System of Equations...3

Discussion in 'Other Pre-University Math' started by nycmathguy, Nov 1, 2021.

  1. nycmathguy

    nycmathguy

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    20211031_200910.jpg
     
    nycmathguy, Nov 1, 2021
    #1
  2. nycmathguy

    MathLover1

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    where did you find that one?
    solutions are a bit complicated, you have two equations and four unknown variables
    it takes a lot to get solutions of two variables expressed in terms of two other variables

    solution looks like this:

    [​IMG]
    or
    [​IMG]
     
    Last edited: Nov 1, 2021
    MathLover1, Nov 1, 2021
    #2
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  3. nycmathguy

    nycmathguy

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    All the system of equations threads come from David Cohen's Precalculus With Unit Circle Trigonometry Edition 4 textbook.

    Instructions:

    Solve each system of equations. If
    there are no solutions in a particular case, say so. In cases in which there are literal (rather than numerical) coefficients, specify any restrictions that your solutions impose on those coefficients.
     
    nycmathguy, Nov 1, 2021
    #3
  4. nycmathguy

    MathLover1

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    (2b)/x-3/y=7ab..........eq.1
    (4a)/x-(5a)/(by)=3a^2......eq.2
    ---------------------------------------

    (2b)/x-3/y=7ab..........eq.1, solve for x

    (2b)/x=7ab+3/y

    x=(2b)/(7ab+3/y)............eq.1a

    substitute x in eq.2
    (4a)/((2b)/(7ab+3/y))-(5a)/(by)=3a^2......eq.2, solve for y

    14a^2 + (6a)/(by)-(5a)/(by)=3a^2
    (6a)/(by)-(5a)/(by)=3a^2-14a^2
    (6a-5a)/(by)=-11a^2
    a/(by)=-11a^2
    a/(-11a^2)= by
    -1/(11a)= by

    y=-1/(11ab)

    go to
    x=(2b)/(7ab+3/y)............eq.1a, substitute y

    x=(2b)/(7ab+3/(-1/(11ab)))
    x=(2b)/(7ab-3(11ab))
    x=(2b)/(7ab-33ab)
    x=(2b)/(-26ab)
    x=-1/(13a)

    restrictions:

    [​IMG]

     
    MathLover1, Nov 1, 2021
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  5. nycmathguy

    nycmathguy

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    This is not your typical system of equations. This challenging problem is perfect to see if students understand the concept of solving system of equations. Well-done.
     
    nycmathguy, Nov 1, 2021
    #5
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