The simplest way to solve problems involving absolute values is to break the domain (#x#-axis) into parts, those where the result inside the absolute value is positive, and those where the result is negative. Then we can explicitly make it positive by multiplying by #+1# or #-1# thus removing the absolute value. We can then solve normally.
In this case we need to know where #0.5x+9# is positive:
#0.5x+9>=0#
#0.5x>=-9#
#x>=-4.5#
And by inspection, it is negative for #x<-4.5#. We now have our two domains. Starting with the negative:
#|0.5x+9|–x+3<0#
# -1*(0.5x+9)–x+3<0# for #x<-4.5#
#-> -0.5x -9 -x +3 <0#
#-> -1.5x -6 <0#
note that when we multiply or divide by a negative value in this step, we must change the direction of the inequality:
#-> x +4 > 0#
#-> x > -4# which cannot happen in the domain #x < -4.5# therefore there is no solution here.
Now for the positive part of the absolute value where #x>=-4.5#
# 0.5x+9–x+3<0# for #x>=-4.5#
#-> -0.5x+12 <0#
# -> -x < -24#
# -> x>24# which is in the domain #x>=-4.5#. So the solution is
#x > 24#
