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Math Help - inequality

  1. #1
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    inequality

    Solve the inequality :

    |x-3|-|x-1|\leq2

    This is what i did :

    squarring both sides .

    (x-3)^2-2(x-3)(x-1)+(x-1)^2\leq4

    x^2-6x+9-2x^2+8x-6+x^2-2x+1-4\leq0

    then

    0\leq0 ???

    What does that mean ? so wat 's the range of x ?
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  2. #2
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    your solution  0 \leq 0 means that the equation is valid for all x.

    notice the x's cancel out no matter what x is. plugging x=200 in to it you will see that the 200's will cancel out giving you -2. so whatever x is you will get -2\leq2 or 2\leq2 because the first absolute value will always be 2 units away from the second absolute value depending on the sign of x.

    The way i would solve these is to solve the equations;

    1. -(x-3)-(x-1)\leq2
    2. (x-3)-(x-1)\leq2
    3. -(x-3)+(x-1)\leq2
    4. (x-3)+(x-1)\leq2

    and take the union of the solutions. because if x solves any one of these then it solves the original inequality.
    Last edited by Krahl; August 27th 2009 at 05:44 AM.
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  3. #3
    MHF Contributor red_dog's Avatar
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    Use the inequality:

    |a|-|b|\leq |a-b|, \ \forall a,b\in\mathbb{R}

    Then |x-3|-|x-1|\leq |x-3-x+1|\Rightarrow |x-3|-|x-1|\leq 2

    Therefore the inequality holds for every real x.
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  4. #4
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    Quote Originally Posted by Krahl View Post
    your solution  0 \leq 0 means that the equation is valid for all x.

    notice the x's cancel out no matter what x is. plugging x=200 in to it you will see that the 200's will cancel out giving you -2. so whatever x is you will get -2\leq2 or 2\leq2 because the first absolute value will always be 2 units away from the second absolute value depending on the sign of x.

    The way i would solve these is to solve the equations;

    1. -(x-3)-(x-1)\leq2
    2. (x-3)-(x-1)\leq2
    3. -(x-3)+(x-1)\leq2
    4. (x-3)+(x-1)\leq2

    and take the union of the solutions. because if x solves any one of these then it solves the original inequality.


    Thanks but how did you get the 4 equations ? Can u explain that to me ?
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  5. #5
    Senior Member nikhil's Avatar
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    Lightbulb correction

    Quote Originally Posted by Krahl View Post
    your solution  0 \leq 0 means that the equation is valid for all x.

    notice the x's cancel out no matter what x is. plugging x=200 in to it you will see that the 200's will cancel out giving you -2. so whatever x is you will get -2\leq2 or 2\leq2 because the first absolute value will always be 2 units away from the second absolute value depending on the sign of x.

    The way i would solve these is to solve the equations;

    1. -(x-3)-(x-1)\leq2
    2. (x-3)-(x-1)\leq2
    3. -(x-3)+(x-1)\leq2
    4. (x-3)+(x-1)\leq2

    and take the union of the solutions. because if x solves any one of these then it solves the original inequality.
    you have not included the interval of x for which you opened the mod and you said to take the union of solution obtained from the inequations. this can be misleading sometimes for example
    |x-1|-|x-3|>=4
    let 1<=x<3 so we have
    x-1+(x-3)>=4 or
    2x-4>=4 or
    x>=4 which contradicts with our interval
    you can check this by putting x=4.you will get LHS=2 which is not >=4 so such solution should be discarded.
    but if you will not write the interval of x for which you opened mod and took union of the intervals obtained then you might get wrong result.
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  6. #6
    Senior Member nikhil's Avatar
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    Lightbulb

    hi thereddevils!!!
    |x|=x if x>=0
    =-x if x<0
    |x-3|-|x-1|>=2
    consider two points 1 and 3
    take case i.e
    1) x>=3
    2) 1<=x<3
    3)x<1
    you will get 3 inequations. solve them and take the union of solution.
    discard the solution if its interval does not match your interval i.e you took x>1 but your corresponding inequation gave x<3.(though you will not have this situation in this problem.)
    Last edited by nikhil; August 27th 2009 at 10:12 AM.
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  7. #7
    Senior Member nikhil's Avatar
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    Lightbulb TIP

    whenever equation is like |x-a|+|x-b|+|x-c|....and so on >=n
    then draw points a,b,c.... on number line.
    now start taking interval i.e x<a then a<=x<b then b<=x<c and so on
    this will help to solve inequality in an organised way
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