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Simultaneous equations with 3 Unknowns

  1. Jul 6, 2012 #1
    Help chaps im having a bad time with this.

    I can appreciate this is basic, however i can not get my head around it.

    Im scrubbing up on my maths as i was thinking of doing a HND in Engineering. I am working my way through K.A. Stroud's Engineering Mathematics and simply cant get to grips with this.

    the exercise question is:-

    2a - 5b + c = 1
    a + c = 2
    b - 3c = -3

    solve the equations for a,b and c.

    Im just not sure on the golden rules? do i use substitution or elimination? If someone can guide me through this it would be of HUGE appreciation.

    Great site by the way:)

    Nick
     
  2. jcsd
  3. Jul 6, 2012 #2
    Last edited: Jul 6, 2012
  4. Jul 6, 2012 #3

    Curious3141

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    The most elementary way of solving a small system like this is to use Gaussian elimination. See: http://en.wikipedia.org/wiki/Gaussian_elimination

    You can just use elimination without putting things in matrix notation, but once you get the hang of it, Gaussian elimination is just much faster and more mechanical.
     
  5. Jul 6, 2012 #4
    Thanks for the quick reply.

    The problem is most examples show 3 equations with the 3 variables in them.

    in my example only 1 equation has all 3 variables. I think this is where i lose it.

    :(
     
  6. Jul 6, 2012 #5

    Curious3141

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    Put 0s where the "missing" coefficients are.
     
  7. Jul 6, 2012 #6
    Still totally stuck. Anychance you can start me in the right direction? I wont give up till ive nailed it.

    :)
     
  8. Jul 6, 2012 #7

    Ray Vickson

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    Use the second eqn to express 'a' in terms of 'c' and use the third eqn to express 'b' in terms of 'c'. Plug those expressions into the first eqn (that is, where you see a or b, use the c-expressions instead). Now you have a single eqn for c alone, and it is easy to solve.

    BTW: when folks say to use Gaussian elimination they just mean: do exactly what I outlined above.

    RGV
     
    Last edited: Jul 6, 2012
  9. Jul 6, 2012 #8

    S.R

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    I haven't learned using matrices to solve systems of equations, however, notice if you find a and b in terms of c, you can substitute these values into the first equation.

    Edit: nevermind, Ray already explained this process!
     
  10. Jul 6, 2012 #9
    Thanks for the replys chaps.

    So this is where im at:-

    1) 2a - 5b + c = 1
    2) a + c = 2
    3) b - 3c = -3

    Express 'a' in terms of 'c' in equation 2)

    a = 2 - c

    Express 'b' in terms of 'c' in equation 3)

    b = -3 + 3c

    substitute values into equation 1)

    2(2-c)-5(-3+3c)+c = 1

    removing brackets:-

    4 - 2c + 15 - 15c + c = 1

    19 - 16c = 1

    -16c = -18

    c = -18/-16

    = 1.125

    is this right?

    Many thanks

    Nick
     
  11. Jul 6, 2012 #10
    so knowing c = 1.125

    substituting into equation 2) to calculate 'a'

    a + 1.125 = 2

    2- 1.125 = a

    a = 0.875

    knowing a and c, substituting into equation 1)

    5 x 0.875 - 5b + 1.125 = 1

    4.375 - 5b + 1.125 = 1

    5.5 - 5b = 1

    -4.5 = -5b

    b = -4.5/-5

    b = 0.9
     
  12. Jul 6, 2012 #11
    hang on............

    if i use equation 2) to calcualte b i get:-

    b = -3 + 3c

    b = -3 + 3 x 1.125

    b = 0.375

    ive gone wrong............
     
  13. Jul 6, 2012 #12
    seen my error its 2a not 5a in equation 1)

    so i have:-

    a = 0.875
    b = 0.375
    c = 1.125

    substituting these value into all 3 equations proves correct e.g

    b - 3c = -3

    0.375 - 3 x 1.125 does indeed = -3

    how does that sound??
     
  14. Jul 6, 2012 #13

    eumyang

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    If you checked the solutions and they worked, then they must be the answers. Good job.
     
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