Let's rearrange the equation
#4/(x-3)-2<0#
#(4-2(x-3))/(x-3)<0#
#(4-2x+6)/(x-3)<0#
#(10-2x)/(x-3)<0#
Let #f(x)=(10-2x)/(x-3)#
The domain of #f(x)# is #D_f(x)=RR-{3}#
Now, we build the sign chart
#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaaa)##3##color(white)(aaaaaaa)##5##color(white)(aaaaaa)##+oo#
#color(white)(aaaa)##x-3##color(white)(aaaaa)##-##color(white)(aaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##10-2x##color(white)(aaaa)##+##color(white)(aa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-#
#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##-##color(white)(aa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-#
Therefore,
#f(x)<0# when #x in ]-oo, 3[uu]5, +oo[#