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Math Help - Simultaneous Equations Help

  1. #1
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    Simultaneous Equations Help

    I'm having some trouble with simultaneous equations when all the letters are not on one side, I'm using the substitution method as I know this will help when moving onto quadratic equations. I'll give you an example question and show some of my workings. If someone could point out where Im going wrong and point me in the right direction i'd appreciate it.


    3x + 2Y - 14 = 0
    2x + 16 = 5Y

    I'd begin by getting rid of the 14 in equation 1 so

    3x + 2Y = 14

    Then taking 5Y off of the right side of equation 2 and 16 off the left giving

    2x - 5Y = -16

    Now I'll try to isolate either the X or the Y, in this case I'll choose the X from equation 1, so:

    X = 3(14-2Y)
    X = 42-6Y

    And put that into equation 2. So,

    2(42-6Y) - 5Y = -16

    84 - 12Y - 5Y = -16

    Then I add 84 to the left and right side of the equation

    Now I dont know what the hell goes on.

    I can do:

    12Y - 5Y = - 100
    7Y = 100

    Y=14.2 - WRONG

    -12Y - 5Y = -17Y

    -17Y = 100 WRONG.

    Im at a loss, I can do all other simultaneous equations. But when the letters start crossing sides I seem to get lost. I'm workin by the rule that whatever you do to one side you do to the other. ITS TWISTING MY MELON MAN.

    I'd be greatful for your help, if you can explain where Im going wrong please.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Simultaneous Equations Help

    You were doing well up to this part:

    Quote Originally Posted by whitedawg View Post
    Now I'll try to isolate either the X or the Y, in this case I'll choose the X from equation 1, so:
    X = 3(14-2Y)
    X = 42-6Y
    Starting with 3x +2y = 14 you first subtract 2y from both sides to get 3x = 14-2y, then divide both sides by 3 (or multiply by 1/3 - it's the same thing) to get:

    x = (1/3)(14-2y)

    Your mistake was to multiply the right side by 3 instead of dividing it by 3, and that error ripples through the rest of your work. Try this and see if it gets you to the right answer.
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  3. #3
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    Re: Simultaneous Equations Help

    Thanks a lot ebaines! I was racking my brain and throwing stationary until I read this! I know where I went wrong. Cheers!
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  4. #4
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    Re: Simultaneous Equations Help

    Quote Originally Posted by whitedawg View Post
    I'm having some trouble with simultaneous equations when all the letters are not on one side, I'm using the substitution method as I know this will help when moving onto quadratic equations. I'll give you an example question and show some of my workings. If someone could point out where Im going wrong and point me in the right direction i'd appreciate it.


    3x + 2Y - 14 = 0
    2x + 16 = 5Y

    I'd begin by getting rid of the 14 in equation 1 so

    3x + 2Y = 14

    Then taking 5Y off of the right side of equation 2 and 16 off the left giving

    2x - 5Y = -16

    Now I'll try to isolate either the X or the Y, in this case I'll choose the X from equation 1, so:

    X = 3(14-2Y)
    X = 42-6Y

    And put that into equation 2. So,

    2(42-6Y) - 5Y = -16

    84 - 12Y - 5Y = -16

    Then I add 84 to the left and right side of the equation

    Now I dont know what the hell goes on.

    I can do:

    12Y - 5Y = - 100
    7Y = 100

    Y=14.2 - WRONG

    -12Y - 5Y = -17Y

    -17Y = 100 WRONG.

    Im at a loss, I can do all other simultaneous equations. But when the letters start crossing sides I seem to get lost. I'm workin by the rule that whatever you do to one side you do to the other. ITS TWISTING MY MELON MAN.

    I'd be greatful for your help, if you can explain where Im going wrong please.
    If you are using substitution, then it will be easier to note that your second equation is nearly y in terms of x (you just need to divide it by 5 first), then you can substitute this expression of y into the first equation.
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