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Thread: simultaneous equations

  1. #1
    Junior Member
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    Smile simultaneous equations

    How would you go about solving:

    6 = e^(1+b) + B
    2 = e^b +B

    where e = euler's number, not an unknown

    Is it something to do with subtracting? I just keep getting myself into a dead end.

    Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by freswood
    How would you go about solving:

    6 = e^(1+b) + B
    2 = e^b +B

    where e = euler's number, not an unknown

    Is it something to do with subtracting? I just keep getting myself into a dead end.

    Thanks
    To solve:

    $\displaystyle
    6=e^{1+b}+B$
    $\displaystyle
    2=e^b+B
    $

    you rewrite them as:

    $\displaystyle
    6=e \times e^{b}+B$
    $\displaystyle
    2=e^b+B
    $,

    which leaves you with a pair of linear simultaneous equations in $\displaystyle e^b$ and $\displaystyle B$,
    which can be solved in the usual manner.

    RonL
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  3. #3
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    How would you finish off what you've started?
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  4. #4
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    Quote Originally Posted by freswood
    How would you finish off what you've started?
    $\displaystyle
    6=e \times e^{b}+B$
    $\displaystyle
    2=e^b+B
    $

    Subtract:

    $\displaystyle
    6-2=(e-1) \times e^{b}+B-B$,

    so:

    $\displaystyle
    e^b=\frac{4}{e-1}
    $,

    and substituting back into the second equation:

    $\displaystyle
    B=2-\frac{4}{e-1}
    $.

    Finally:

    $\displaystyle
    b=\log_e(4/(e-1))
    $

    RonL
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  5. #5
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    Thanks so much! It kinda confused me because I wasn't expecting the answer to be that messy. Whole numbers all the way!
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  6. #6
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    Hello, freswood!

    Solve: .$\displaystyle \begin{array}{cc}(1)\;e^{b+1} + B \:= \:6 \\ (2) \;e^b +B \:= \:2\end{array}$
    This is a variation of Captain Black's solution . . .

    We have: .$\displaystyle \begin{array}{cc}e^{b+1}\:=\:6 - B \\ e^b\:=\:2 - B\end{array}$

    Divide the equations: .$\displaystyle \frac{e^{b+1}}{e^b}\;=\;\frac{6 - B}{2 - B}\quad\Rightarrow\quad e\;=\;\frac{6 - B}{2 - B}$

    . . Then: .$\displaystyle e(2 - B)\:=\:6 - B\quad\Rightarrow\quad 2e - Be\:=\:6 - B$

    . . And: .$\displaystyle 2e - 6 \:=\:Be - B\quad\Rightarrow\quad 2(e - 3)\:=\:B(e - 1)$

    . . Therefore: .$\displaystyle B\:=\:\frac{2(e-3)}{e-1}$


    Substitute into (2): .$\displaystyle e^b - \frac{2(e-3)}{e - 1}\:=\:2\quad\Rightarrow\quad e^b \:=\:2 - \frac{2(e - 3)}{e - 1}$

    . . . $\displaystyle e^b\:=\:\frac{2(e - 1) - 2(e - 3)}{e - 1} \quad\Rightarrow\quad e^b \:=\:\frac{4}{e - 1}$

    . . Therefore: .$\displaystyle b\:=\:\ln\left(\frac{4}{e-1}\right)$
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