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Math Help - Simultaneous Equation: Help please!

  1. #1
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    I need help to find the values of "p" & "c" if:

    pc=120
    (p+2)(c-3)=120

    My first step was to expand the brackets to get:

    pc=120
    pc-6-3p+2c=120

    Which simplifies to:

    pc=120 A
    pc-3p+2c=126 B

    I then take A from B which leaves:

    -3p+2c=6

    I am not sure what to do from this point nor am i confident it is correct up to this point. I would be very greatful for any pointers.

    Thanks in advance
    Last edited by mr fantastic; January 28th 2009 at 12:16 PM. Reason: Deleted question restored
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  2. #2
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    Simultaneous equations

    Hello tomh381

    Your working is perfect up to this point. Thanks for showing us what you've done so far.

    All you need to do now is to make c the subject of your last equation:

    c = \tfrac{1}{2}(6 + 3p) (1)

    and substitute into your equation
    A to eliminate c. You'll then get a quadratic equation in p, which will factorise to give you two values. Substitute back into equation (1) here, and you're done.

    Can you complete this now?

    Grandad
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  3. #3
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    I am not sure I can, this is what i have so far:

    cp=120 A
    c=(6+3p)/2 B

    I then minus B from A to leave:

    p=120-((6+3p)/2)

    I am unsure on what to do from here!

    Thanks_381
    Last edited by mr fantastic; January 28th 2009 at 12:17 PM. Reason: Deleted question restored
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  4. #4
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    cp=120 --------- Eqn 1
    c=(6+3p)/2 ----- Eqn 2

    Sub Eqn 2 into Eqn 1

    (6+3p)/2 )p=120
    (6p+3p^2)/2=120
    6p+3p^2 = 240
    3p^2+6p-240=0

    Can you take it from here and solve for p. Hope it helps.

    Quote Originally Posted by tomh381 View Post
    I am not sure I can, this is what i have so far:

    cp=120 A
    c=(6+3p)/2 B

    I then minus B from A to leave:

    p=120-((6+3p)/2)

    I am unsure on what to do from here!

    Thanks_381
    Follow Math Help Forum on Facebook and Google+

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