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Math Help - Algebraic fraction?

  1. #1
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    Algebraic fraction?

    2x/x-1 - 7x-3/x^2-1

    Thanks for any help on this.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Algebraic fraction?

    Without bracketing symbols, i.e., 2x/(x - 1) - (7x - 3)/(x^2 - 1), I am assuming we are given to simplify:

    \frac{2x}{x-1}-\frac{7x-3}{x^2-1}

    A good first step is to factor the denominators, so that we can easily determine what our lowest common denominator should be. When you perform this step, what do you have?
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  3. #3
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    Re: Algebraic fraction?

    First you get a common denominator which would be (x-1)(x^2-1)

    then you apply what's missing to the numerator while cancelling the denominator.

    you should get:

    2x((x^2)-1)-(7x-3)(x-1)
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  4. #4
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    Re: Algebraic fraction?

    Quote Originally Posted by bakinbacon View Post
    First you get a common denominator which would be (x-1)(x^2-1)

    then you apply what's missing to the numerator while cancelling the denominator.

    you should get:

    2x((x^2)-1)-(7x-3)(x-1)
    That would be a foolish thing to do. It is much simpler to use the least common denominator as MarkFL suggested.

    (And, I assume you mean that last result is the numerator of the fraction.)
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  5. #5
    Junior Member alane1994's Avatar
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    Re: Algebraic fraction?

    Quote Originally Posted by MarkFL2 View Post
    Without bracketing symbols, i.e., 2x/(x - 1) - (7x - 3)/(x^2 - 1), I am assuming we are given to simplify:

    \frac{2x}{x-1}-\frac{7x-3}{x^2-1}

    A good first step is to factor the denominators, so that we can easily determine what our lowest common denominator should be. When you perform this step, what do you have?
    Assuming that we got the equation right, as it was rather difficult to read... I believe the least common denominator would be x^2-1.
    To get that you would multiply the left side by \frac{x+1}{x+1}.
    \frac{2x}{x-1}\times\frac{x+1}{x+1}
    I could be wrong because I haven't done this sort of stuff in a while.
    If I am incorrect, someone correct me.
    Once you get the common denominator, go from there using what you know. If you need more help come back!
    Last edited by alane1994; October 10th 2012 at 04:40 AM. Reason: Silly Errors
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  6. #6
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    Re: Algebraic fraction?

    Quote Originally Posted by MarkFL2 View Post
    Without bracketing symbols, i.e., 2x/(x - 1) - (7x - 3)/(x^2 - 1), I am assuming we are given to simplify:

    \frac{2x}{x-1}-\frac{7x-3}{x^2-1}

    A good first step is to factor the denominators, so that we can easily determine what our lowest common denominator should be. When you perform this step, what do you have?
    That's right.

    that leaves us with 2x/x-1 - 7x-3/(x+1)(x-1). I'm unsure of the next step.
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  7. #7
    Junior Member alane1994's Avatar
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    Re: Algebraic fraction?

    You should really learn to use latex, it makes it much easier on the people trying to help you.
    \frac{2x}{x-1}-\frac{7x-3}{(x+1)(x-1)}
    And that isn't a common denominator... You want the same denominator on both sides of the subtraction sign.
    Which in this case should be x^2-1 on the bottom on both sides.
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  8. #8
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    Re: Algebraic fraction?

    Quote Originally Posted by alane1994 View Post
    You should really learn to use latex, it makes it much easier on the people trying to help you.
    \frac{2x}{x-1}-\frac{7x-3}{(x+1)(x-1)}
    And that isn't a common denominator... You want the same denominator on both sides of the subtraction sign.
    Which in this case should be x^2-1 on the bottom on both sides.
    Where can I learn it?

    Oh, right.

    So multiplying the first denominator by 2 to get a common denominator you also multiply the numerator by itself, so it's 4x^2, so the answer will be -3x^2-3/x^2-1, am I right?
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  9. #9
    MHF Contributor MarkFL's Avatar
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    Re: Algebraic fraction?

    Quote Originally Posted by Ashir View Post
    That's right.

    that leaves us with 2x/x-1 - 7x-3/(x+1)(x-1). I'm unsure of the next step.
    Yes, we now have:

    \frac{2x}{x-1}-\frac{7x-3}{(x+1)(x-1)}

    We can now easily see that if we multiply the first term by 1 in the form of \frac{x+1}{x+1} then both terms will have the same denominator:

    \frac{2x}{x-1}\cdot\frac{x+1}{x+1}-\frac{7x-3}{(x+1)(x-1)}

    Now, we may combine the terms using the property \frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}:

    \frac{2x(x+1)-(7x-3)}{(x+1)(x-1)}

    Now simplify the numerator. You will get a quadratic that nicely factors, with one of the factors being in the denominator that you can then divide out.
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  10. #10
    Junior Member alane1994's Avatar
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    Re: Algebraic fraction?

    This is the original right?
    \frac{2x}{x-1}-\frac{7x-3}{x^2-1}
    The common denominator should be x^2-1
    To get it you would do this.
    \frac{2x}{x-1}\times\frac{x+1}{x+1}
    You would FOIL the bottom to get that part. As for the top, you distribute.
    Try that, and see what happens.
    This is how I was taught, get the same thing on the bottom on both sides. Add the numerators, and put what you get from that over the common denominator.
    Last edited by alane1994; October 10th 2012 at 09:54 AM.
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  11. #11
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    Re: Algebraic fraction?

    didn't get that
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  12. #12
    MHF Contributor MarkFL's Avatar
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    Re: Algebraic fraction?

    What didn't you get? If you just say "didn't get that" then we are left wondering what it is you didn't get. It is best to cite exactly which step you have trouble with, so we can address that specifically.
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  13. #13
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    Re: Algebraic fraction?

    Why did you do that?
    Where did x+1 come from?
    Why both numerator and denominator?
    Why can't we just treat it as a normal fraction?
    Why wasn't my method correct? Or was it?
    What do you mean 'foil'?
    What do you mean by distribute?
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  14. #14
    Junior Member alane1994's Avatar
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    Re: Algebraic fraction?

    I did that because if you did what I said to the left side of the problem you would get the same denominator... which you need for the problem because it is a subtraction problem. The x+1, it came from me finding something to multiply the left side by to make the denominators the same. When you expanded the right side, you got (x+1)(x-1). Well on the other side there is only x-1, so you have to multiply by x+1 to make them the same. Honestly I am not sure why exactly you do it to both numerator and denominator, that was just how I was taught and it works. Your method works, you just stopped about halfway through a step basically. FOIL means First Outer Inner Last. It is a method of multiplying terms like (x+1)(x-1). First x*x, outer x*-1, inner 1*x, last 1*-1. Combine like terms after that. Distribute is when you multiply a single outer term by multiples in parenthises. 7(x+1). 7*x and 7*1.
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  15. #15
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    Re: Algebraic fraction?

    still confused, that method just doesn't work for me. Is -5x-3 the right answer?
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