3 < |x+2| < 8

Case 1:  x+2 >= 0 which is equialent to x >= -2

3 < x+2 < 8

Subtract 2 from all three sides:

1 < x < 6 which is compatible with x >= -2

1 < x < 6 in interval notation is (1,6)

Part of the graph is


----------o===================o----------------------------------------------------
-13 -12 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7   8


Case 2:  x+2 < 0 which is equivalent to x < -2

3 < -x-2 < 8

Add 2 to all three sides:

5 < -x < 10

Divide all three sides by -1 which causes
the inequality signs to reverse:

-5 > x > -10

which is the same as

-10 < x < -5 which is compatible with x < -2

-10 < x < -5 in interval notation is (-10,-5)

Graph of solution:


----------o===================o-----------------------o===================o--------
-13 -12 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7   8


Solution in set builder notation {x|-10 < x < -5 OR 1 < x < 6}
 
Solution in interval notation: (-10,-5) U (1,6) 

Edwin