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Solving A System of Equations

  1. Jul 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve for a, b, and c.

    4c+6b+14a=25
    6c+14b+36a=21
    14c+36b+98a=105

    2. Relevant equations



    3. The attempt at a solution

    I need to solve this without using matrices.

    The easiest way would be to get it into row echelon form and back-substitute. However, I'm not sure how you can get this system of equations into row echelon form.

    I can subtract 6 times the first equation from the third equation:

    14c+36b+48a=105
    - 24c+36b+84a=150
    ------------------------
    -10c+0b+14a=-45

    so you get:

    -10c+0b+14a=-45
    6c+14b+36a=21
    14c+36b+98a=105

    Next the only thing you could do would be subtract 7 times the first equation from the third equation, but if you did that you would be getting rid of the a term but reintroducing the b term which I just eliminated.

    How do you get this into Row Echelon form to solve?
     
  2. jcsd
  3. Jul 2, 2010 #2

    LCKurtz

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    First step is divide the first equation by 4. Then subtract appropriate multiples of it from the second and third rows to eliminate the c's from them.

    Then divide the second equation by whatever you need to get the coefficient of b = 1 and subtract appropriate multiples of it to eliminate the b's in the other equations. This won't affect the c terms. Continue this process.
     
  4. Jul 2, 2010 #3
    Thanks! Now I understand.
     
  5. Jul 2, 2010 #4

    HallsofIvy

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    Hold off with that new first equation until you have eliminated b from the other two. 14= 2(7) and 36= 4(9)= 2(2)(3)(3) so multiplying 14 by 18 gives 252 as does multiplying 36 by 7: multiply the second equation by 18 and the third equation by 7 and subtract. That will eliminate b again and now you have two equations in a and c. Combine them to eliminate one of those.
     
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