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Math Help - is there only one way to simplify this?

  1. #1
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    is there only one way to simplify this?

    1/v+1 - v/v-4 + v^2+4/v^2-3v-4
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  2. #2
    Bar0n janvdl's Avatar
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    The layout of this problem is really not clear.
    Could you try to use a few brackets please?
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by rj2001 View Post
    1/v+1 - v/v-4 + v^2+4/v^2-3v-4
    For future reference, if you have a series of terms being divided by a series of terms, write the numerator and denominator in parenthesis to show they are ALL being divided.

    Your problem,
    1/v+1 - v/v-4 + v^2+4/v^2-3v-4
    Is more easily understood written as:

    1/(v + 1) - v/(v - 4) + (v^2 + 4)/(v^2 - 3v - 4)


    We need to find the lowest common denominator. To do this, factor each denominator:
    (v + 1) is factored already
    (v - 4) is factored already
    (v^2 - 3v - 4) = (v - 4)(v + 1) is now in factored form

    Next we need to use the LCD to rewrite each fraction:

    1/(v + 1)*(v - 4)/(v - 4) - v/(v - 4)*(v + 1)/(v + 1) + (v^2 + 4)/[(v - 4)(v + 1)]

    Adding each numerator, we get:

    [(v - 4) - v(v + 1) + (v^2 + 4)]/[(v - 4)(v + 1)]
    (v - 4 - v^2 - v + v^2 + 4)/[(v - 4)(v + 1)]
    (0)/[(v - 4)(v + 1)] = 0
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  4. #4
    Bar0n janvdl's Avatar
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    Mathgeek, your method differs from mine.

    1/(v + 1) - v/(v - 4) + (v^2 + 4)/(v^2 - 3v - 4)

    1/(v + 1) - v/(v - 4) + (v^2 + 4)/((v - 4)(v + 1))

    [multiply every term with (v - 4)(v + 1)]

    then we get:

    v - 4 - v^2 - v + v^2 + 4

    which equals zero.
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by janvdl View Post
    Mathgeek, your method differs from mine.

    1/(v + 1) - v/(v - 4) + (v^2 + 4)/(v^2 - 3v - 4)

    1/(v + 1) - v/(v - 4) + (v^2 + 4)/((v - 4)(v + 1))

    [multiply every term with (v - 4)(v + 1)]

    then we get:

    v - 4 - v^2 - v + v^2 + 4

    which equals zero.
    The problem with this is that you cannot legally multiply every term by the LCD. In essence, you are changing the value of the expression by doing so. (That method is only legal when it is an "equation" where there is an equal sign and some terms on the other side of it. In any algibraic operation you do, you MUST ALWAYS make sure you preserve the identity of the original problem (its value or its relationship).)
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  6. #6
    Bar0n janvdl's Avatar
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    Ah, ok.
    Sorry bout that.
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