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    Anyone know how to solve
    |x+1| is more than or equal to -3?
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    (Original post by crystalong)
    Anyone know how to solve
    |x+1| is more than or equal to -3?
    Weird question. |x + 1| is surely greater than -3 for all real x since |x + 1| is greater than or equal to zero by definition
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    This inequality is true for all rational x actually, as an absolute value is always positive?
    (If you meant to write lx+1l>=3 then you can rewrite this inequality as -3>x+1>3, which you can then divide into -3>x+1 and x+1>3 to solve for x)
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    (Original post by crystalong)
    Anyone know how to solve
    |x+1| is more than or equal to -3?
    true for all x
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    Why is the answer -1<=x<=0
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    (Original post by crystalong)
    Why is the answer -1<=x<=0
    Well, it's actually not. If you are referring to the solution I gave you for 'more then or equal to positive 3', then the solution is x=>2 and x<=-4. If you plug in the values you would get 2+1=3 and l-4+1l=3
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    (Original post by crystalong)
    Why is the answer -1<=x<=0
    That is not the answer to the question you've asked. Can you post a picture of it or something?
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    The actual qn is
    -3<=|x+1|<=1
    Ans:
    -1<x<=0 and -2<=x<-1
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    The actual qn is
    -3<=|x+1|<=1
    Ans:
    -1<x<=0 or -2<=x<-1[/QUOTE]
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    (Original post by crystalong)
    The actual qn is
    -3<=|x+1|<=1
    Ans:
    -1<x<=0 and -2<=x<-1
    In this case, as -3<=lx+1l is always true, you only need to consider the right side:
    As lx+1l<=1 is basically -1<=x+1<=1 you can see, by substracting -1 on each side, that -2<=x<=0.
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    then why is the ans
    -1<x<=0 and -2<=x<-1 ?
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    If you try to visualise your solution, you will see that my solution includes both of your answers. If x is between -2 and 0, that implies that it's also between -1 and 0 and -2 and -1
 
 
 
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