
We can solve this system of equations by using the method of substitution or elimination. 

Let's use elimination. Multiply the first equation by 3 and the second equation by 4 to eliminate the variable $b$.

$3(a + 4b) = 3(33) \implies 3a + 12b = 99$
$4(6a + 3b) = 4(51) \implies 24a + 12b = 204$

Now we can subtract the first equation from the second equation to eliminate $b$.

$(24a + 12b) - (3a + 12b) = 204 - 99$
$24a - 3a = 105$
$21a = 105$
$a = 5$

Substitute this value of $a$ into one of the original equations to solve for $b$.

$a + 4b = 33$
$5 + 4b = 33$
$4b = 28$
$b = 7$

Finally, we can find the value of $a + b$.

$a + b = 5 + 7 = \boxed{12}$

The answer is: 12
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Given that $a + 4b = 33$ and $6a + 3b = 51$, what is the value of $a + b$?

### Answer:
