If $a + 4b = 33$ and $6a + 3b = 51$, what is the value of $a + b$?
To solve this problem, I will use the elimination method to eliminate one of the variables and solve for the other. I notice that the coefficients of $a$ in the two equations are 1 and 6, so I can multiply the first equation by -6 and add it to the second equation to eliminate $a$. Multiplying the first equation by -6 gives $-6a - 24b = -198$. Adding this to the second equation gives $-21b = -147$. Solving for $b$, I get $b = 7$. Now that I have the value of $b$, I can plug it into either equation to solve for $a$. I will use the first equation, since it has smaller coefficients. Substituting $b = 7$ into the first equation gives $a + 4(7) = 33$. Simplifying, I get $a + 28 = 33$. Solving for $a$, I get $a = 5$. Now that I have both $a$ and $b$, I can find the value of $a + b$. Adding $a$ and $b$ gives $a + b = 5 + 7$. Simplifying, I get $a + b = 12$. 