Given
#color(white)("XXXX")##3abs(x+2)-2 >7#
Adding 2 to both sides maintains the inequality, so
#color(white)("XXXX")##3abs(x+2) > 9#
Dividing both sides by a positive amount also maintains the inequality, so
#color(white)("XXXX")##abs(x+2) > 3#
There are two possibilities.
Possibility 1: #(x+2) < 0 rarr x<-2#
In this case
#color(white)("XXXX")##abs(x+2) = -(x+2)#
So
#color(white)("XXXX")##abs(x+2) > 3#
becomes
#color(white)("XXXX")##-x-2 > 3#
Adding 2 to both sides (which maintains the inequality)
#color(white)("XXXX")##-x > 5#
Multiplying by #(-1)# (remembering that multiplication by a negative reverses the inequality)
#color(white)("XXXX")##x < -5#
The two conditions for Possibility 1: #x < -2# and #x < -5# simplify to
#color(white)("XXXX")##x<-5#
Possibility 2: #(x+2)>=0 rarr x>=-2#
In this case
#color(white)("XXXX")##abs(x+2) = x+2#
So
#color(white)("XXXX")##abs(x+2) >3#
becomes
#color(white)("XXXX")##x+2 > 3#
Subtracting 2 from both sides (inequality maintained)
#color(white)("XXXX")##x>1#
The two conditions for Possibility 2: #x>= -2# and #x > 1# simplify to
#color(white)("XXXX")##x > 1#
Combining
Either Possibility 1 or Possibility 2
so
#color(white)("XXXX")##x < -5# or #x > 1#
