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)#