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Math Help - Linear equations

  1. #1
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    Linear equations

    I have to determine for which value of parameter m the system has
    a) indefinite solutions
    b) no solution.


    System is:

    2x + (m-1)y = 3
    (m+1)x + 4y = -3


    From first equation x = (3 - y(m-1)) / 2x and
    pluging that x into second equation I have get that (if I am right)

    y = (3(3+m)) / (m-9)(m+1)

    In book stands that a) m = -3 and b) m = 3.

    Can someone explain me is that right solution and if it is, why it is?
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  2. #2
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    Consider this system:
    2x + (m-1)y = 3
    (m+1)x + 4y = -3
    If the determinant is not zero then by Cramer's Rule there exist a unique solution, thus the determinant has to be zero:
    (m^2-1)-8=0 thus m^2-9=0 thus m=-3 or 3.
    When, m=3 we have:
    2x+2y=3
    4x+4y=-3
    Which is equivalent to:
    4x+4y=6
    4x+4y=-3
    An impossibility thus at m=3 there are no solutions.
    If m=-3
    2x - 4y = 3
    -2x + 4y = -3
    Which is equivalent to:
    2x - 4y = 3
    2x - 4y = 3
    Which has infinitely many solutions.
    Thus at m=-3 there are infinitely mant solutions.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Boban
    I have to determine for which value of parameter m the system has
    a) indefinite solutions
    b) no solution.


    System is:

    2x + (m-1)y = 3
    (m+1)x + 4y = -3


    From first equation x = (3 - y.(m-1)) / 2x
    This looks OK except for that extraneous trailing x.

    ..and pluging that x into second equation I have get that (if I am right)

    y = (3(3+m)) / (m-9)(m+1)

    In book stands that a) m = -3 and b) m = 3.

    Can someone explain me is that right solution and if it is, why it is?

    Plugging your expression for x into the second equation gives:

    (m+1) \frac{3-y(m-1)}{2}+4.y=-3,

    expanding a bit and replacing (m+1)(m-1) by m^2-1:

    \frac{3.(m+1)}{2}-\frac{y.(m^2-1)}{2}+4y=-3.

    Which can be rearranged to:

    y.(4-\frac{m^2-1}{2})=-3+\frac{3.(m+1)}{2},

    or:

    y.(9-m^2)=3m-3.

    and finally:

    y=\frac{3.m-3}{9-m^2}.

    This last expression is well behaved except at m=\pm 3 and
    will give a sensible solution. So we need only investigate
    the cases m=3 and m=-3.

    In the original equations put m=3, and they become:

    2.x+2.y=3
    4.x+4.y=-3,

    which are clearly inconsistent and so there is no solution.

    In the original equations put m=-3, and they become:

    2.x-4.y=3
    -2.x+4.y=-3,

    which are clearly just multiples of one another and so any point
    satisfying 2.x-4.y=3 is a solution to the equations and so
    there are an infinite number of solutions (choose any value of x, and the
    equation gives a value of y which satisfies the system)

    (if you know about Cramer's rule and determinants ThePerfectHackers
    solution should be prefered)

    RonL.
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