jmc

Let the number of acrobats in the show be $a$ and the number of elephants be $e$. We are looking for the value of $a$. Assuming that each acrobat has 2 legs and 1 head, and that each elephant has 4 legs and 1 head, we can set up the following system of equations:

$$\begin{array}{rl}2a+4e& =40\\ a+e& =15\end{array}$$To solve for $a$, we need to eliminate $e$ from the equations above. We can rewrite the second equation above as $e=15-a$, and substituting this into the first equation to eliminate $e$ gives $2a+4(15-a)=40$, or $a=10$. Thus, there are $\boxed{10}$ acrobats in the circus show.