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Math Help - Graphic-solution

  1. #1
    Super Member dhiab's Avatar
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    Graphic-solution

    Give the Graphic-solution of this inequality :
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  2. #2
    MHF Contributor red_dog's Avatar
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    (2x-y-3)(2x-y+3)\leq 0

    The solution is the region between the lines 2x-y-3=0 and 2x-y+3=0.
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  3. #3
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    Quote Originally Posted by dhiab View Post
    Give the Graphic-solution of this inequality :
    Alternatively:

    (2x - y)^2 \leq 9

    |2x - y| \leq 3

    -3 \leq 2x - y \leq 3


    Case 1: -3 \leq 2x - y

    y \leq 2x + 3.


    Case 2: 2x - y \leq 3

    y \geq 2x - 3.


    So putting them together gives

    2x - 3 \leq y \leq 2x + 3.


    You should be able to graph these.
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  4. #4
    Super Member dhiab's Avatar
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    Quote Originally Posted by red_dog View Post
    (2x-y-3)(2x-y+3)\leq 0

    The solution is the region between the lines 2x-y-3=0 and 2x-y+3=0.
    Hello , Thank you, Here are the details

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  5. #5
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    Just to give yet another way:

    Look at (2x-y)^2= 9. 2x- y= \pm 3. Taking the positive sign, 2x- y= 3. Taking the negative sign, 2x-y= -3. Those are the two parallel lines red dog gives. I notice that x= 0 gives y= 3 and y= -3 for the two lines so (0,0) is between the two lines. Also, (2(0)- 0)= 0< 9 so that point, between the two lines satisfies the inequality. That tells us that every point between the two lines satisfies the inequality, again, as red dog said.
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