Suppose $a$ is a real number such that $3a + 6$ is the greatest integer less than or equal to $a$ and $4a + 9$ is the least integer greater than or equal to $a$. Compute $a$.
Given the problem, we need to find the real number \( a \) such that:
\[ 3a + 6 \leq a < 4a + 9 \]

We can rewrite the problem using the floor and ceiling functions:
\[ \begin{cases} 
3a + 6 = \lfloor a \rfloor \\
4a + 9 = \lceil a \rceil 
\end{cases} \]

### Case 1: \( a \in \mathbb{Z} \)
If \( a \) is an integer, then \( \lfloor a \rfloor = \lceil a \rceil = a \). Substituting into the equations, we get:
\[ 3a + 6 = a \]
\[ 4a + 9 = a \]

Solving the first equation:
\[ 3a + 6 = a \]
\[ 3a - a = -6 \]
\[ 2a = -6 \]
\[ a = -3 \]

Substituting \( a = -3 \) into the second equation to verify:
\[ 4(-3) + 9 = -12 + 9 = -3 \]
This is consistent, so \( a = -3 \) is a valid solution.

### Case 2: \( a \notin \mathbb{Z} \)
If \( a \) is not an integer, then \( \lceil a \rceil = \lfloor a \rfloor + 1 \). Let \( \lfloor a \rfloor = n \) where \( n \) is an integer, then:
\[ 3a + 6 = n \]
\[ 4a + 9 = n + 1 \]

Solving these equations:
From the first equation:
\[ 3a + 6 = n \]
\[ 3a = n - 6 \]
\[ a = \frac{n - 6}{3} \]

From the second equation:
\[ 4a + 9 = n + 1 \]
\[ 4a = n - 8 \]
\[ a = \frac{n - 8}{4} \]

Equating the two expressions for \( a \):
\[ \frac{n - 6}{3} = \frac{n - 8}{4} \]

Cross-multiplying:
\[ 4(n - 6) = 3(n - 8) \]
\[ 4n - 24 = 3n - 24 \]
\[ 4n - 3n = -24 + 24 \]
\[ n = 0 \]

Substituting \( n = 0 \) back into the expression for \( a \):
\[ a = \frac{0 - 6}{3} = -2 \]

However, we need to check if \( a = -2 \) satisfies the original conditions:
\[ 3(-2) + 6 = -6 + 6 = 0 \]
\[ 4(-2) + 9 = -8 + 9 = 1 \]

Since \( \lfloor -2 \rfloor = -3 \) and \( \lceil -2 \rceil = -2 \), this does not satisfy the conditions. Therefore, \( a = -2 \) is not a valid solution.

Conclusion:
The only valid solution is when \( a \) is an integer, specifically \( a = -3 \).

The final answer is \( \boxed{-3} \)