We can remove the absolute value sign by splitting the expression into more than one range according to when and when it doesn't change the sign of what's inside.
#|t/2+4|+3>2t#
First case - positive
If #t/2+4>=0#, then the inequality asks for #t/2+4+3>2t#, which rearranges to #-3/2t> -7#. Divide through by the #t# coefficient, noticing that the change of sign needs a matching switch of inequality direction:
#t<14/3#.
When #t/2+4>=0#, #t>=-8#. So for this first case we have specified a range of possible #t# values:
#-8<=t<14/3#.
Second case - negative
If #t/2+4<0#, then the inequality asks for #-t/2-4+3>2t#, which rearranges to #-5/2t> 1#. Divide through by the #t# coefficient, noticing that the change of sign needs a matching switch of inequality direction:
#t<-2/5#.
When #t/2+4<0#, #t<-8#. So for this second case we've specified two conditions that both point in the same direction. For them both to hold we take the strongest one, #t<-8#.
Combine cases
We now have two ranges of #t# that fit the inequality, #t<-8# and #-8<=t<14/3#. In general it is perfectly acceptable to finish an answer to such a question by finding multiple ranges of validity. However, in this case, the two ranges can be combined because they touch each other at one end. So our final answer is:
#t<14/3#
Sanity check the answer by comparing graphs of both sides of the inequality:
graph{(y-(|x/2+4|+3))(y-2x)=0 [-40, 40, -20, 20]}
