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Simplifying algebraic fractions (c3) need help!

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    (Original post by lollage123)
    This is the part which did my head in. Does noticing this sort of thing just become a knack after time? As all I did was just blindly carry on (and wrongly cancel down)..
    In questions where you know the answer look to working towards it

    For example this one ... you knew what the numerator need to cancel to so you needed to change the one you had to the answer * (x-1)


    The fact that you cancelled as you did is just poor maths

    Consider this \dfrac{2+3}{4} = 1\frac{1}{4}

    You have effectively cancelled the 2 and the 4 giving

    \dfrac{2+3}{4} = \dfrac{1+3}{2} = 2

    This is clearly not possible but is what you did when you cancelled your (x-1)s
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    It does, fortunately.
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    (Original post by lollage123)
    This is the part which did my head in. Does noticing this sort of thing just become a knack after time? As all I did was just blindly carry on (and wrongly cancel down)..

    Thank you all for your help
    If it is a knack then the knack is in reading the question - the question says show that [...] = something divided by (x+3) and you have something divided by (x+3)(x-1) therefore (unless the question wrong) the (x-1) must cancel. As this is not initially obvious but one term contains (x-1) then the other must as well. It is then merely showing this.

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Updated: July 31, 2012
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