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Math Help - Systems of linear equations

  1. #1
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    Systems of linear equations

    I need to determine a value of K for which the system has no solutions

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

    So,

    y=-3/2x
    k=10/y
    x=-3/2
    y=6

    Therefore, K=-5/3

    With all these values, I found that the system has a unique solution of K=-5/3 at the point (-3/2,-6)

    How would I go about finding a value of K for which there is no solution? Is it just any value on the real number line that does not equal -5/3?
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  2. #2
    MHF Contributor
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    Talking

    Quote Originally Posted by cognoscente View Post
    I need to determine a value of K for which the system has no solutions

    2x-y=3
    4x+ky=4
    Solving the first equation for "y=" and substituting, you should get:

    . . . . .4x + 2kx - 3k = 4

    Solving for "x=", you should get:

    . . . . .x = (4 + 3k) / [2(2 + k)]

    For which value of k does this expression not provide a valid numerical value?
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  3. #3
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    -2 is undefined.
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