    Last Post: October 28th 2010, 06:00 AM
  3. Simultaneous Literal Equation Help
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 30th 2009, 04:29 AM
  4. Literal Equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 18th 2009, 08:34 PM
  5. Literal Equations
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: January 8th 2006, 05:34 PM

Search Tags


/mathhelpforum @mathhelpforum