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Math Help - Combine the fractions

  1. #1
    Newbie
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    Dec 2008
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    Question Combine the fractions

    I'm having trouble finding the answer to this problem. Here is how I have been trying to solve it:

    First, here is the problem:

    \frac{x}{x+1} + \frac{4}{1-x} + \frac{x^2-5x-8}{x^2-1} =

    The denominator of the second fraction is arranged wrong (x should be first, like the others) so I multiply it by -1.

    \frac{x}{x+1} - \frac{4}{x-1} + \frac{x^2-5x-8}{x^2-1} =

    Next, I factor the third fraction and I see it's factors are (x+1)(x-1), so that must be the common denominator.

    <br />
\frac{x(x-1)-4(x+1)+x^2-5x-8}{(x+1)(x-1)}

    Next, I multiply:

    \frac{x^2-x-4x-4+x^2-5x-8}{(x+1)(x-1)}

    Last, I combine like terms and I have the answer:

    \frac{2x^2-10x-12}{(x+1)(x-1)}

    I think I did something wrong though, because the book says this fraction here is the correct answer:

    \frac{2(x-6)}{x-1}

    But it doesn't show how to get that answer, so I have no way to figure out what I'm doing wrong. I have done it three time and I still come up with the same incorrect answer.

    What am I doing wrong?
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  2. #2
    Super Member
    Joined
    Dec 2008
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    Scotland
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    Quote Originally Posted by bryang View Post
    I'm having trouble finding the answer to this problem. Here is how I have been trying to solve it:

    First, here is the problem:

    \frac{x}{x+1} + \frac{4}{1-x} + \frac{x^2-5x-8}{x^2-1} =

    The denominator of the second fraction is arranged wrong (x should be first, like the others) so I multiply it by -1.

    \frac{x}{x+1} - \frac{4}{x-1} + \frac{x^2-5x-8}{x^2-1} =

    Next, I factor the third fraction and I see it's factors are (x+1)(x-1), so that must be the common denominator.

    <br />
\frac{x(x-1)-4(x+1)+x^2-5x-8}{(x+1)(x-1)}

    Next, I multiply:

    \frac{x^2-x-4x-4+x^2-5x-8}{(x+1)(x-1)}

    Last, I combine like terms and I have the answer:

    \frac{2x^2-10x-12}{(x+1)(x-1)}

    I think I did something wrong though, because the book says this fraction here is the correct answer:

    \frac{2(x-6)}{x-1}

    But it doesn't show how to get that answer, so I have no way to figure out what I'm doing wrong. I have done it three time and I still come up with the same incorrect answer.

    What am I doing wrong?
    \frac{2x^2-10x-12}{(x+1)(x-1)}

    \frac{2(x^2-5x-6)}{(x+1)(x-1)}

    \frac{2(x-6)(x+1)}{(x+1)(x-1)}

    \frac{2(x-6)}{x-1}
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  3. #3
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
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    Medgidia, Romania
    Posts
    1,252
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    You did nothing wrong. You must continue.

    2x^2-10x-12=2(x^2-5x-6)=2(x+1)(x-6)

    Then \frac{2x^2-10x-12}{(x+1)(x-1)}=\frac{2(x+1)(x-6)}{(x+1)(x-1)}=\frac{2(x-6)}{x-1}
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