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Math Help - How do you simplify this algebraically?

  1. #1
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    How do you simplify this algebraically?

    Algebraically simplify: -x^(-1) +1-(x-1)(x^-2) to ((1/x)-1)^2
    Last edited by mr fantastic; September 16th 2010 at 08:31 PM. Reason: Re-posted in lower case, edited title.
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  2. #2
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    You have \displaystyle \frac{-1}{x}+1-\frac{x-1}{x^2}

    I would suggest starting by making a common denominator of x^2
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  3. #3
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    Remember that multiplying by negative powers is the same as dividing by that positive power.

    So you would have the equation of:

    \dfrac{-1}{x}+1-\dfrac{x-1}{x^2}

    Make each of them have a common denominator by multiplying by corresponding values of x:

    = \dfrac{-x}{x^2}+\dfrac{x^2}{x^2}-\dfrac{x-1}{x^2}

    =\dfrac{-x+x^2-x+1}{x^2}

    =\dfrac{x^2-2x+1}{x^2}

    =\dfrac{(1-x)^2}{x^2}

    Note, x^2-2x+1 can be factorised to (x-1)^2 or (1-x)^2. I used (1-x)^2 because it helps in getting the final answer you gave.

    =\left(\dfrac{1-x}{x}\right)^2

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

    =(\frac{1}{x}-1)^2
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  4. #4
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    Hello, yess!

    \text{Simplify }\,-x^{-1} +1-(x-1)x^{-2}\:\text{ to }\:\left(\dfrac{1}{x}-1\right)^2

    We have: . -\dfrac{1}{x} + 1 - \left(\dfrac{x-1}{x^2}\right) \;\;=\;\;-\dfrac{1}{x} + 1 - \left(\dfrac{x}{x^2} - \dfrac{1}{x^2}\right)

    . . . . . . =\;\;-\dfrac{1}{x} + 1 - \dfrac{1}{x} + \dfrac{1}{x^2} \;\;=\;\;\dfrac{1}{x^2} - \dfrac{2}{x} + 1


    Factor: . \left(\dfrac{1}{x} - 1\right)^2
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  5. #5
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    Quote Originally Posted by yess View Post
    Algebraically simplify: -x^(-1) +1-(x-1)(x^-2) to ((1/x)-1)^2
    -x^{-1}+1-(x-1)x^{-2}

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

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

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

    =-\left(\frac{1}{x}-1\right)+\left(\frac{1}{x}-1\right)x^{-1}

    =\left(\frac{1}{x}-1\right)\left(x^{-1}-1\right)=\left(\frac{1}{x}-1\right)\left(\frac{1}{x}-1\right)
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