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Math Help - Figuring out x and y values in a constraint

  1. #1
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    Figuring out x and y values in a constraint

    I need to know how to find an x and y value to graph for the constraint

    3 < 3x + y < 12

    I know how to do something like: 3x -2y > -6

    where I set either x or y = to 0 and end up with x = 2, y = 3

    However, I do not know how to do 3 < 3x + y < 12


    Help, please!
    Last edited by mr fantastic; May 4th 2010 at 05:17 PM. Reason: Restored deleted question.
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  2. #2
    Junior Member
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    Take a look at this image.



    It shows all possible values (well, not ALL, that would require infinite space) of X and Y that satisfy the equation

    3 < 3x+y < 12

    From there I'm not sure what you are supposed to answer? Your question isn't very clear.

    Notice that the equations of the two red lines are

    y=-3x+3

    and

    y=-3x+12

    so

    -3x+3 < y < -3x+12
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  3. #3
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    That is all I needed to know.

    Thank you so much!
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  4. #4
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    Hello, pungakitty!

    Sorry, the work you showed makes no sense . . .


    Graph the inequality: .  3 \:<\: 3x + y \:<\: 12

    We have two inequalties:

    . . \begin{array}{ccc}3x + y &<& 12  \\ 3x + y &>& 3 \end{array}


    Solve for y\!:\;\;\begin{array}{cccc}y &<& -3x + 12  & [1] \\ y & > &-3x+3 & [2] \end{array}

    Then graph the two inequalties.



    [1]\;\;y \;<\;-3x+12

    Graph the line: . y \:=\:-3x+12
    . . and shade the region below the line.


    Code:
              |
         :\   |
         ::\  |
         :::\ |
         ::::\|
         :::::*12
         :::::|\
         :::::|:\
         :::::|::\
         :::::|:::\
         :::::|::::\
         :::::|:::::\
         :::::|::::::\
         :::::|:::::::\
         :::::|::::::::\
      - -:-:-:+:-:-:-:-:\ - - -
         :::::|::::::::::\
         :::::|:::::::::::\
         :::::|::::::::::::\
              |



    [2]\;\;y \;>\;-3x+3

    Graph the line: . y \:=\:-3x+3
    . . and shade the region above the line.


    Code:
              |
         :::::|::::::::::::::::
         :::::|::::::::::::::::
         :::::|::::::::::::::::
         :::::|::::::::::::::::
         :::::|::::::::::::::::
         :::::|::::::::::::::::
          \:::|::::::::::::::::
           \::|::::::::::::::::
            \:|::::::::::::::::
             \|::::::::::::::::
             3*::::::::::::::::
              |\:::::::::::::::
              | \::::::::::::::
        - - - + -\-:-:-:-:-:-:- - -
              |   \::::::::::::
              |    \:::::::::::
              |     \::::::::::
              |


    The solution is the region common to both shadings.


    Code:
              |
         :\   |
         ::\  |
         :::\ |
         ::::\|
         :::::*12
         :::::|\
         :::::|:\
          \:::|::\
           \::|:::\
            \:|::::\
             \|:::::\
             3*::::::\
              |\::::::\
              | \::::::\
      - - - - + -\-:-:-:\ - - -
              |   \::::::\
              |    \::::::\
              |     \::::::\
              |      \::::::\
              |
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