abs (n) < abs ( n -3) < 9 we can write
sqrt (n^2) < sqrt [ ( n -3)^2 ] < 9
Taking the first part
sqrt (n^2) < sqrt [ (n -3)^2 ] square both sides
n^2 < (n -3)^2
n^2 < n^2 -6n + 9
0 < -6n + 9
6n < 9
n < 9/6
n < 3/2 so..... n takes on all integer values from -infinity to 1 → (1)
Taking the second part
sqrt [ (n -3)^2 ] < 9 square both sides
(n -3)^2 < 81
n^2 - 6n + 9 < 81
n^2 - 6n - 72 < 0
(n - 12) ( n + 6) < 0
This will be true when n is on the imterval of (-6 ,12).....so n takes on all integer values from -5 to 11 → (2)
However the intersection of (1) and (2) are the integers -5, -4, -3 , -2, -1 , 0 , 1
The sum of these = -14
