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Math Help - simultaneous equations

  1. #1
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    simultaneous equations

    I am trying to solve the simultaneous equations by algebraic method, I know the solutions for "a" and "b" are 0.46 and -5.2, I have used a few different methods and the latest is close but not correct.

    Using elimination method I came up with;

    12a + 3b = - 10
    3a - 9b = 17

    36a + 9b = - 30

    9a - 9b = 51

    36 - ( - 9)a + (9 - 9)b = -30 + 51

    41a = 21

    a = 0.512

    3a - 3b = 17

    3(0.512) - 3b = 17

    6.15 - 3b = 17

    3b = 17 - 6.15

    b = -5.38

    12(0.512) + 3(-5.38) = -10

    3(0.512) - (-16.14) = 17.69

    As can be seen by the solutions they are not a million miles out, but somewhere I am making an error?
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  2. #2
    Super Member Quacky's Avatar
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    Re: simultaneous equations

    This is very promising work! With the first situation, you start off by correctly multiplying the first equation by three, which leads to the following equations:

    3a - 9b = 17
    36a + 9b = - 30

    You then get confused. You multiply the first equation by 3, which isn't necessary, and forget to multiply the 9b by 3. Instead, if at this stage you simply add the equations, you get that 39a=-13. This leads to a=\frac{-13}{39}=\frac{-1}{3}, which is not the answer you've been presented with. Are you sure you've written the question correctly?
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  3. #3
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    Re: simultaneous equations

    Yes I have made an error as you say with the multiplication, but the equations (Original) are correct, I will try again?

    Tried again a different method but can't get closer to the right solution at the moment?
    Last edited by David Green; January 7th 2012 at 04:12 PM. Reason: back to the drawing board on this one?
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  4. #4
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    Re: simultaneous equations

    If the original equations are:

    12a + 3b = - 10
    3a - 9b = 17

    ...then the given answer is incorrect.
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  5. #5
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    Re: simultaneous equations

    Quote Originally Posted by David Green View Post
    I am trying to solve the simultaneous equations by algebraic method, I know the solutions for "a" and "b" are 0.46 and -5.2, I have used a few different methods and the latest is close but not correct.

    Using elimination method I came up with;

    12a + 3b = - 10
    12(0.46)+ 3(-5.2)= 5.52- 15.6= -10.08, not -10

    3a - 9b = 17
    3(0.46)- 9(-5.2)= 1.38+ 46.8= 48.18, not 17.

    The solutions you say you "know" are incorrect.

    36a + 9b = - 30

    9a - 9b = 51

    36 - ( - 9)a + (9 - 9)b = -30 + 51

    41a = 21

    a = 0.512

    3a - 3b = 17

    3(0.512) - 3b = 17

    6.15 - 3b = 17

    3b = 17 - 6.15

    b = -5.38

    12(0.512) + 3(-5.38) = -10

    3(0.512) - (-16.14) = 17.69

    As can be seen by the solutions they are not a million miles out, but somewhere I am making an error?
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  6. #6
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    Re: simultaneous equations

    OK I have had another attemp, so far I can only get one of the two equations to work?

    The previous thread points out that the solutions I have are incorrect, these solutions were calculated using a software program for simutaneous equations, therefore if they are indeed incorrect, then the software or a mathematical technique (Idea) is different to solving these equations?

    12a + 3b = -10
    3a - 9b = 17

    I have multiplied out the first equation;

    36a + 9b = -30

    next I have subtracted the second equation;

    36a + 9b = -30
    3a - 9b = 17

    a = 1.42

    Using equation (1) I have worked out the value for b;

    12(1.42) + 3b = -10
    17 + 3b = -10
    3b = -10 - 17

    b = -9

    substitute this value into equation (1)

    12(1.42) + 3(-9) = 17 - 27 = -10

    Up to here the values for a and b work, however I am either doing something wrong, or there is another mathematical method for solving these equations that I am not aware of, the values will not work for the second equation, and I am struggling to find another method?

    any help now would be much appreciated.

    P.S.

    I have read in a Algebra book that the laws of basic arithmetic + and - do not equal - ?

    Is this true or can I take it that the book has a typo error?

    Thanks
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  7. #7
    Super Member Quacky's Avatar
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    Re: simultaneous equations

    Remember that as I showed in post#2, your given answer is wrong. You are correct up to:

    36a + 9b = -30
    3a - 9b = 17

    But if you subtract the equations, you get this:

    36a+9b-(3a-9b)=-30-17
    33a+18b=-47 ...which doesn't help us.

    Rule: If the terms you are trying to eliminate have the same sign, subtract the equations. If the terms you are trying to eliminate have opposing signs, add the equations.

    Here, the 9b is positive in 36a + 9b = -30 but it's negative in 3a - 9b = 17. This means, using previously stated rule, we add the equations.
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  8. #8
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    Re: simultaneous equations

    following on from what you said in the previous thread, the equations;

    12a + 3b = -10
    3a - 9b = 17

    are added.

    36a + 9b = -30
    3a - 9b = 17

    This is where the confusion starts;

    -30 + 17 = -13
    + 9 - 9 = 0
    36 + 3 = 39

    a = -0.33

    b = -1.99

    = -21.87, which is of course completely wrong.

    The arithmetic I think is were I am getting this wrong.

    I was always told that the following rules to arithmetic applied;

    + and + = +
    - and - = +
    + and - = -

    here in the equations I am now unsure, no matter which way I try to solve this I get one answer correct and one incorrect as follows;

    12a + 3b = -30
    3a - 9b = 17

    36a + 9b = -30
    3a - 9b = 17

    33a = -47

    a = -1.42

    12(-1.42) + 3b = -10

    -17 + 3b = -10

    3b = -10 + 17

    b = 2.33

    12(-1.42) + 3(2.33) = -17 + 6.99 = -10

    The next equation gives a solution completely wrong at -16.71 using those same values, so at the moment I am either way off track or these equations cannot be solved using integers?
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  9. #9
    Super Member Quacky's Avatar
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    Re: simultaneous equations

    When you said that

    a = -0.33
    b = -1.99, you had the correct solution but for a rounding error. Remember where appropriate to leave your answers in an exact form. a=\frac{-1}{3}, which gives b=-2, which is the correct solution when we back substitute:

    12(\frac{-1}{3}) + 3(-2)

    =\frac{-12}{3}-6

    =-4-6

    =-10, as expected.
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