Algebraic Fraction Help.

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    Hi, would somebody be able to demonstrate how these type of questions are done. I don't mean to be needy, but could you annotate so I can get a proper understanding.

    Thanks!

    Simplify  \frac{6(x+5)^2}{2 (x+5)}

    Simplify  \frac{x^2 - 9}{x^2 + 3x}
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    You look for common factors

    With ordinary numbers you know that

    \dfrac{12}{42} = \dfrac{6\times2}{6\times7} = \dfrac{2}{7}


    You know that you can cancel the 6 x because it is in the numerator and the denominator



    In the same way

    \dfrac{(x+2)(x-5)}{(x+2)(2x+1)}

    has (x+2) multiplying in both the numerator and denominator so it can be cancelled

    \dfrac{(x+2)(x-5)}{(x+2)(2x+1)} = \dfrac{(x+2)\times(x-5)}{(x+2)\times(2x+1)} = \dfrac{x-5}{2x+1}
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    First one, cancel out common factor of x + 5

    Second one factorise the numerator(difference of two squares) and the denominator and then cancel out the common factor of x + 3
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    (Original post by TenOfThem)
    You look for common factors

    With ordinary numbers you know that

    \dfrac{12}{42} = \dfrac{6\times2}{6\times7} = \dfrac{2}{7}


    You know that you can cancel the 6 x because it is in the numerator and the denominator



    In the same way

    \dfrac{(x+2)(x-5)}{(x+2)(2x+1)}

    has (x+2) multiplying in both the numerator and denominator so it can be cancelled

    \dfrac{(x+2)(x-5)}{(x+2)(2x+1)} = \dfrac{(x+2)\times(x-5)}{(x+2)\times(2x+1)} = \dfrac{x-5}{2x+1}
    (Original post by Mr M)
    First one, cancel out common factor of x + 5

    Second one factorise the numerator(difference of two squares) and the denominator and then cancel out the common factor of x + 3
    For the first one, does that become   \frac{6(x+5)}{2}
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    (Original post by TheMysteryMan)
    For the first one, does that become   \frac{6(x+5)}{2}
    Apart from the fact that 2 goes into 6, yes
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    (Original post by TheMysteryMan)
    For the first one, does that become   \frac{6(x+5)}{2}
    Yes, but it can be made even simpler. Expand the brackets and try to simplify further.
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    (Original post by CharlieBoardman)
    Yes, but it can be made even simpler. Expand the brackets and try to simplify further.
     \frac{6(x+5)}{2} = \frac{6x+30}{2}
    3x+15 ?
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    (Original post by TheMysteryMan)
     \frac{6(x+5)}{2} = \frac{6x+30}{2}
    3x+15 ?
    Correct
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    Remember to exclude values of x that make the denominator zero - from the domain:
    first one: x =/= -5
    second one: x =/= 0, -3 in your final answer!
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    (Original post by TheMysteryMan)
     \frac{6(x+5)}{2} = \frac{6x+30}{2}
    3x+15 ?
    Yes but I would suggest dividing the 6 by 2 would be a better move.
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    (Original post by tomctutor)
    Remember to exclude values of x that make the denominator zero - from the domain:
    first one: x =/= -5
    second one: x =/= 0, -3 in your final answer!
    At this level, that is not necessary.
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    (Original post by tomctutor)
    Remember to exclude values of x that make the denominator zero - from the domain:
    first one: x =/= -5
    second one: x =/= 0, -3 in your final answer!
    You lost me
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    For the 2nd one, am I factorising top and bottom first?
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    (Original post by TheMysteryMan)
    You lost me
    don't worry ... he was answering a more complex question
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    (Original post by TheMysteryMan)
    You lost me
    Don't worry, you do not need to know that yet.
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    (Original post by TenOfThem)
    don't worry ... he was answering a more complex question

    (Original post by CharlieBoardman)
    Don't worry, you do not need to know that yet.
    Ok, would you be able to explain the 2nd one a little more to me, not sure where to start. Am I factorising out the top and bottom?
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    (Original post by TheMysteryMan)
    Ok, would you be able to explain the 2nd one a little more to me, not sure where to start. Am I factorising out the top and bottom?
    Yes, but try to look for the 'difference of two squares' somewhere. Once the numerator and denominator have been factorised correctly, they should have a term in common which you will be able to cancel.
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    I don't think i've been taught that, only the quadratic formula, factorising and graphs.
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    (Original post by TheMysteryMan)
    I don't think i've been taught that, only the quadratic formula, factorising and graphs.
    You've not been taught the difference of two squares?

    x^2-4=x^2-2^2=(x+2)(x-2)

    Hence difference of squares?
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    Is the answer -3?
 
 
 
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