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Math Help - Using Parameters

  1. #1
    Senior Member Stroodle's Avatar
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    Using Parameters

    A question from my text is:

    The simultaneous equations x+2y+3z=13,\;-x-3y+2z=2 and -x-4y+7z=17 have infinitely many solutions. Describe these solutions through the use of a parameter.

    I understand how to do this. I just don't understand the purpose of using a parameter?

    Thanks for your help
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Stroodle View Post
    A question from my text is:

    The simultaneous equations x+2y+3z=13,\;-x-3y+2z=2 and -x-4y+7z=17 have infinitely many solutions. Describe these solutions through the use of a parameter.

    I understand how to do this. I just don't understand the purpose of using a parameter?

    Thanks for your help
    The parameter (let's say t) is used to define x, y, and z solutions. Since t\in\mathbb{R}, that implies that t can take on any value (hence, the system has infinite solution). Its just a nice way of writing the solution, because you can't possibly list all the solutions (you can even call it a general solution for any t)!
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  3. #3
    Senior Member Stroodle's Avatar
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    Actually. I don't quite understand how to do these. Can someone show working for:

    Using a parameter find all the solutions for:

    2x-y+z=0 and y+2z=2

    Thanks
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  4. #4
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    Quote Originally Posted by Stroodle View Post
    Actually. I don't quite understand how to do these. Can someone show working for:

    Using a parameter find all the solutions for:

    2x-y+z=0 and y+2z=2

    Thanks
    Let x = t where t is any real number.

    Then:

    -y + z = -2t .... (1)

    y + 2z = 2 .... (2)

    Now solve (1) and (2) simultaneously for y and z in terms of t.
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  5. #5
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    Hello, Stroodle!

    Using a parameter find all the solutions for: . \begin{array}{cc}2x-y+z\:=\:0 & [1] \\ y+2z\:=\:2& [2]\end{array}
    Here's one solution . . . [There are many others.]

    Solve one of the equations for one of its variables.
    . . Solve [2] for y\!:\;\;{\color{blue}y \:=\:2-2z}

    Substitute into the other equation.
    . . 2x - (2 - 2z) + z \:=\:0
    And solve for the third variable.
    . . {\color{blue}x \:=\:1 - \tfrac{3}{2}z}

    So we have: . \begin{Bmatrix}x &=& 1 - \frac{3}{2}z \\ y &=& 2 - 2z \\ z &=& z \end{Bmatrix}


    On the right side, replace z with a parameter t.

    Therefore, we have: . \boxed{\begin{array}{ccc} x &=& 1 - \frac{3}{2}t \\ y &=& 2 - 2t \\ z &=& t \end{array}}

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  6. #6
    Senior Member Stroodle's Avatar
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    Awesome. Thanks for your help guys!!
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