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Algebra with absolute value. Please help me to to solve.

  1. Jul 23, 2013 #1
    I came across this algebra problem, can someone please help me solve this problem? Please show the steps as well. Much appreciated.

    |(x-1)| + |(y-3)| = 11
    |(x- 3)| + |(y-17)| = 3

    Find the nearest/possible x and y
     
  2. jcsd
  3. Jul 23, 2013 #2

    Fredrik

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    I moved the thread to a homework forum, because it's a textbook-style problem, and we treat all textbook-style problems as homework. We only give hints here, not complete solutions. You are required to post your own thoughts on how to solve the problem, up to the point where you're stuck.

    I can give you one hint right now. You can get rid of the absolute value signs by considering several possibilities separately.
     
    Last edited: Jul 23, 2013
  4. Jul 23, 2013 #3

    HallsofIvy

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    |x- 1| is 1- x for x less than 1, x- 1 for x greater than or equal to 1. Similarly for |x-3|. So there are 3 possiblities to consider: x< 1, 1< x< 3, and x> 3.

    For y, we have the same situation: y< 3, 3< y< 17, and y> 17. Since x and y are independent, you have to consider each of the three x situations with all three of the y situations, a total of 3x3= 9 cases. You had better get busy!
     
  5. Jul 23, 2013 #4

    LCKurtz

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    @Jim_ I suggest you plot those two graphs in the xy plane and have a look.
     
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