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Math Help - Solving system of linear equations

  1. #1
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    Solving system of linear equations

    Find the parameter m in order that the system of linear equations:

    x+y+z=2
    2x+3y+2z=5
    2x+3y+(2m-1)z=2m+1

    a) has one unique solution
    b) has no solutions
    c) has infinitely many solutions

    Thank you
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  2. #2
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    Okay, what have you done? What happens if you try to solve x, y, and z? I suspect that you will get x, y, or z, whichever you solve for first, equal to a fraction. There will be one solution if the denominator is not 0, no solution if the denominator is 0 but the numerator is not, and an infinite number of solutions if both numerator and denominator are 0 . What values of a make each of those things happen?

    (Look closely at the last two equations. Tthere is an obvious first step that will simplify things very fast!)
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  3. #3
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    Hi HallsofIvy,

    Using matrices, I came to an equivalent system of linear equations and found that for m=3/2 the system is impossible. For any other value other than m=3/2 the equation has an unique solution. (I don't know if this statement is completely right, and if it's an answer to the first two questions).

    I don't know how to find a value of m in order that the system can have infinitely many solutions.

    If you can write me the complete solution and required information, I would be really grateful.

    Thank you.
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  4. #4
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    Then you have done very well.

    If you subtract the second equation from the third, both x and y are eliminated and you have
    (2m- 3)z= 2m- 4

    If m 2m- 3 is NOT 0, that is, if m is NOT 3/2, then you can solve for z by dividing- if m is not 3/2, then there exist a single solution for z. If m= 3/2, the the equation becomes 0z= 3- 4= -1 which is impossible. There is no solution.

    Because there is no m in either of the first two equations, if m is not 3/2, we can put z= (2m-4)/(2m-3) into the other two equations and find no new restriction on x and y. There is NO value of m for which there are an infinite number of solutions.

    (If the equations had reduced to, say, (m-2)(m+1)z= m+1, then we could say that, for m= 2, we get 0z= 3 which is impossible but for m=-1, it reduces to 0z= 0 which is true for all z. That would give an infinite number of solutions but that does not happen in the given problem.)
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  5. #5
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    Thank you very much for confirmation.
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