Since #abs(4/3 x)=4/3abs(x)#, we have the following two expressions:

**If #x# is positive:**

#4/3x<2x+5#

We can multiply both sides by #3# to get

#4x<6x+15#

And finally #2x> -15 \to x> -15/2#

Anyway, we can take the interval #(-15/2, infty)#, because we are assuming that #x>0#. So, the two conditions together yield

#x>0 \and x> -15/2 \implies x>0#

**If #x# isnegative:**

#-4/3x<2x+5#

We can multiply both sides by #3# to get

#-4x<6x+15#

And finally #10x> -15 \to x> -15/10= -3/2#

Again, we can't take the whole interval #(-3/2, infty)# because we are assuming that #x<0#, so

#x<0 \and x> -3/2 \implies x \in (-3/2,0)#

So, the inequality is verified from #-3/2# to #0#, and from #0# onward.

Moreover, when #x=0#, we have #0<5#, which is true. So, the global solution is #x> -3/2#

Here you can see the graph of the two functions, and as you can see, from #-3/2# on, #4/3abs(x)# is below #2x+5#.