1. ## Absolute Value Inequality

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).

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

Can anyone confirm this for me?

2. I'm assuming this is wrong. Having trouble reading that because of the lack of latex or whatever it is called.

3. Originally Posted by JaysFan31
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

4. Originally Posted by JaysFan31

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).

-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

5. Originally Posted by Quick
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

6. Yeah the inequality is the last one. Sure it's 5/2 and not 3/2?

7. Originally Posted by OReilly
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

8. Originally Posted by JaysFan31
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.

9. Here is a graphical solution.
If you want me to do it formally you can ask.

10. Originally Posted by ThePerfectHacker
Here is a graphical solution.
If you want me to do it formally you can ask.
That's not for an inequality though...

11. Originally Posted by Quick
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.

12. Originally Posted by ThePerfectHacker
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.

13. Originally Posted by Quick
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

14. Originally Posted by ThePerfectHacker
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