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Thread: Simultaneous equations

  1. #1
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    Simultaneous equations

    Solve for a, b and c given that:
    • 2a+4b+5c= 2
    • 5a+3b-2c= 13
    • 3a-2b-3c= 0
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by oldschool999 View Post
    Solve for a, b and c given that:
    • 2a+4b+5c= 2
    • 5a+3b-2c= 13
    • 3a-2b-3c= 0
    One way is called the substitution method. Pick any one of the equations, say the bottom one, and solve it for one of the variables:
    $\displaystyle 3a - 2b - 3c = 0$

    $\displaystyle c = a - \frac{2}{3}b$

    Insert this value of c into the other two equations:
    $\displaystyle 2a + 4b + 5 \left ( a - \frac{2}{3}b \right ) = 2 \implies 7a + \frac{2}{3}b = 2$

    $\displaystyle 5a + 3b - 2 \left ( a - \frac{2}{3}b \right ) = 13 \implies 3a + \frac{13}{3}b = 13$

    So now we need to solve:
    $\displaystyle 7a + \frac{2}{3}b = 2$

    $\displaystyle 3a + \frac{13}{3}b = 13$

    Solve one of these for one of the other unknowns, etc.

    For verification I get that $\displaystyle a = 0, ~ b = 3, ~ c = -2$

    -Dan
    Last edited by topsquark; Aug 7th 2007 at 08:09 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by oldschool999 View Post
    Solve for a, b and c given that:
    • 2a+4b+5c= 2
    • 5a+3b-2c= 13
    • 3a-2b-3c= 0
    What methods have you been shown?

    RonL
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