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Math Help - Combining, simplifying and factoring fractions.

  1. #1
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    Combining, simplifying and factoring fractions.

    I have no idea what I'm doing as I haven't touched this topic in months. If anyone can help me, that would be great.

    Edit: For some reason, all of the fractions are being messed up so I'm going to post a picture of how they should look.

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  2. #2
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    Quote Originally Posted by Lemmy View Post
    I have no idea what I'm doing as I haven't touched this topic in months. If anyone can help me, that would be great.

    Edit: For some reason, all of the fractions are being messed up so I'm going to post a picture of how they should look.

    For 1)
    \frac{3x + 1}{x^2 - 1} - \frac{1}{x + 1} notice that x^2 - 1 = (x - 1)(x + 1), so in order to subtract the two fractions you need the same denominator which means multiplying the second fraction by \frac{x - 1}{x - 1} to get the difference \frac{3x + 1}{x^2 - 1} - \frac{x - 1}{x^2 - 1} and then computing the difference as \frac{4x}{x^2 - 1}.
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  3. #3
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    Quote Originally Posted by icemanfan View Post
    For 1)
    \frac{3x + 1}{x^2 - 1} - \frac{1}{x + 1} notice that x^2 - 1 = (x - 1)(x + 1), so in order to subtract the two fractions you need the same denominator which means multiplying the second fraction by \frac{x - 1}{x - 1} to get the difference \frac{3x + 1}{x^2 - 1} - \frac{x - 1}{x^2 - 1} and then computing the difference as \frac{4x}{x^2 - 1}.
    Thanks. That's pretty much the only one of the four that I was on the right track for.
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  4. #4
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    Quote Originally Posted by Lemmy View Post
    Thanks. That's pretty much the only one of the four that I was on the right track for.
    The second problem on the list is actually the easiest. What happens if you multiply the fraction on the right by \frac{-1}{-1}?
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  5. #5
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    Quote Originally Posted by icemanfan View Post
    The second problem on the list is actually the easiest. What happens if you multiply the fraction on the right by \frac{-1}{-1}?
    Yes, I was trying to figure out which way to go about doing that. Guess I haven't done these in a while.
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  6. #6
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    Quote Originally Posted by Lemmy View Post
    I have no idea what I'm doing as I haven't touched this topic in months. If anyone can help me, that would be great.

    Edit: For some reason, all of the fractions are being messed up so I'm going to post a picture of how they should look.
    to #4:

    Keep in mind that

    \frac14-y^2 = -\left(y^2-\frac14\right)=-\left(y-\frac12\right) \left(y+\frac12\right)

    Now your fraction becomes:

    \frac{y-\frac12}{-\left(y-\frac12\right) \left(y+\frac12\right)} = \frac1{-\left(y+\frac12\right)} = -\frac1{y+\frac12}
    Last edited by earboth; April 28th 2008 at 12:27 PM. Reason: found a typo
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  7. #7
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    Quote Originally Posted by Lemmy View Post
    I have no idea what I'm doing as I haven't touched this topic in months. If anyone can help me, that would be great.

    Edit: For some reason, all of the fractions are being messed up so I'm going to post a picture of how they should look.
    to #3:

    \frac4{x-1} = \frac5{2x-2} + \frac{3x}4~,~x \neq 1

    The common denominator of all fractions is 4(x-1)
    Multiply both sides of the equation by the denominator. You should get:

    16 = 10 + 3x(x-1)~\implies~3x^2-3x-6=0

    For confirmation only: x = -1 or x = 2
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