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Maths C3 - Cancelling down Algebraic Fractions...Help with Q? Watch

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    So I'm stuck on Exercise 1A Question q from the Edexcel C3 Modular textbook.

    \frac{3x^2-x-2}{\frac{1}{2}x+\frac{1}{3}}
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    (Original post by Philip-flop)
    So I'm stuck on Exercise 1A Question q from the Edexcel C3 Modular textbook.
    Could you please post the question? Most of us don't have access to the textbook.

    Edit: I see you have done that now. Just to check, on the bottom is it \frac{1}{2x}?
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    (Original post by rayquaza17)
    Could you please post the question? Most of us don't have access to the textbook.

    Edit: I see you have done that now. Just to check, on the bottom is it \frac{1}{2x}?
    I'm not quite sure how to type this properly. But the numbers in the denominator of the main fraction are fractions themselves e.g. 1x/2 is a half, 1/3 is a third...
    \frac{3x^2-x-2}{1x/2+1/3}

    Edit: \frac{3x^2-x-2}{\frac{1x}{2}+\frac{1}{3}}
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     \frac{3x^2 - x - 2}{\frac{1}{2}x +\frac{1}{3}}

    First you want to get rid of the 2 and 3, so that it's a normal looking fraction, so multiply top and bottom by 2\times 3, put the bottom level of the fraction over a common denominator first if that makes it easier.
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    Ok so I managed to write the question out properly to Exercise 1A Question q from the C3 Modular textbook..

    \frac{3x^2-x-2}{\frac{1}{2}x+\frac{1}{3}}
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    (Original post by NotNotBatman)
     \frac{3x^2 - x - 2}{\frac{1}{2}x +\frac{1}{3}}

    First you want to get rid of the 2 and 3, so that it's a normal looking fraction, so multiply top and bottom by 2\times 3, put the bottom level of the fraction over a common denominator first if that makes it easier.
    Thank you. I will attempt to follow your steps now
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    (Original post by rayquaza17)
    Could you please post the question? Most of us don't have access to the textbook.

    Edit: I see you have done that now. Just to check, on the bottom is it \frac{1}{2x}?
    No I finally fixed my question. I meant it like this...
    \frac{3x^2-x-2}{\frac{1}{2}x+\frac{1}{3}}
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    (Original post by Philip-flop)
    No I finally fixed my question. I meant it like this...
    \frac{3x^2-x-2}{\frac{1}{2}x+\frac{1}{3}}
    Multiply top and bottom by 6 giving  \displaystyle \frac{6(3x^2-x-2)}{3x+2} .
    What can you do from here?
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    I'm still stuck...

    I ended up with...
    \frac{(3x-2)(x+1)}{\frac{1}{6}(3x+2)}
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    (Original post by B_9710)
    Multiply top and bottom by 6 giving  \displaystyle \frac{6(3x^2-x-2)}{3x+2} .
    What can you do from here?
    Oh yeah, I think I'm following now. How did you know whether to multiply top and bottom by 6? I'm terrible at LCM's
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    (Original post by Philip-flop)
    Oh yeah, I think I'm following now. How did you know whether to multiply top and bottom by 6? I'm terrible at LCM's
    Gets rid of the half and third in the denominator. Horrible with fractions in the denominator.
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    I managed to get the answer now. Thank you everyone so much.

    My next steps were..
     \displaystyle \frac{6(3x^2-x-2)}{3x+2}

    \frac{6(3x+2)(x-1)}{3x+2}

    6(x-1)
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    Yeah I realised that I factoried the numerator incorrectly that's when I managed to completely figure it out.

    Thank you everyone that helped. I feel so silly for such a noob mistake!!
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    6X-6, I believe.
 
 
 
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