Let #f(x)=(3-x)/(x+5)#
We build a sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##-5##color(white)(aaaaaaa)##3##color(white)(aaaaaa)##+oo#
#color(white)(aaaa)##x+5##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##3-x##color(white)(aaaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-#
#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-#
Therefore,
#f(x)<=0# when #x in (-oo,-5) uu [3, +oo)# in interval notation or
#x<-5# and #x>=3# as inequality