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  1. #1
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    equations

    Solve the following system of simultaneous equations:
    6x1 + 4x2 = 40
    2x1 + 3x2 = 20

    I'm a little confused, can one help.
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  2. #2
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    Quote Originally Posted by sweettea331
    Solve the following system of simultaneous equations:
    6x1 + 4x2 = 40
    2x1 + 3x2 = 20

    I'm a little confused, can one help.
    I'm a Physicist, so I usually don't do the fancier methods...

    Substitution method. Solve one equation for one of the unknowns, then plug it into the other equation. I'll solve the top equation first:
    $\displaystyle 6x_1+4x_2=40$
    Thus $\displaystyle x_2=1/4*(40-6x_1)=10-3/2*x_1$

    Now use this value for $\displaystyle x_2$ in the second equation. This leaves us an equation in 1 unknown ($\displaystyle x_1$).
    $\displaystyle 2x_1+3(10-3/2*x_1)=20$
    $\displaystyle 2x_1+30-9/2*x_1=20$
    $\displaystyle -5/2*x_1+30=20$

    So, solve this equation for $\displaystyle x_1$. Use this value in either of the original equations to get a value for $\displaystyle x_2$. (I get $\displaystyle x_1=4$ and $\displaystyle x_2=4$).

    There are other methods to use. One of them is to multiply the first equation by a constant and the second by another constant such that you can subtract the two equations and cancel out one of the variables. In this case, you don't need to do anything to the first equation. Multiply the second equation by 3:
    $\displaystyle 6x_1+4x_2=40$
    $\displaystyle 6x_1+9x_2=60$

    Now we subtract the two equations from each other:
    $\displaystyle 6x_1+4x_2=40$
    $\displaystyle -6x_1-9x_2=-60$
    ----------------------------------
    $\displaystyle 0-5x_2=-20$
    This gives a value for $\displaystyle x_2$. Then use this $\displaystyle x_2$ in either of the original equations and solve for $\displaystyle x_1$.

    -Dan
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