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  1. #1
    Member helloying's Avatar
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    Simply help.

    The question is simplfy:  \frac{t^2-9}{6-2t} \times \frac{6}{t-3}


    i simply till \frac{3(t-3)}{3-t)} then i mulitpy by -1 for both denominator and numerator so i get -3 in the end. is this correct? the ans given is  \frac{3(t-3)}{3-t)} . but i can get -3. which shld be the correct ans?
    Last edited by CaptainBlack; August 31st 2009 at 01:21 AM.
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  2. #2
    MHF Contributor red_dog's Avatar
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    \frac{t^2-9}{6-2t}\cdot\frac{6}{t-3}=\frac{6(t-3)(t+3)}{-2(t-3)^2}=\frac{3(t+3)}{3-t}
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by helloying View Post
    The qn is simply  \frac{t^2-9}{6-2t} \times \frac{6}{t-3}

    i simply till \frac{3(t-3)}{3-t)} then i mulitpy by -1 for both denominator and numerator so i get -3 in the end. is this correct? the ans given is \frac{3(t-3)}{3-t)}. but i can get -3. which shld be the correct ans?
     \frac{t^2-9}{6-2t} \times \frac{6}{t-3}=\frac{(t-3)(t+3)}{2(3-t)}\times \frac{2 \times 3}{t-3}

    Now cancel where possible to get the answer.

    CB
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  4. #4
    Senior Member
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    Hello helloying!

    Quote Originally Posted by helloying View Post
    The qn is simply  \{frac{\t^2-9}{6-2t} x \frac{\6}{t-3}[\math]<br /> <br />
i simply till \frac{\3(t-3)}{3-t)}  then i mulitpy by -1 for both denominator and numerator so i get -3 in the end. is this correct? the ans given is \frac{3(t-3)}{3-t)}. but i can get -3. which shld be the correct ans?
    Edit: whoops - wasn't the right the solution, Sorry!

    Rapha
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