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Math Help - fractions with variables

  1. #1
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    fractions with variables

    2x/x^2-1 - 1/x+1

    i got x+1 as my final answer
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  2. #2
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    Re: fractions with variables

    How have you got  x + 1 ? I don't understand what you're trying to achieve.

    I'll presume the denominator on the left side is  x^2 - 1

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

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

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

    \frac{x^2 + 2x + 1}{x^3 + x^2 - x -1}
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  3. #3
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    Re: fractions with variables

    Wouldn't it be better to use the least common denominator?

    \frac{2x}{x^2- 1}- \frac{1}{x+1}
    Since x^2- 1= (x+ 1)(x- 1), that is the least common denominator and we can just multiply numerator and denominator of the second fraction by x- 1:
    \frac{2x}{x^2- 1}- \frac{x- 1}{(x+1)(x-1)}= \frac{2x}{x^2- 1}- \frac{x- 1}{x^2- 1}= \frac{x+ 1}{x^2- 1}= \frac{1}{x- 1}

    AStartledDeer, your \frac{x^2+ 2x+ 1}{x^3+ x^2- x- 1}= \frac{(x+ 1)^2}{(x+1)(x^2- 1)}= \frac{(x+1)^2}{(x+1)^2(x- 1)}= \frac{1}{x- 1}

    That is still not "x+1".
    Last edited by HallsofIvy; July 18th 2012 at 03:40 PM.
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  4. #4
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    Re: fractions with variables

    Quote Originally Posted by HallsofIvy View Post
    Wouldn't it be better to use the least common denominator?

    \frac{2x}{x^2- 1}- \frac{1}{x+1}
    Since x^2- 1= (x+ 1)(x- 1), that is the least common denominator and we can just multiply numerator and denominator of the second fraction by x- 1:
    \frac{2x}{x^2- 1}- \frac{x- 1}{(x+1)(x-1)}= \frac{2x}{x^2- 1}- \frac{x- 1}{x^2- 1}= \frac{x+ 1}{x^2- 1}= \frac{1}{x- 1}

    AStartledDeer, your \frac{x^2+ 2x+ 1}{x^3+ x^2- x- 1}= \frac{(x+ 1)^2}{(x+1)(x^2- 1)}= \frac{(x+1)^2}{(x+1)^2(x- 1)}= \frac{1}{x- 1}

    That is still not "x+1".
    I don't believe it! Thanks for pointing that out; I completely missed that one.

    P.S, how do I change my username from 'astartleddeer' to 'AStartledDeer'?
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  5. #5
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    Re: fractions with variables

    "that is the least common denominator and we can just multiply numerator and denominator of the second fraction by x- 1:"

    why did you only multiple the second fraction? can you even do that?

    what i did was mutiple both fractions by (x+1)(x-1) THUS making 2x - x-1= x-1

    i'm aware i originally said my answer was x+1. this has been revised to x-1
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  6. #6
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    Re: fractions with variables

    Quote Originally Posted by ReginaldCuthbert View Post
    i'm aware i originally said my answer was x+1. this has been revised to x-1
    NO; it's 1 / (x-1).
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  7. #7
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    Re: fractions with variables

    Quote Originally Posted by ReginaldCuthbert View Post
    "that is the least common denominator and we can just multiply numerator and denominator of the second fraction by x- 1:"

    why did you only multiple the second fraction? can you even do that?

    what i did was mutiple both fractions by (x+1)(x-1) THUS making 2x - x-1= x-1

    i'm aware i originally said my answer was x+1. this has been revised to x-1
    The most efficient method is to find the LOWEST common denominator. To do this you need to factorise each denominator and see which factors are missing from each.

    In this case we have \displaystyle \begin{align*}x^2 - 1 = (x - 1)(x + 1)  \end{align*} as the first denominator, and \displaystyle \begin{align*} x + 1 \end{align*} as the second denominator. Can you see that you already have a factor of \displaystyle \begin{align*} x + 1 \end{align*} in your first denominator? So all you need to get the lowest common denominator is that extra factor of \displaystyle \begin{align*} x - 1 \end{align*}.
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    Re: fractions with variables

    Quote Originally Posted by Prove It View Post
    The most efficient method is to find the LOWEST common denominator. To do this you need to factorise each denominator and see which factors are missing from each.

    In this case we have \displaystyle \begin{align*}x^2 - 1 = (x - 1)(x + 1)  \end{align*} as the first denominator, and \displaystyle \begin{align*} x + 1 \end{align*} as the second denominator. Can you see that you already have a factor of \displaystyle \begin{align*} x + 1 \end{align*} in your first denominator? So all you need to get the lowest common denominator is that extra factor of \displaystyle \begin{align*} x - 1 \end{align*}.
    multiple the numerator and denominator by x-1?

    you mean.. multiply by x-1, then immediately divide by x-1? because i thought multiplying the denominator is the same as division.

    what i did. cancelling out the denominators. why does that not work?
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  9. #9
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    Re: fractions with variables

    You don't seem to understand what's "typed" at you...
    Why don't you ask your teacher...at least, he'll have a blackboard and chalk
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  10. #10
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    Re: fractions with variables

    It is basically the same thing you learned in arithmetic: in order to add or subtract fractions, you need to get the same denominator.

    If, for example, you want to add \frac{1}{3} and \frac{1}{6} you could get a common denominator by multiplying, both numerator and denominator of the first fraction by 6, and of the second fraction by 3: \frac{1}{3}+\frac{1}{6}= \frac{6}{18}+ \frac{3}{18}= \frac{9}{18} which reduces to \frac{1}{2}.

    But since 6= 2(3), we can get the "least common denominator" of 6 by multiplying numerator and denominator of the first fraction by 2: \frac{1}{3}+ \frac{1}{6}= \frac{2}{6}+ \frac{1}{6}= \frac{3}{6}= \frac{1}{2}.
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