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3 variable equation

  1. Dec 17, 2012 #1
    I've been looking on youtube to try and learn how to do these more complex equations but all the examples use small numbers that easily fall into place.

    Ive got
    Equation One: 48A - 32B - 2C = 0
    Equation Two: -32A +46B - 11C = 0
    Equation Three: -2A - 11B + 13C = 54

    Ive mulitplied equation three by -16 which gives me:

    NewThree: 32A + 176B - 208C = -864

    Then I added equations Two and NewThree:
    Equation Two: -32A + 46B - 11C = 0
    NewThree: 32A + 176B - 208C = -864

    Which cancelled out the 32A and gave me:
    222B -219C = -864

    I then took Equation Three and mulitplied it by 24 which gave me:
    Anothernewthree: -48A - 264B + 312C = 1296

    Then added Equation One and Three together:

    Equation One: 48A - 32B - 2C = 0
    Anothernewthree: -48A - 264B + 312C = 1296

    Which cancelled out the 48A leaving me with:

    -296B + 310C = 1296

    Which left me with:
    222B -219C = -864
    -296B + 310C = 1296

    And it's at this point that I get stuck :/

    Is this the best route to take? Is there another method that is better to take for these larger numbers? The examples on youtube use examples that have small numbers and are easily solved, but I don't know what step to take next for this one :/

    If anyone could help me I'ld be very grateful
     
  2. jcsd
  3. Dec 17, 2012 #2

    Dick

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    Divide 222B -219C = -864 by 222 and -296B + 310C = 1296 by 296 and add them. It's nice to stick to the whole numbers for small numbers but you are going to get fractions in the end anyway. May as well start now.
     
  4. Dec 17, 2012 #3

    Ray Vickson

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    The standard way to solve such problem is also the simplest way: just solve successively for some variables, one-by-one, until you have just one variable left. Let me illustrate: your original equations are:
    (1) 48A - 32B - 2C = 0
    (2) -32A +46B - 11C = 0
    (3) -2A - 11B + 13C = 54
    Pick one of the equations and one of the variables; solve for that variable in terms of the others. It does not matter in the end which equation and variable you pick, as long as you do exact arithmetic. (It does matter when using a computer using floating-point computations---but that's another story for another time.)

    So, let's pick eq (3) and solve for A in terms of B and C: (3) --> 2A = -54 -11B + 13C, so

    (3a) A = -27 - (11/2)B +(13/2)C.
    Substitute this expression for A in eqs (1) and (2). For example, eq(1) becomes 48[ -27 - (11/2)B +(13/2)C] = 32B - 2C = 0, or -1296 - 296B + 310C. Do the same type of operations on eq (2). Our new equations are:

    (1a) -1296 - 296 B + 310 C = 0
    (2a) 864 + 222 B - 219 C = 0

    Now do the same to (1a)-(2a): pick an equation and a variable and solve. For example, let's solve for B from eq (1a): 296 B = -1296 + 310 C, or B = -(162/37) + (155/148) C.

    Substitute this into (2a) to get -108 + (27/2) C = 0.
    Our new system of equations is:

    (1b) B = -(162/37) + (155/148) C
    (2b) -108 + (27/2) C = 0
    (3a) A = -27 - (11/2)B +(13/2)C

    Now solve for C from (2b) to get C = 8. Substitute that into (1b) to get B = 4. Substitute those two values into (3a) to get A = 3.

    It is annoying to carry along all those fractions, but it is usually unavoidable. Keeping the exact fractions allows us to obtain an exact final solution; if we rounded off decimals we would be making some errors that could result in some errors in the final solution.
     
  5. Dec 17, 2012 #4
    I'ld just like to say many thanks to those who were kind enough to take time out of their day to help me.

    You guys have made the world a little bit brighter for me :)

    Thanks again
     
  6. Dec 17, 2012 #5
    It hurts my eyes to see your first equation, which has a factor of 2 that can be cancelled. After doing this, you immediately have C = 24A - 16B. If you substitute this into the second equation to eliminate C, you get 111B = 148A. If you reduce this to lowest terms, you quickly get A = 3B/4. Substituting this back for C gives you C = 2B. Those zeros on the right hand sides of equations 1 and 2 should have driven you to this approach. Finally, you can substitute for A and C in the third equation to solve explicitly for B.
     
  7. Dec 18, 2012 #6
    Yeah I threw away the method I used in my original post.

    I haven't done it the same way as you guys, (I got values for A,B,C after about 9 steps while you guys seemed to have done it in 5 or 6)

    However battling the logic out for myself has helped me understand the process better, and I did think about going back and simplifying my method but considering it's the first time I've had to do one (without any tuition on how to do 3 variable equations from my lecturer) so I'm going to keep my 9 step original :)

    Thanks again everyone :)
     
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