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Math Help - Is this allowed?

  1. #1
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    Is this allowed?

    \frac{x^2-3x+2}{4x}\times\frac{12x^2}{x^2-2x}\times\frac{x}{x-1}

    \frac{(x-2)(x-1)}{4x}\times\frac{12x^2}{(x-1)(x+2)}\times\frac{x}{x-1}

    I know there are many things I can do here but am I allowed to cancel out all of the x-1 or can I only cancel out the first and the second, or the first and the third occurrence? Because after cancelling and reducing everything I got this:

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

    But wolfram alpha tells me the answer is 3x so what did I do wrong or miss?

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  2. #2
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    Re: Is this allowed?

    Quote Originally Posted by uperkurk View Post
    \frac{x^2-3x+2}{4x}\times\frac{12x^2}{x^2-2x}\times\frac{x}{x-1}

    \frac{(x-2)(x-1)}{4x}\times\frac{12x^2}{(x-1)(x+2)}\times\frac{x}{x-1}

    I know there are many things I can do here but am I allowed to cancel out all of the x-1 or can I only cancel out the first and the second, or the first and the third occurrence? Because after cancelling and reducing everything I got this:

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

    But wolfram alpha tells me the answer is 3x so what did I do wrong or miss?

    Thanks
    You've made one mistake in factoring the denominator of the 2nd fraction....
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  3. #3
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    Re: Is this allowed?

    Middle fraction,

    x^{2} - 2x is not equal to (x-1)(x+2).
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  4. #4
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    Re: Is this allowed?

    Thanks I see my mistake.
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  5. #5
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    Re: Is this allowed?

    Just a quick question without making a new topic.

    \frac{9x-9}{(x-3)(x-1)(x+3)} = \frac{9}{(x-3)(x+3)}

    What happened here? The book doesn't explain where the 9x or the (x-1) goes
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  6. #6
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    Re: Is this allowed?

    Quote Originally Posted by uperkurk View Post
    Just a quick question without making a new topic.

    \frac{9x-9}{(x-3)(x-1)(x+3)} = \frac{9}{(x-3)(x+3)}

    What happened here? The book doesn't explain where the 9x or the (x-1) goes
    \frac{9x-9}{(x-3)(x-1)(x+3)} = \frac{9(x-1)}{(x-3)(x-1)(x+3)}
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  7. #7
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    Re: Is this allowed?

    Quote Originally Posted by Plato View Post
    \frac{9x-9}{(x-3)(x-1)(x+3)} = \frac{9(x-1)}{(x-3)(x-1)(x+3)}
    omg I'm so stupid I feel so so stupid lol, these stupid mistakes happen when it gets late
    Last edited by uperkurk; July 28th 2013 at 12:45 PM.
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  8. #8
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    Re: Is this allowed?

    Thanks for this thread...
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