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Math Help - Absolute Value Inequality

  1. #1
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    Absolute Value Inequality

    Just had a homework problem:

    Find all x in R that satisfy 4 less than abs(x+2)+abs(x-1) less than 5.

    I did it all out assuming abs(x+2) can be written as x+2 or -x-2 and the same for abs(x-1).

    I got an answer of:

    -3 less than x less than (-5/2)
    and
    (3/2) less than x less than 2.

    Can anyone confirm this for me?
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  3. #3
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    I'm assuming this is wrong. Having trouble reading that because of the lack of latex or whatever it is called.
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  4. #4
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    Quote Originally Posted by JaysFan31 View Post
    I'm assuming this is wrong. Having trouble reading that because of the lack of latex or whatever it is called.
    before I help, is this your inequality?

    |x+2|+|x-1|-4<5

    or:

    5-|x+2|+|x-1|-4
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  5. #5
    Senior Member OReilly's Avatar
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    Quote Originally Posted by JaysFan31 View Post
    Just had a homework problem:

    Find all x in R that satisfy 4 less than abs(x+2)+abs(x-1) less than 5.

    I did it all out assuming abs(x+2) can be written as x+2 or -x-2 and the same for abs(x-1).

    I got an answer of:

    -3 less than x less than (-5/2)
    and
    (3/2) less than x less than 2.

    Can anyone confirm this for me?
    I have come up with:

    5/2 < x < 2
    and
    -3 < x < -5/2
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  6. #6
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Quick View Post
    before I help, is this your inequality?

    |x+2|+|x-1|-4<5

    or:

    5-|x+2|+|x-1|-4
    I think inequality is:

    4 < |x+2| + |x-1| < 5
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  7. #7
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    Yeah the inequality is the last one. Sure it's 5/2 and not 3/2?
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    Quote Originally Posted by OReilly View Post
    I think inequality is:

    4 < |x+2| + |x-1| < 5
    okay,

    if x is greater than or equal to zero, than we get: 4<x+2+x-1<5

    thus: 3<2x<4

    thus: 3/2<x<2

    if x is less than zero we get: 4<-x-2+-x+1<5

    thus: 5<-2x<6

    thus: -3<x<-5/2
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  9. #9
    Senior Member OReilly's Avatar
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    Quote Originally Posted by JaysFan31 View Post
    Yeah the inequality is the last one. Sure it's 5/2 and not 3/2?
    Sorry, you are right.
    It's 3/2. My mistake. Typed in hurry.
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  10. #10
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    Here is a graphical solution.
    If you want me to do it formally you can ask.
    Attached Thumbnails Attached Thumbnails Absolute Value Inequality-picture3.gif  
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  11. #11
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    Quote Originally Posted by ThePerfectHacker View Post
    Here is a graphical solution.
    If you want me to do it formally you can ask.
    That's not for an inequality though...
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  12. #12
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    Quote Originally Posted by Quick View Post
    That's not for an inequality though...
    Yes that is, below the green line is the interval for which it is true.

    You are probably thinking about the points of intersection.
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  13. #13
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    Quote Originally Posted by ThePerfectHacker View Post
    Yes that is, below the green line is the interval for which it is true.

    You are probably thinking about the points of intersection.
    what is your equation for the red graph then? I would like to see how to solve this graphically.
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  14. #14
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    Quote Originally Posted by Quick View Post
    what is your equation for the red graph then? I would like to see how to solve this graphically.
    y1=abs(x+2)+abs(x-1)
    y2=5
    And you seen when,
    y1<=y2
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  15. #15
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    Quote Originally Posted by ThePerfectHacker View Post
    y1=abs(x+2)+abs(x-1)
    y2=5
    And you seen when,
    y1<=y2
    okay, I see how that works, you did forget to add y3=4 and then find y3<y1<y2
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