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Math Help - more help needed

  1. #1
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    more help needed

    Is this right? I'm dividing complex fractions....

    8x^2 divided by 4x
    over
    x^2 -25 divided by 3x+15

    I flipped the other fraction and changed the problem to multiplication

    8x^2 ovwer x^2 -25 divided by 3x +15 divided by 4x

    I factored x^2 -25 and got x+5 and x-5 and divided 8 by 4 and got two and 4 divided by 4 is 1 or x

    and now I have 2 times 3x plus 15 over x...is that correct?
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  2. #2
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    Quote Originally Posted by Laura901 View Post
    Is this right? I'm dividing complex fractions....

    8x^2 divided by 4x
    over
    x^2 -25 divided by 3x+15

    I flipped the other fraction and changed the problem to multiplication

    8x^2 ovwer x^2 -25 divided by 3x +15 divided by 4x

    I factored x^2 -25 and got x+5 and x-5 and divided 8 by 4 and got two and 4 divided by 4 is 1 or x

    and now I have 2 times 3x plus 15 over x...is that correct?
    You are getting confused with constant and variables ( x). They cannot cancel each other out. Constants can only cancel out constants and variables can only cancel variables. Let me explain the procedure:

    \frac{8x^2}{4x} \div \frac{x^2-25}{3x+15}

    You correctly said to flip a fraction and multiply, hence:

    \frac{8x^2}{4x} \times \frac{3x+15}{x^2 - 25}

    Simplify the expressions. \frac{8x^2}{4x} can be simplied to 2x as the denominator has 4x^1 and rules of indices says \frac{x^a}{x^1} = x^{a-1} so we have \frac{8x^2}{4x} = 2x^{2-1} = 2x^1 = 2x (Notice, we only let constants work with constant \left(\frac84=2\right) and variables work with variable \left(\frac{x^2}{x} = x\right)). For \frac{3x+15}{x^2 - 25}, the numerator and the denominator can be factorised to get \frac{3(x+5)}{(x+5)(x-5)}. We have (x+5) on the numerator and the denominator so we can cancel it out to get \frac{3}{(x-5)} So we get:

    2x \times \frac{3}{(x-5)}

    For fractions, when we have to multiply, we multiply the numerator (top) together and we multiply the denominator (bottom) togather. So we get:

    \frac{2x}{1} \times  \frac{3}{(x-5)} = \frac{(3)(2x)}{(1)(x+5)} = \frac{6x}{x+5}
    Last edited by Simplicity; July 21st 2008 at 09:21 AM. Reason: Latex Corrections
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