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Math Help - Simultaneous Eqn 4 unknowns

  1. #1
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    Unhappy Simultaneous Eqn 4 unknowns

    Hi..

    I have the following problem...

    -207 x 10^6 = A - B/0.01

    0 = C - D/0.04

    A - B/0.0225 = C - D/0.0225

    C + D/0.0225 - ( A + B/0.0225 ) = 138 x 10^6


    I've stick on this problem for some time now, so any help would be really helpful..

    Regards...
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  2. #2
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    Quote Originally Posted by Bailey View Post
    Hi..

    I have the following problem...

    -207 x 10^6 = A - B/0.01

    0 = C - D/0.04

    A - B/0.0225 = C - D/0.0225

    C + D/0.0225 - ( A + B/0.0225 ) = 138 x 10^6


    I've stick on this problem for some time now, so any help would be really helpful..

    Regards...
    Rewrite the system of equations:

    \left|\begin{array}{rcl}-207 \cdot 10^6 &=& A-100B \\ 0&=&C-25D \\ A-\dfrac{400}{9} B &=& C-\dfrac{400}{9} D \\ C+\dfrac{400}{9} D - \left(A+\dfrac{400}{9}B \right)&=& 138 \cdot 10^6\end{array}\right.

    1. Calculate A from the 1st equation
    2. Calculate C from the 2nd equation
    3. Plug in these terms into the 3rd equation. Calculate D from this equation.
    4. Plug in A and C from the first two equations into equation #4. Plug in D from #3 into this new equation #4 which now contains only the variable A. Calculate A. (28,750,000)
    5. Resubstitute this value to calculate D. (3,910,000)
    6. Resubstitute these values to calculate C (97,750,000) and B (2,357,500)
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  3. #3
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    Awsome !!!

    I have a few more problems that are similar to this, so i may be back. Again thank you !!!!
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  4. #4
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    Me again...

    How did you manage to calc A from Eqn 1 ??

    Your answers right on the money & your maths is obviously far better than mine !!

    Cheers

    Bailey
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  5. #5
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    Quote Originally Posted by Bailey View Post
    Me again...

    How did you manage to calc A from Eqn 1 ??

    Your answers right on the money & your maths is obviously far better than mine !!

    Cheers

    Bailey
    Add 100B on both sides of the equation:

    -207 \cdot 10^6 = A-100B~\implies~-207 \cdot 10^6 +100B= A

    Maybe the -207 \cdot 10^6 is a little bit confusing. But that's nothing but a constant number

    -207 \cdot 10^6 = -207,000,000
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  6. #6
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    Hi..

    I can solve for A & C but when i substitute these into eqn 3 to calc D, I seem to be making a mistake..I still have more than 1 vairable left in the eqn..

    Cheers,,,
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  7. #7
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    Quote Originally Posted by Bailey View Post
    Hi..

    I can solve for A & C but when i substitute these into eqn 3 to calc D, I seem to be making a mistake..I still have more than 1 vairable left in the eqn..

    Cheers,,,

    I think it's the best I publish my complete calculations, add a few comments so you can use my notes to control your own calculations.

    <br />
\left|\begin{array}{rcl}-207 \cdot 10^6 &=& A-100B \\ 0&=&C-25D \\ A-\dfrac{400}{9} B &=& C-\dfrac{400}{9} D \\ C+\dfrac{400}{9} D - \left(A+\dfrac{400}{9}B \right)&=& 138 \cdot 10^6\end{array}\right.<br />

    -207 \cdot 10^6 = A-100B ~\implies~ \boxed{A=-207 \cdot 10^6+100B}

    0=C-25D ~\implies~\boxed{C=25D}

    Plug in the terms of A and C into equation #3:

     -207 \cdot 10^6+100B - \dfrac{400}{9} B = 25D - \dfrac{400}{9} D  ~\implies~ -207,000,044.444B = - \dfrac{175}9 D ~\implies~ \boxed{D = 10,645,716.57 B }

    Therefore

    \boxed{C=266,142,914.3 B}

    Now the variables A, C and D are expressed by the variable B. Plug in these terms into the equation #4:

    266,142,914.3 B+\dfrac{400}{9} 10,645,716.57 B - \left(-207 \cdot 10^6+100B+\dfrac{400}{9}B \right)= 138 \cdot 10^6

    This equation contains B only. Collect all coefficients of B and calculate the value of B. Then re-substitute to get A, C, D.
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  8. #8
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    Smile

    Thanks very much,, thats excellent

    You are a math god !!!
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  9. #9
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    Hi..

    I've tried & tried but i still can't solve for B.. Could you please show me the transposition of the formula so I get to grips with it...

    I know i keep saying this, but again thanks for the help

    Regards
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  10. #10
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    Quote Originally Posted by Bailey View Post
    Hi..

    I've tried & tried but i still can't solve for B.. Could you please show me the transposition of the formula so I get to grips with it...

    I know i keep saying this, but again thanks for the help

    Regards
    <br />
266,142,914.3 B+\dfrac{400}{9} 10,645,716.57 B - \left(-207 \cdot 10^6+100B+\dfrac{400}{9}B \right)= 138 \cdot 10^6<br />

    Collect all terms containing B at the LHS oof the equation, the constants at the RHS (I'm using rounded decimal numbers now!):

    <br />
266,142,914.3 B+473,142,958.7 B + 207 \cdot 10^6-100B-44.4444B= 138 \cdot 10^6<br />

    <br />
739,285,728.6B= -69,000,000~\implies~\boxed{B = -0.9333333}<br />

    hmmmm ... I know now why you don't have any success with my calculations. Somewhere in my previous posts is one of those silly mistakes which happen quite often to me. .... I'm going to check my previous posts and when I've found my error I'll be back.
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  11. #11
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    OK, once again:

    equ. 1:

    [math-207\cdot 10^6=A-\dfrac B{0.01}~\implies~ A=100B-207,000,000[/tex]

    equ. 2:
    0=C-\dfrac D{0.04}~\implies~ C=25D

    equ. 3:
    A-\dfrac B{0.0225}=C - \dfrac D{0.0225} Plug in A and C:

    100B-207,000,000-\dfrac B{0.0225}=25D - \dfrac D{0.0225}~\implies~D=-\dfrac{20}7 B+ \dfrac{74,520,000}7

    and consequently C = 25 \left(-\dfrac{20}7 B+ \dfrac{74,520,000}7 \right) = -\dfrac{500}7 B +\dfrac{1,863,000,000}7

    Now plug in into equ. 4 the Terms for A, C and D which only depends on B:

    C+\dfrac D{0.0225}-\left(A+\dfrac B{0.0225}\right)=138 \cdot 10^6

    \left( -\dfrac{500}7 B +\dfrac{1,863,000,000}7  \right)+\dfrac {-\dfrac{20}7 B+ \dfrac{74,520,000}7}{0.0225} -\left(100B-207,000,000+\dfrac B{0.0225}\right)=138 \cdot 10^6

    \dfrac{2400(2760000-B)}7 = 138,000,000

    2760000 - B = \dfrac7{2400}\cdot 138,000,000= 402500

    -B= -2,357,500

    \boxed{B= 2,357,500}

    Uff!

    Re-substitute this value at the equations for A, C and D.

    Remark: I did the complete calculations again because I wasn't able to find where I slipped at my first attempt. But now I guess everything is OK.
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