Simultaneous Equation techniques

Discussion in 'Math' started by Gears, Feb 2, 2007.

  1. Gears

    Thread Starter New Member

    Feb 2, 2007
    4
    0
    Hello, I was wondering if someone could help explain a couple of techniques envolved in solving simultaneous equations because I think I'm getting confused on which method to use and if I'm doing that method correctly.

    On this question for example

    7a + 8b = 17 (1)
    4a + 6b = 9 (2)

    I mutilply the coeffecient of b in equation (2) to the top equation and get

    42a + 48b = 102

    then multiply the coeffecient of b in equation (1) to the bottom equation

    32a + 48b = 72
    ____________________________


    the b gets cancelled out so then I subtract right? 42a - 32a = 10a
    102 - 72 = 30

    so does that mean 10a gets subtracted from 30 to leave a = 20?

    Is this the correct method for this type of question? If not could someone perhaps go over where I've gone wrong. Thanks for your help.
     
  2. paul9619

    New Member

    Sep 18, 2006
    8
    0
    What your are trying to do with the original equation is make either a or b equal in size.

    7a + 8b = 17 (1)
    4a + 6b = 9 (2)

    I choose to make b equal so

    (1) x 3 = 21a + 24b = 51
    (2) x 4 = 16a + 24b = 36

    Then you want to cancel out b, so (1) - (2) gives 21a - 16a = 5a
    and 51 -36 = 15

    So you now have 5a = 15, now just divide both sides by 5 to give a = 15/5 = 3

    Now you can substitute this value back into any of your original equations I will use equation (1)

    (1) 7a + 8b = 17
    (7x3) + 8b = 17
    21 + 8b = 17
    8b = 17 -21
    8b = -4
    b = -4/8 = -0.5

    So your answer should be a = 3 and b= -0.5

    Your method from above would have worked but you had 10a = 30 therefore divide both sides by 10 gives a= 3
     
  3. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    Another method you could employ (particularly if you have only two variables) is to rearrange one of the equations with one of the variables as the subject, for example the first equation rearranged with a as the subject would be:

    a = (17 - 8b)/7

    You can then substitute this into the second equation such that the second equation is solely in terms of the variable b. Solve for b and then substitute this back into the first equation to solve for a. If this sounds confusing, say so, and I will show you this techniques using your above equations.

    Dave
     
  4. Gears

    Thread Starter New Member

    Feb 2, 2007
    4
    0
    Hi Dave, I would be grateful if you could illustrate that method for me. Thanks.
     
  5. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    Start with your initial equations:

    7a + 8b = 17 (1)
    4a + 6b = 9 (2)

    Rearrange (1) to get a as the subject:

    a = (17 - 8b)/7 (3)

    Sub (3) into (2):

    4((17 - 8b)/7) + 6b = 9

    Multiply the equation by 7 to remove the fraction in the bracket:

    4(17 - 8b) + 42b = 63

    Then multiply out the bracket:

    68 - 32b + 42b = 63

    Solve for b:

    10b = -5

    b = -0.5

    Sub back into (1):

    7a + 8(-0.5) = 17

    Solve for a:

    7a - 4 = 17

    7a = 21

    a = 3

    Check in both your original equations (1) and (2):

    7(3) + 8(-0.5) = 17
    4(3) + 6(-0.5) = 9

    This technique is simple and works for all simultaneous equations with two variables. If you have some imagination you can also get it to work for simultaneous equations with three or more variables, but requires a bit more work.

    Can I bring you attention to the Simultaneous Equations Section in the AAC e-book.

    Dave
     
  6. DrNick

    Active Member

    Dec 13, 2006
    110
    2
    I think one of the easyest ways to solve N equations and N unknowns with cramers rule. Here is a 2 equation example.

    Lets say you have

    5x + 2y = 5
    3x - y = 10

    put it in to matrix form:

    [5 2 ][x] [ 5 ]
    [3 -1][y] = [10]

    now the 5, 3 column corrisponds to x, and the 2, -1 column corrisponds to y.

    to solve for x, replace the x column with what is on the other side of the equals sign, now the matrix should be

    [5 2 ]
    [10 -1]

    take the determinant: 5*(-1) - 2(10) = -25

    take the determinant of the original matrix 5*-1 - 2*3 = -11

    devide det(new matrix)/det(old matrix) = x = 25/11

    to get y, do the same thing execpt replace the y column, to get:

    [5 5]
    [3 10]

    take determinant: 5*10 - 5*3 = 35
    devide by original matricies determinant 35/-11 = y = -35/11

    so the solution (x, y) = (25/11, -35/11)

    lets check to make sure this is correct.

    plugging x,y in to the original equation 5x + 2y = 5
    => 5(25/11) + 2(-35/11) = 125/11 - 70/11 = 55/11 = 5 (good times)
    and plugging in to 3x -y = 10

    3(25/11) - (-35/11) = 75/11 + 35/11 = 110/11 = 10 (good times).

    this way there is no messy algebra to get lost in and it works for any sized matrix.
     
  7. antseezee

    Active Member

    Sep 16, 2006
    45
    0
    Matrices are a good manner in solving these problems, however, I am not a big fan of them. The column alignment must be essentially perfect, and the column alignments vary for 3 or 4 variable situations. I always use substitution, even if it gets ridiculously messy. It's easier for me to solve for one variable (using quadratic equation when necessary), but it's all up to the solver.

    Don't forget the TI-89 has a solve function and also matrices on it.
     
  8. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    Indeed matrices are highly appropriate for such problems and would be my recommended approach for equations of many variables, hence a matrix-based application like Matlab or Octave is highly suitable. The problem comes in when these kinds of questions are done as part of a homework assignment - i.e. you must show all workings, and Matlab and Octave hide this level from the end user.

    Dave
     
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