18. As shown in the figure, given $A B=2, B G=3, G E=4, D E=5, \triangle B C G$ and $\triangle E F G$ have a combined area of $24, \triangle A G F$ and $\triangle C D G$ have a combined area of 51. Therefore, the combined area of $\triangle A B C$ and $\triangle D E F$ is $\qquad$.
【Solution】Solution: Draw $C M \perp A D$, with the foot of the perpendicular at $M$, and draw $F N \perp A D$, with the foot of the perpendicular at $N$. Let $C M=x, F N=y$. According to the problem, we get the system of equations $\left\{\begin{array}{l}\frac{1}{2} \times 3 x+\frac{1}{2} \times 4 y=24 \\ \frac{1}{2} \times 9 x+\frac{1}{2} \times 5 y=51\end{array}\right.$, solving the system of equations yields $\left\{\begin{array}{l}x=8 \\ y=6\end{array}\right.$,

Therefore, the sum of the areas of $\triangle A B C$ and $\triangle D E F$ is:
$$
\frac{1}{2} A B \cdot C M+\frac{1}{2} D E \cdot F N=\frac{1}{2} \times 2 \times 8+\frac{1}{2} \times 5 \times 6=8+15=23 \text {. }
$$

Hence, the answer is: 23.