We cannot do crossing over.
Let's simplify the inequality
#6/(x+1)<-3#
#6/(x+1)+3<0#
#(6+3(x+1))/(x+1)<0#
#(6+3x+3)/(x+1)<0#
#(3x+9)/(x+1)<0#
#(3(x+3))/(x+1)<0#
Let #f(x)=(3(x+3))/(x+1)#
We can build the sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaaaaa)##-1##color(white)(aaaaaaa)##+oo#
#color(white)(aaaa)##x+3##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##+#
#color(white)(aaaa)##x+1##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+#
Therefore,
#f(x)<0# when #x in (-3,-1)#