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Math Help - I little bit of help needed

  1. #1
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    I little bit of help needed

    I have a fraction that needs simplifying and I think that I have it right, I just want to make sure my method is right.

    [anything that is underlined is canceled]

    The fraction is: (2x^2+x-6)/(2x+4)

    My simplification is: (3x+3)/2

    My method is: (2x^2+x-6)/(2x+4) >> (2x-3)(x-2)/2x+4 >> (2x-3)(x+2)/(2x+4) >> (3x+3)/2

    thank you in advance!!!!

    Drakmord
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by drakmord View Post
    I have a fraction that needs simplifying and I think that I have it right, I just want to make sure my method is right.

    [anything that is underlined is canceled]

    The fraction is: (2x^2+x-6)/(2x+4)

    My simplification is: (3x+3)/2

    My method is: (2x^2+x-6)/(2x+4) >> (2x-3)(x-2)/2x+4 >> (2x-3)(x+2)/(2x+4) >> (3x+3)/2

    thank you in advance!!!!

    Drakmord
    how you make this (2x^2+x-6)/(2x+4) >> to this (2x-3)(x-2)/2x+4 ??

    how you cancel like this (2x-3)(x+2)/(2x+4) >> (3x+3)/2
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by drakmord View Post
    I have a fraction that needs simplifying and I think that I have it right, I just want to make sure my method is right.

    [anything that is underlined is canceled]

    The fraction is: (2x^2+x-6)/(2x+4)

    My simplification is: (3x+3)/2

    My method is: (2x^2+x-6)/(2x+4) >> (2x-3)(x-2)/2x+4 >> (2x-3)(x+2)/(2x+4) >> (3x+3)/2

    thank you in advance!!!!

    Drakmord
    You can't cancel like that, you cannot cancel where addition is involved.

    In this case 2x^2+x-6 = (2x-3)(x+2) and 2x+4 = 2(x+2)

    This means your fraction becomes \frac{2x^2+x-6}{2x+4} = \frac{(2x-3)(x+2)}{2(x+2)} = \frac{2x-3}{2} = x - \frac{3}{2}

    As (x+2) is on the top and the bottom it may be cancelled.
    Last edited by e^(i*pi); July 19th 2009 at 05:38 AM. Reason: because I'm an idiot and cancelled the wrong factor
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  4. #4
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    Ah.....I see, that is why my answer made so much trouble for me, I did not work it out properly.

    Let me make sure I have this straight,what you are saying is:
    (2x^2+x-6)/(2x+4) >> (2x-3)(x+2)/2(x+2) NOT "(2x-3)(x-2)/2x+4" like I said before. Therefore, (2x-3)(x+2)/2(x+2) >> (2x-3)(x+2)/2(x+2) >> (2x-3)/2 >> x-3.
    Is that correct?

    Thanks again.

    Drakmord
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  5. #5
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    Quote Originally Posted by drakmord View Post
    Ah.....I see, that is why my answer made so much trouble for me, I did not work it out properly.

    Let me make sure I have this straight,what you are saying is:
    (2x^2+x-6)/(2x+4) >> (2x-3)(x+2)/2(x+2) NOT "(2x-3)(x-2)/2x+4" like I said before. Therefore, (2x-3)(x+2)/2(x+2) >> (2x-3)(x+2)/2(x+2) >> (2x-3)/2 >> x-3.
    No, your very last step:
    \frac{2x - 3}{2} = x - 3
    is not correct.

    It should be this:
    \frac{2x - 3}{2} = \frac{2x}{2} - \frac{3}{2} = x - \frac{3}{2}.


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  6. #6
    Senior Member Stroodle's Avatar
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    Quote Originally Posted by drakmord View Post
    Ah.....I see, that is why my answer made so much trouble for me, I did not work it out properly.

    Let me make sure I have this straight,what you are saying is:
    (2x^2+x-6)/(2x+4) >> (2x-3)(x+2)/2(x+2) NOT "(2x-3)(x-2)/2x+4" like I said before. Therefore, (2x-3)(x+2)/2(x+2) >> (2x-3)(x+2)/2(x+2) >> (2x-3)/2 >> x-3.
    Is that correct?

    Thanks again.

    Drakmord
    That's correct, except that \frac{2x-3}{2}=\frac{2x}{2}-\frac{3}{2}=x-\frac{3}{2}

    Edit* yeongil beat me to it
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