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Solutions to a Linear System

  1. Aug 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Find all solutions of the linear system
    x + 2y + 3z = a
    x + 3y + 8z = b
    x + 2y + 2z = c
    where a,b, and c are arbitrary constants.
    2. Relevant equations


    3. The attempt at a solution
    Using elimination, I managed to set the coefficients on the diagonal equal to 1, which then allowed me to solve for z, which was z = -c + a. Substituting z into the other equations to obtain x and y, I ended up with the following solution:
    x= -6a - 2b + 13c
    y = b + 4a -5c
    z = a - c
    I was wondering if my method of solving is valid, and if the answer I obtained seems reasonable. Thanks.
     
  2. jcsd
  3. Aug 29, 2015 #2

    Student100

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    Education Advisor
    Gold Member

    Your z looks correct, but I get something different for y and x. Recheck your work, and I'll recheck mine.
     
  4. Aug 29, 2015 #3
    I plugged my values back into the original system and the equalities don't match up. Now I'm just trying to find out where my mistake was.
     
  5. Aug 29, 2015 #4

    Student100

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    Education Advisor
    Gold Member

    I just redid it, your Z is correct, so start back from there. I'm assuming you just made a calculation error with y. Your a is positive (and smaller than it should be), while it should be negative. Same deal with your C.
     
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