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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
