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    How do you solve these?!?!

    1) y + 4 = [-1/[4x(x-2)(x+2)]] [x-1]

    2) [-160/v^2] + [v/50] = 0

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    (Original post by Blush Babe)
    How do you solve these?!?!

    1) y + 4 = [-1/[4x(x-2)(x+2)]] [x-1]

    2) [-160/v^2] + [v/50] = 0

    BB

    errr............................ ...????
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    (Original post by xAngelx)
    errr............................ ...????
    It's AS pure 1 maths
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    (Original post by Blush Babe)
    How do you solve these?!?!

    1) y + 4 = [-1/[4x(x-2)(x+2)]] [x-1]

    2) [-160/v^2] + [v/50] = 0

    BB
    well,
    1) this will be only valid for x=1, for which, y=-4;

    2) taking lcm.....(8000+v^3)/(50*v^3) = 0
    So, v^3=-8000.
    so, v=-20
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    (Original post by prince_capri)
    well,
    1) this will be only valid for x=1, for which, y=-4;

    2) taking lcm.....(8000+v^3)/(50*v^3) = 0
    So, v^3=-8000.
    so, v=-20


    er............i'm still trying to read it
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    (Original post by Blush Babe)
    How do you solve these?!?!

    1) y + 4 = [-1/[4x(x-2)(x+2)]] [x-1]

    2) [-160/v^2] + [v/50] = 0

    BB
    In the 1st number, I think you have to use the method in solving for line equations. In the second, since there is an exponent in the variable, use the quadratic equation. Hope that helps!
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    (Original post by prince_capri)
    well,
    1) this will be only valid for x=1, for which, y=-4;

    2) taking lcm.....(8000+v^3)/(50*v^3) = 0
    So, v^3=-8000.
    so, v=-20
    Cheers , except it's a wee bit too late cuz I just did my maths exam (and failed)...the first question was so easy it got my hopes up, until I turned to next page.
    BB
 
 
 
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