Let's rewrite the inequality

#5+1/x-16/x>0#

#5-15/x>0#

#(5(x-3))/x>0#

Let #f(x)=(5(x-3))/x#

The domain of #f(x)# is #D_f(x)=RR-{0}#

Now we can do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##0##color(white)(aaaaaaa)##3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aa)##∥##color(white)(aa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-3##color(white)(aaaaa)##-##color(white)(aa)##∥##color(white)(aa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aa)##∥##color(white)(aa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)>0#, when #x in ] -oo,0 [ uu ] 3, +oo[ #