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Linear System Question

  1. Jul 11, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm trying to solve #1 here:
    http://www.student.math.uwaterloo.ca/~math115/Exams/M115.FE.pdf [Broken]

    The problem is:

    3. The attempt at a solution
    (a) At a=2, the slope is the same. I figure any value but 2 gives unique solutions. But I don't know if this is right.
    (b) At a=2, the slopes are the same, but lines are different. I cannot find any value of a that would give the same slope and same outputs for inputs of x. DNE is the answer?
    (c) At a=2, there are no solutions for the system because the lines are parallel to each other. a=2 is the answer.
    (d) I think this part was done correctly. I used substitution and solved for the x and y values at a=1 and resulted with (2/3, 1/3)

    I would like to know about a, b, and c. I don't know if I did these right.

    Thanks.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 11, 2012 #2

    eumyang

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    Ignoring (a) and (b) for the moment:
    (d) is right, but (c) is wrong. At a = 2, you have this system:
    x + 2y = 1
    2x + 4y = 2
    Parallel lines have the same slope, but different y-intercepts. Is that the case here?
     
  4. Jul 11, 2012 #3

    chiro

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    Hey JettyZ.

    For c), there are many solutions corresponding to x = t and y = 1 - 2t for any value of t.
    For a) and b) you need to use determinants and check for possibilities of inconsistent solutions.
     
  5. Jul 11, 2012 #4

    HallsofIvy

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    If a= 2, your equations are x+ 2y= 1 and 2x+ 4y= 2. If you multiply the first equation by 2, what happens? What does that tell you?
     
  6. Jul 11, 2012 #5

    Ray Vickson

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    Do you know about determinants and their relationship to such questions? If so, use a determinant; you will see that there is a critical value of 'a' that you have missed.

    RGV
     
    Last edited by a moderator: May 6, 2017
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