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A quick way to solve a system of 3 equations?

  1. Dec 14, 2009 #1
    1. The problem statement, all variables and given/known data
    I need to quickly be able to solve something like this:

    A + B + C = -3
    6A + 3B + C = -4
    8A - 4B - 2C = 22

    My calculus prof set some things to zero and was able to solve it really fast that way. He didn't use matrices. I can't figure out now what he did... can anyone help?


    2. Relevant equations



    3. The attempt at a solution
    I think he was making each of the variables 0 at one point, but I tried that and it definitely didn't work, haha.
     
  2. jcsd
  3. Dec 15, 2009 #2

    Mark44

    Staff: Mentor

    Your description of what your prof did wasn't too clear, but he might have done this:
    Add -6 times the first equation to the second equation, then add -8 times the first equation to the third equation. These two operations eliminate the A terms from the second and third equation.
    The system now looks like this:
    A + B + C = -3
    - 3B - 5C = 14
    - 12B - 10C = 46

    Now add -4 times the second equation to the third, which eliminates B from the third equation, which allows you to solve for C, which you can use in the 2nd equation to find B, and which you can use in the 1st equation to find A.

    If that wasn't it, here's another possibility -- Add the 1st and 2nd equations to the 3rd, which gives you 15A = 15, or A = 1. Then substitute that value for A in the 1st and 2nd equation and subtract the 2nd from the first, from which you can solve for B, and eventually C.
     
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