We cannot do crossing over, we rearrange the inequality
#(3x)/(x+2)<5#
#(3x)/(x+2)-5<0#
#(3x-5(x+2))/(x+2)<0#
#(3x-5x-10)/(x+2)<0#
#(-2x-10)/(x+2)<0#
#(-2(x+5))/(x+2)<0#
Let #f(x)=(-2(x+5))/(x+2)#
Let's build a sign chart
#color(white)(aaaa)##x##color(white)(aaaaaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaaaaa)##-2##color(white)(aaaaaaa)##+oo#
#color(white)(aaaa)##-(x+5)##color(white)(aaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##-#
#color(white)(aaaa)##(x+2)##color(white)(aaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##-#
Therefore,
#f(x)<0#, when #x in (-oo,-5) uu (-2, +oo)#
graph{(3x)/(x+2)-5 [-36.53, 36.52, -18.28, 18.27]}