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Thread: Solving this system of equation with two unknowns.

  1. #1
    Newbie Archytas's Avatar
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    Solving this system of equation with two unknowns.

    Hello,

    I am trying to solve:

    a0 = 2 = a 30 + b 20 = a + b
    a1 = 6 = a 31 + b 21 = 3a + 2b
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  2. #2
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    Re: Solving this system of equation with two unknowns.

    Hey Archytas.

    Hint - Try getting one equation in terms of one variable and substituting it into the other one.
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  3. #3
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    Re: Solving this system of equation with two unknowns.

    Or- multiply the second equation by 2 and subtract that from the first equation.

    Then multiply the second equation by 3 and subtract the first equation from that.
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    Re: Solving this system of equation with two unknowns.

    Quote Originally Posted by Archytas View Post
    a0 = 2 = a 30 + b 20 = a + b
    a1 = 6 = a 31 + b 21 = 3a + 2b
    Looks to me that above is simply this:
    a + b = 2
    3a + 2b = 6
    ...which evidently means b = 0.

    Do I need another coffee?
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  5. #5
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    Re: Solving this system of equation with two unknowns.

    I am no so sure that is it "evident". yes, if a= 2 then 3a= 6 so a= 2, b= 0 satisfy the equation.

    Or, as chiro suggested, from a+ b= 2, b= 2- a so the second equation becomes 3a+ 2(2- a)= a+ 4= 6 so a= 2 and then b= 2- 2= 0.

    Or, as I suggested, multiplying the first equation by 2, 2a+ 2b= 4 and, subtracting that from 3a+ 2b= 6, a= 2. Multiplying the first equation by 3, 3a+ 3b= 6 and, subtracting that from 3a+ 2b= 6 from that b= 0.
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    Re: Solving this system of equation with two unknowns.

    Quote Originally Posted by HallsofIvy View Post
    I am no so sure that is it "evident".
    .........
    Multiplying the first equation by 3, 3a+ 3b= 6 and,
    subtracting that from 3a+ 2b= 6 from that b= 0.
    That's why it's evident (more or less).
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  7. #7
    Newbie Archytas's Avatar
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    Re: Solving this system of equation with two unknowns.

    It is evident thanks to your clarification, thank you!
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