4. As shown in Figure $2, D, E$ are points on the sides $AC, AB$ of $\triangle ABC$, respectively, and $BD, CE$ intersect at point $O$. If $S_{\triangle OCD}=2, S_{\triangle OBE}=3, S_{\triangle OBC}=4$, then, $S_{\text {quadrilateral } A O D E}=$ $\qquad$
4. $\frac{39}{5}$.

As shown in Figure 5, let $S_{\triangle O O D}=x$, $S_{\triangle \text { SOE }}=y$. Then we have
$$
\frac{x}{3+y}=\frac{2}{4}, \frac{y}{2+x}=\frac{3}{4} .
$$