We note first that division by zero is not allowed and that for this reason $5x+1$ may not be equal to zero and that therefore $x=-{{1}\over{5}}$ is not a solution.

We now distinguish two cases, namely $5x+1>0$ and $5x+1<0$.
In both cases we multiply the inequality on both sides by $5x+1$ because we then get a linear inequality, for which we know there is a solution method.

Suppose $5x+1>0$, i.e. $x> -{{1}\over{5}}$. Then we get $8<-8(5x+1)$.
When we move everything with $x$ to the left and all constant terms to the right, we get $40x<-16$.
Then, dvision by the coefficient of $x$gives $x < -{{2}\over{5}}$.
So we have the following system of inequalities: $x> -{{1}\over{5}}\,\wedge\; x < -{{2}\over{5}}$
and this simplifies to $\text{an empty solution set}$.

Suppose $5x+1<0$, i.e. $x< -{{1}\over{5}}$. Then we get $8>-8(5x+1)$.
When we move everything with $x$ to the left and all constant terms to the right, we get $40x>-16$.
Then, division by the coefficient of $x$ gives $x > -{{2}\over{5}}$.
So we have the following system of inequalities: $x< -{{1}\over{5}}\,\wedge\; x > -{{2}\over{5}}$
and this simplifies to $-{{2}\over{5}}\lt x\lt -{{1}\over{5}}$.

The solution of the original inequality is $-{{2}\over{5}}\lt x\lt -{{1}\over{5}}$.