Assume there exists a line $l$ passing through the point $(-5, -4)$,
such that it forms a triangle with the two coordinate axes with an area of 5.
Let the equation of line $l$ be: $$\frac{x}{a} + \frac{y}{b} = 1,$$
then $$\frac{-5}{a} + \frac{-4}{b} = 1.$$ This simplifies to $4a + 5b + ab = 0.$ The area $S = \frac{1}{2}|ab| = 5,$ which simplifies to $|ab| = 10.$
Combining $$\begin{cases} 4a + 5b + ab = 0 \\ |ab| = 10 \end{cases},$$
we get $$\begin{cases} a = -\frac{5}{2} \\ b = 4 \end{cases}$$ or $$\begin{cases} a = 5 \\ b = -2 \end{cases}.$$
Therefore, there exists a line $l$ with the equations $8x - 5y + 20 = 0$ or $2x - 5y - 10 = 0.$

Thus, the final answer is $\boxed{8x - 5y + 20 = 0 \text{ or } 2x - 5y - 10 = 0}$.