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#