We cannot do crossing over
#(x+10)/(x+3)<2#
Let's rewrite the inequality
#(x+10)/(x+3)-2<0#
#((x+10)-2(x+3))/(x+3)<0#
#((x+10-2x-6))/(x+3)<0#
#((4-x))/(x+3)<0#
Let #f(x)=(4-x)/(x+3)#
We build the sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##-3##color(white)(aaaaaaaa)##4##color(white)(aaaa)##+oo#
#color(white)(aaaa)##x+3##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##4-x##color(white)(aaaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aa)##0##color(white)(aa)##-#
#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aa)##0##color(white)(aa)##-#
Therefore,
#f(x)<0# when #x in (-3,4)#
graph{(x+10)/(x+3)-2 [-16.02, 16.02, -8.01, 8.02]}