# Absolute within an absolute

1. Oct 18, 2012

### HerroFish

Solve:

| |x+1| +2| - | x-2 | = 3

Relevant equations:

if |x| = a, then x = a; x = -a

My attempt:
|x+1| +2 - (x-2) = 3 ; |x+1| + 2 - (x-2) = -3 (by theorem provided by teacher above)
|x+1| = x- 1 ; |x+1| = x-7

if |x+1| < 0:

-(x+1) = x -1
-x - 1 = x - 1
-2x = 0
x = 0
------------------------
-(x+1) = x-7
-x-1 = x-7
-2x = -6
x = 3

If |x+1| => 0:

x+1 = x-1
1≠ -1
-------------------------
x+1 = x-7
1≠ -7

The answer is 1 and I'm not sure if the theorem my teacher provided applies...

Last edited: Oct 18, 2012
2. Oct 18, 2012

### HallsofIvy

Staff Emeritus
Are you giving this any thought or just trying to copy what you have seen before?

x= -1 is clearly NOT the solution, it does not satisfy the equation:
|x+ 1|= 0 so ||x+1|+ 2|= |2|= 2 while |x- 2|= |-1- 2|= |-3|= 3. 2- 3 is NOT equal to 3.

I would start off doing this is a series of cases:

case 1) x> 2
Then x> -1 so |x+1|= x+ 1 and then |x+1|+ 2= x+ 1+ 2= x+ 3> 0 so ||x+1|+ 2|= x+ 3.

case 2) $-1\le x\le 2$
|x+ 1| is still equal to x+ 1 so ||x+1|+ 2|= |x+ 3|= x+ 3 but now |x- 2|= 2- x. The equation becomes x+ 3- (2- x)= 2x+ 1= 3. 2x= 2, x= 1.
If x= 1, |x+ 1|= 2 so ||x+1|+ 2|.

case 3) x< -1.
|x+ 1|= -x- 1 so ||x+ 1|+ 2|= |-x-1+ 2|= |-x+ 1|. With x< -1, that is positive so ||x+ 1|+ 2|= -x+ 1.

3. Oct 18, 2012

### HerroFish

Oh i meant to type 1, sorry for the typo :(

4. Oct 18, 2012

### SammyS

Staff Emeritus
Supposing that the theorem provided by teacher is:
If |x| = a, where a ≥ 0, then x = a, or x = -a .​
That theorem cannot be directly applied to the equation
| |x+1| +2 | - | x-2 | = 3​
in the manner which you applied it.

Another way to state that theorem is:
If |x| = a, where a ≥ 0, then x = a, or -x = a .​
If you carefully consider the first term in your equation, the one with the absolute value inside the absolute value, you can simplify the equation a bit.

Is |x+1| + 2 ever negative? ... No !

Then | |x+1| +2 | = |x+1| +2 .