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Math Help - Why some solutions of system of linear equations create contradictions?

  1. #1
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    Why some solutions of system of linear equations create contradictions?

    Lets take a simple system of two equations:
    \begin{cases} a_{1}*q^{2}-a_{1}=15 \\  a_{1}*q^{3}-a_{1}*q=6 \end{cases}

    after evaluating a_{1} from the first equation and putting it into the second we get
    15q^3-6q^2-15q+6=0

    this gives
    (15q-6)(q-1)(q+1)=0
    so, solutions are: q=1, q=-1 and q= \frac{2}{5}

    but if you try to put this into the original system of equations, only q= \frac{2}{5} works, why?

    thanks
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  2. #2
    Super Member Aryth's Avatar
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    You're first evaluation is incorrect... If you solve the first equation for a_1 you get:

    a_1(q^2 - 1) = 15

    a_1 = \frac{15}{q^2 - 1}

    Plug this into the second equation to get:

    \frac{15}{q^2 - 1}q^3 - \frac{15}{q^2 -1}q = 6

    \frac{15q}{q^2 - 1}(q^2 - 1) = 6

    15q = 6

    q = \frac{2}{5}

    That's why the other answers don't work, because they are not solutions to the system.
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  3. #3
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    <blushing> Thanks
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  4. #4
    Super Member Aryth's Avatar
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    No problem.
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