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Math Help - Hi not sure if this right area but really need help with this

  1. #1
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    Hi not sure if this right area but really need help with this

    ive not done maths for many years, been given 2 papers to learn. ive been trying for ages to work these out but reallt have no idea,
    can someone help me with this 1 and then i will try second on own..

    Ive uploaded the paper via PDF file please help me







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  2. #2
    Junior Member Barioth's Avatar
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    Re: Hi not sure if this right area but really need help with this

    Hi, I'll help you with the Task 5.

    x^2 - 2xy +4y = -60\\ x+y = 18 \\ y = 18-x
    from there on we'll replace every y with (18-x)
    x^2-2x(18-x)+4(18-x)=-60\\x^2 -2x*18-2x*(-x)+4*18+4*(-x)\\ = x^2 -36x +2x^2+72-4x\\=3x^2-40x+72 = -60

    Then
    3x^2-40x+132 = 0

    From there on you can use the quadratic formula

    end up with  x = 6 \\ x=\frac{22}{3}

    as for the graph part, Draw the 2 fonction they'll meet in these value of X. (*Unless I messed up*)
    Last edited by Barioth; January 17th 2013 at 10:58 AM.
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  3. #3
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    Re: Hi not sure if this right area but really need help with this

    thank you so much for your time and trying to help.. could you possible explain each step and how you do it.. that way i can learn and possible do the other paper myself
    thanks again
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  4. #4
    Junior Member Barioth's Avatar
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    Re: Hi not sure if this right area but really need help with this

    mhm after replacing y, I just added everything together. then I used the quadratic formula.

    The quadratic formula is
     \frac{-b(\pm)\sqrt(b^2-4ac)}{2a}
    for  ax^2+bx+c =0

    Maybe you could tell me witch part you didn't understand?
    Last edited by Barioth; January 17th 2013 at 11:20 AM.
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  5. #5
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    Re: Hi not sure if this right area but really need help with this

    Quote Originally Posted by Barioth View Post
    mhm after replacing y, I just added everything together. then I used the quadratic formula.

    The quadratic formula is
     \frac{-b(+-)\sqrt(b^2-4ac)}{2a}
    for  ax^2+bx+c =0

    Maybe you could tell me witch part you didn't understand?

    [TEX]\pm[/TEX] gives \pm.


    Also [TEX] \frac{-b\pm)\sqrt{b^2-4ac}}{2a}[/TEX] gives  \frac{-b\pm\sqrt{b^2-4ac}}{2a}
    Last edited by Plato; January 17th 2013 at 11:37 AM.
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    Re: Hi not sure if this right area but really need help with this

    Thanks from Barioth
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