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Math Help - rational functions

  1. #1
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    rational functions

    x+3/(x+6)(x+4) + 5/4-x=x+2/3-x

    this is what I have done so far...

    I know to put a negative 1 from the 4-x and the 3-x in order to get -1(x-4) and -1(x-3)

    Now I have -1(x+3)(x-3) this is where I am stuck at..if someone could
    (x+6)(x-4)

    please walk me through this step by step, I would greatly appreciate it.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by getnaphd View Post
    x+3/(x+6)(x+4) + 5/4-x=x+2/3-x

    this is what I have done so far...

    I know to put a negative 1 from the 4-x and the 3-x in order to get -1(x-4) and -1(x-3)

    Now I have -1(x+3)(x-3) this is where I am stuck at..if someone could
    (x+6)(x-4)

    please walk me through this step by step, I would greatly appreciate it.
    please use parentheses. is the equation \frac {x + 3}{(x + 6)(x + 4)} + \frac {5}{4 - x} = \frac {x + 2}{3 - x} ?

    what exactly are we supposed to do? solve for x? prove one side is equivalent to the other?
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  3. #3
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    rational functions

    Yes this is exactly what the equation looks like.

    It says solve the equation.
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  4. #4
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    rational functions

    I don't know where to go once the -1 is removed from the fraction in the denominator.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by getnaphd View Post
    I don't know where to go once the -1 is removed from the fraction in the denominator.
    please double check the question. the problem boils down to solving a quartic equation which has no nice solutions. it would not be reasonable to ask you to solve such a question without technology
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by getnaphd View Post
    I don't know where to go once the -1 is removed from the fraction in the denominator.
    To answer your question directly
    \frac{a}{-b} = \frac{-a}{b} = -\frac{a}{b}

    So:
    \frac {x + 3}{(x + 6)(x + 4)} + \frac {5}{4 - x} = \frac {x + 2}{3 - x} <-- Factor a -1 from the 4 - x and the 3 - x terms

    \frac {x + 3}{(x + 6)(x + 4)} + \frac {5}{-(x - 4)} = \frac {x + 2}{-(x - 3)} <-- The -1 factor changes the sign of the whole term

    \frac {x + 3}{(x + 6)(x + 4)} + -\left ( \frac {5}{x - 4} \right )  = - \left ( \frac {x + 2}{x - 3} \right )

    \frac {x + 3}{(x + 6)(x + 4)} - \frac {5}{x - 4} = - \frac {x + 2}{x - 3}

    You can take it from here. But Jhevon is right, this one's a monster without numerical approximation.

    -Dan
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