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|x+1| is more than or equal to -3? watch

1. Anyone know how to solve
|x+1| is more than or equal to -3?
2. (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
3. 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)
4. (Original post by crystalong)
Anyone know how to solve
|x+1| is more than or equal to -3?
true for all x
5. Why is the answer -1<=x<=0
6. (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
7. (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?
8. The actual qn is
-3<=|x+1|<=1
Ans:
-1<x<=0 and -2<=x<-1
9. The actual qn is
-3<=|x+1|<=1
Ans:
-1<x<=0 or -2<=x<-1[/QUOTE]
10. (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.
11. then why is the ans
-1<x<=0 and -2<=x<-1 ?
12. 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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