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Math Help - Solving the System

  1. #1
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    Solving the System

    A + C = 3 (1)
    11 = A e^(B(4)) + C (2)
    22 = (Ae^(4B)-A)/B + 4C (3)

    I pretty much only know how you would approach this by substitution on.

    i keep getting massive answers with ln everywhere. Im not getting anywhere.

    Is there a more logical method or do i just need to practice with my simplifying in MAD MATPS
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  2. #2
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    Re: Solving the System

    Quote Originally Posted by asaver View Post
    A + C = 3 (1)
    11 = A e^(B(4)) + C (2)
    22 = (Ae^(4B)-A)/B + 4C (3)

    I pretty much only know how you would approach this by substitution on.

    i keep getting massive answers with ln everywhere. Im not getting anywhere.

    Is there a more logical method or do i just need to practice with my simplifying in MAD MATPS
    Equation 1: \displaystyle A+C=3 \implies C=3-A.

    Substituting into Equation 2:

    \displaystyle \begin{align*} 11 &= Ae^{4B} + C \\ 11 &= Ae^{4B} + 3 - A \\ 8 &= Ae^{4B} - A \\ 8 &= A\left(e^{4B} - 1\right) \\ \frac{8}{e^{4B}-1} &= A \end{align*}

    Substituting into Equation 3:

    \displaystyle \begin{align*} 22 &= \frac{Ae^{4B} - A}{B} + 4C \\ 22 &= \frac{A\left(e^{4B}-1\right)}{B} + 4C \\ 22 &= \frac{\frac{8}{e^{4B} - 1}\left(e^{4B} - 1\right)}{B} + 4C \\ 22 &= \frac{8}{B} + 4C \\ 22 &= \frac{8}{B} + 4\left(3-A\right) \\ 22 &= \frac{8}{B} + 12 - 4A \\ 10 &= \frac{8}{B} - 4A \\ 10 &= \frac{8}{B} - 4\left(\frac{8}{e^{4B}-1}\right) \\ 10 &= \frac{8}{B} - \frac{32}{e^{4B}-1} \end{align*}

    From here, get a common denominator and then use an iterative method to solve for \displaystyle B.
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