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Math Help - [SOLVED] what is an equation to find 2 variables simultaneously?

  1. #1
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    [SOLVED] what is an equation to find 2 variables simultaneously?

    Ok so this is more for a program to do general solving of 2 variables with 2 equations.
    I know how to solve simultaneously on paper but need to know how with one equation that can use variables.

    eg. I want to solve for x and y when i have two equations:
    x + 3y = 34
    &
    2x + 4y = 50
    I can find that x=7 & y=9 on paper but I want to get this within a program
    All I need to know is if:
    *(in that example above a=1 b=3 c=34 d=2 e=4 f=50)

    ax + by = c
    dx + ey = f

    .: x=
    .: y=

    what are the equations that x= and y= in terms of a, b, c, d, e, & f?

    Note: I'm not trying to solve the example, I just thought it might help to visualize with an example.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by DarkReviver View Post
    Ok so this is more for a program to do general solving of 2 variables with 2 equations.
    I know how to solve simultaneously on paper but need to know how with one equation that can use variables.

    eg. I want to solve for x and y when i have two equations:
    x + 3y = 34
    &
    2x + 4y = 50
    I can find that x=7 & y=9 on paper but I want to get this within a program
    All I need to know is if:
    *(in that example above a=1 b=3 c=34 d=2 e=4 f=50)

    ax + by = c
    dx + ey = f

    .: x=
    .: y=

    what are the equations that x= and y= in terms of a, b, c, d, e, & f?

    Note: I'm not trying to solve the example, I just thought it might help to visualize with an example.
    I'll change it just slightly for you and make your two equations look like:

    ax+by=e
    cx+dy=f

    Solve for x using elimination:

    adx+bdy=de Multiply the first equation by d
    bcx+bdy=bf Multiply the second equation by b

    Subtract these two equations:

    adx-bcx=de-bf

    Factor:

    (ad-bc)x=de-bf

    \boxed{x=\frac{de-bf}{ad-bc}}

    Solving for y in the same way produces:

    \boxed{y=\frac{af-ce}{ad-bc}}
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  3. #3
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    Quote Originally Posted by masters View Post
    \boxed{x=\frac{de-bf}{ad-bc}}

    Solving for y in the same way produces:

    \boxed{y=\frac{af-ce}{ad-bc}}
    thanks that's perfect, and I can see how it made more sense to have the equation arranged the way you put it.
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