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Math Help - linear system

  1. #1
    Senior Member euclid2's Avatar
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    linear system

    without graphing or using any algebraic methods determine whether or not there is a solution to the following

    3x+y=2
    6x-2y=3
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  2. #2
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    That is an utterly silly request. What's left, Astrology?

    Easiest determination, in my view 3(-2) - 6(1) = -6 - 6 = -12 this is not zero (0). There is a unique solution. However, addition and subtraction seem awfully "algebraic" to me.
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  3. #3
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    Quote Originally Posted by euclid2 View Post
    without graphing or using any algebraic methods determine whether or not there is a solution to the following

    3x+y=2
    6x-2y=3
    If you have asystem of linear equations like:

    \left|\begin{array}{l} a_1 x+b_1 y = c_1 \\a_2 x + b_2 y = c_2\end{array}\right.

    then there exist a unique solution if the determinant

    D=\left|\begin{array}{cc}a_1 & b_1 \\a_2 & b_2 \end{array} \right| \neq 0

    With your example D = -12 \neq 0 and therefore there must be an unique solution.
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  4. #4
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    3x+y=2
    6x-2y=3

    3x + y would be half of 6x-2y except for the operator. You are left with 6x=4 and 6x=3. Again, this does use algebraic concepts, but I just did it without the equations.
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  5. #5
    Senior Member euclid2's Avatar
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    i was thinking if you went
    (3x+y=2)2
    6x+2y=4
    6x-2y=3
    Since the X are the same when multiplied by two i thought there would be no solution
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  6. #6
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    thats close
    but if you end up with

    6x+2y=4
    6x-2y=3

    then one of the 6x would have to be -6x for them to cancel
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