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Math Help - System of equations: Infinitely many, unique, or no solution.

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    System of equations: Infinitely many, unique, or no solution.

    While doing problem sets, I came across this question:

    Given the linear system:
    x + 2y = 10
    3x + (6+t)y = 30

    a) Determine a particular value of t so that the system has infinitely many solutions.
    b) Determine a particular value of t so that the system has a unique solution.

    According to the book, the value for t so that the system has infinitely many solutions is t = 0, and for a unique solution any real number that isn't 0.

    That would imply that:
    x + 2y = 10
    3x + 6y= 30
    has infinitely many solutions, but if you multiply the first equation by 3 and then proceed to eliminate, nothing would remain. So, wouldn't this be a linear system with no solution (inconsistent system) rather than infinitely many? And for any nonzero number, wouldn't it likewise have no solution?
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  2. #2
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    Re: System of equations: Infinitely many, unique, or no solution.

    Quote Originally Posted by Pupil View Post
    ...
    That would imply that:
    x + 2y = 10
    3x + 6y= 30
    has infinitely many solutions, but if you multiply the first equation by 3 and then proceed to eliminate, nothing would remain. So, wouldn't this be a linear system with no solution (inconsistent system) rather than infinitely many? And for any nonzero number, wouldn't it likewise have no solution?
    The solutions given is correct. You would have 0 = 0 if you eliminated everything, and this is typical of a consistent (dependent) system. Now, if you had something like

    0 = 2 (an always false statement), that would imply "no solution".

    For your second issue, let's solve the first for x...
    x = 10 - 2y
    sub into the second.
    3(10 - 2y) + (6 + t)y = 30
    30 - 6y + 6y + ty = 30.
    Everything cancels except for
    ty = 0
    If t = 0 (as in the above), there are infinite solutions, since 0*y is true for all y.
    If t is NOT zero, then ty = 0 has the UNIQUE solution when y = 0 (since we're in a field)
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    Re: System of equations: Infinitely many, unique, or no solution.

    Quote Originally Posted by Pupil View Post
    While doing problem sets, I came across this question:

    Given the linear system:
    x + 2y = 10
    3x + (6+t)y = 30

    a) Determine a particular value of t so that the system has infinitely many solutions.
    b) Determine a particular value of t so that the system has a unique solution.

    According to the book, the value for t so that the system has infinitely many solutions is t = 0, and for a unique solution any real number that isn't 0.

    That would imply that:
    x + 2y = 10
    3x + 6y= 30
    has infinitely many solutions, but if you multiply the first equation by 3 and then proceed to eliminate, nothing would remain. So, wouldn't this be a linear system with no solution (inconsistent system) rather than infinitely many? And for any nonzero number, wouldn't it likewise have no solution?
    Hi Pupil,

    The two equations, x+2y=10\mbox{ and }3x+6y=30 are both equivalent. That is when you multiply the first one with 3 you get the second one. I suppose you have mentioned "elimination" without seeing this. Since there is only one equation; x or y cannot be eliminated as you do in solving simultaneous equations. Now if you take any real number for x you can find the value of y. For example, (1,4.5), (2,4), (3,3.5) are some of the solutions of x+2y=10 So there are infinity many solutions.

    A inconsistent system does not have a solution. For example consider the system,

    2x+4y=3\mbox{ and }x+2y=1

    Using the first equation you can get, x+2y=\frac{3}{2}. And if you compare this with the second equation it is clear that no values which satisfies both equations. If there is, \frac{3}{2}=1 which is contradictory.
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    Re: System of equations: Infinitely many, unique, or no solution.

    Thanks both of you for all the help. I learned more in those two posts than a 2 hour lecture with my prof.
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