6. In the chicken and rabbit cage, there are a total of 40 heads, the number of rabbit feet is 8 less than 10 times the number of chicken feet, so the number of rabbits is $\qquad$.
Analysis: Using the method of enumeration by approximation, since the number of rabbit feet is 9 times more than the number of chicken feet, and when the number of chickens and rabbits is the same, the number of rabbit feet is twice that of chicken feet, it follows that there are more rabbits than chickens. We can assume there are 35 rabbits, and adjust this number up or down to find the solution.
\begin{tabular}{|l|l|l|l|}
\hline Rabbits & 35 & 34 & 33 \\
\hline Rabbit feet & 140 & 136 & 132 \\
\hline Chicken feet & 10 & 12 & 14 \\
\hline Multiple of rabbit feet to chicken feet & $>10$ times & $>10$ times & \\
\hline
\end{tabular}

We can set up equations to solve the problem. Let the number of rabbits be $x$, then the number of chickens is (40-x). Based on the relationship of the number of feet, we can set up the equation:
$$
4 x + 8 = 10 \times 2 \times (40 - x)
$$

Solving for $x$ gives $x = 33$.
Method 3: The number of rabbit feet is 8 less than 10 times the number of chicken feet, and the number of rabbits is 2 less than 5 times the number of chickens. Therefore, the number of chickens is $(40 + 2) \div (5 + 1) = 7$, and the number of rabbits is $40 - 7 = 33$.
Method 4: Let the number of chicken feet be 1 unit, then the number of rabbit feet is 10 units minus 8. If we consider chickens as rabbits, since one rabbit has twice the number of feet as one chicken, the total number of feet increases by 1 unit. The total number of feet is $40 \times 4 = 160$. One unit is $(160 + 8) \div (10 + 1 + 1) = 14$, so the number of chickens is $14 \div 2 = 7$, and the number of rabbits is $40 - 7 = 33$.