Let's rewrite the inequality
5+1/x-16/x>05+1x−16x>0
5-15/x>05−15x>0
(5(x-3))/x>05(x−3)x>0
Let f(x)=(5(x-3))/xf(x)=5(x−3)x
The domain of f(x)f(x) is D_f(x)=RR-{0}
Now we can do a sign chart
color(white)(aaaa)xcolor(white)(aaaa)-oocolor(white)(aaaaaa)0color(white)(aaaaaaa)3color(white)(aaaa)+oo
color(white)(aaaa)xcolor(white)(aaaaaaaa)-color(white)(aa)∥color(white)(aa)+color(white)(aaaa)+
color(white)(aaaa)x-3color(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[