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Math Help - Simultaneous Equations

  1. #1
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    Simultaneous Equations

    Hi

    I am trying to solve the following equation, but I'm not sure if I'm right, can anybody advise?

    2x + 3y = 19
    6x - 7y = -7

    So far I have done the following:

    4x + 6y = 38

    Added the second and third equation, to get:

    10x + 13y = 31

    31 / 10 = 3.1
    3.1 * 2 = 6.2

    19 - 6.2 = 12.8 / 3 = 4.3

    x = 3.1
    y = 4.3

    I appreciate any advice.
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  2. #2
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    Quote Originally Posted by Gall1987 View Post
    Hi

    I am trying to solve the following equation, but I'm not sure if I'm right, can anybody advise?

    2x + 3y = 19
    6x - 7y = -7

    So far I have done the following:

    4x + 6y = 38

    Added the second and third equation, to get:

    10x + 13y = 31
    The goal of adding two equations is to get one equation in one unknown. I'd recommend multiplying both sides of your first equation by -3, giving
    -6x - 9y = -57

    Now add that to the second equation. You will have an equation in y only that you can solve.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    The goal of adding two equations is to get one equation in one unknown. I'd recommend multiplying both sides of your first equation by -3, giving
    -6x - 9y = -57

    Now add that to the second equation. You will have an equation in y only that you can solve.

    -Dan
    Can you tell me why I need to multiply the first equation by a negative number? I have originally tried the following, but now I'm stuck:

    6x + 9y = 57 +6x -7y = -7 = 12x + 2y = 50.

    Can you advise me how to resolve it? This is just revision for an upcoming exam, so I just need to know the concept to solve them and I should hopefully then be OK.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Gall1987 View Post
    Can you tell me why I need to multiply the first equation by a negative number? I have originally tried the following, but now I'm stuck:

    6x + 9y = 57 +6x -7y = -7 = 12x + 2y = 50.

    Can you advise me how to resolve it? This is just revision for an upcoming exam, so I just need to know the concept to solve them and I should hopefully then be OK.
    When you are trying to solve a system of equations the goal is to find some way of writing a single equation in a single unknown. In this case we compare the two equations:
    2x + 3y = 19
    6x - 7y = -7

    We want to get rid of, say, x. So if we multiply the first equation by -3:
    -6x + -18y = -57
    6x - 7y = -7

    And add the two equations:
    (-6x + 6x) + (-18y - 7y) = (-57 - 7)

    -25y = -64
    which you can immediately solve for y.

    The other way is to
    2x + 3y = 19
    6x - 7y = -7

    multiply the top equation by 7 and the bottom by 3:
    14x + 21y = 133
    18x - 21y = -21

    (14x + 18x) + (21y - 21y) = (133 - 21)

    32x = 112
    which you can, again, solve.

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Gall1987 View Post
    Can you tell me why I need to multiply the first equation by a negative number? I have originally tried the following, but now I'm stuck:

    6x + 9y = 57 +6x -7y = -7 = 12x + 2y = 50.

    Can you advise me how to resolve it? This is just revision for an upcoming exam, so I just need to know the concept to solve them and I should hopefully then be OK.
    To address your method how did you get
    6x + 9y = 57 +6x -7y

    -Dan
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  6. #6
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    Quote Originally Posted by Gall1987 View Post
    Hi

    Added the second and third equation, to get:

    10x + 13y = 31
    6x+4x=10x - right
    -7+38=31 - right
    -7y+6y=-1y and you`ve written 13y
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    Using google can save you lotsa time(!):
    Google
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  8. #8
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    Quote Originally Posted by Gall1987 View Post
    Hi

    I am trying to solve the following equation, but I'm not sure if I'm right, can anybody advise?

    2x + 3y = 19
    6x - 7y = -7

    So far I have done the following:

    4x + 6y = 38
    Okay, you multiplied the first equation by 2- any specific reason for doing that?

    Added the second and third equation, to get:

    10x + 13y = 31
    Again, yes, that's true but why did you do that? You still have two unknowns- there are an infinite number of x,y pairs that will satify that equation.

    31 / 10 = 3.1
    Are you saying that x= 3.1? That would be true if x satisfied 10x= 31 which would be true if y= 0.

    3.1 * 2 = 6.2

    19 - 6.2 = 12.8 / 3 = 4.3

    x = 3.1
    y = 4.3
    And now you are saying that, with this value of x, y is not 0 so that x= 31/10 was wrong.

    I appreciate any advice.
    You seem to be just doing things at random. Also, you said "I'm not sure if I'm right". Did you consider checking your answer?
    If x= 3.1 and y= 4.3 then 3x+ 3y= 2(3.1)+ 3(4.3)= 6.2+ 12.9= 19.1, not 19. And 6x- 7y= 18.6- 30.1= --11.5 not -7.
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