8. (6 points) Schools A, B, and C jointly held an art exhibition. Among the paintings on display, 41 were not from School A, 38 were not from School B, and the total number of paintings from Schools A and B was 43. Therefore, the number of paintings from School C is $\qquad$.

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【Analysis】41 pieces are either from School A or Schools B and C, 38 pieces are either from School B or Schools A and C, where the number of pieces from School C is included in both 41 and 38. Therefore, $41+38=79$ (pieces) is twice the total number of pieces from Schools A, B, and C. Subtracting the total number of pieces from Schools A and B gives twice the number of pieces from School C, and dividing by 2 gives the number of pieces from School C.

【Solution】Solution: $(41+38-43) \div 2$
$$
\begin{array}{l}
=(79-43) \div 2 \\
=36 \div 2 \\
=18 \text { (pieces) }
\end{array}
$$

Answer: School C has 18 pieces of art on display.
The answer is: 18.