I notice that the equation involves an absolute value function, which can have different behaviors depending on the sign of the argument inside the brackets. So I want to split the equation into two cases: one where $y-6 \geq 0$ and one where $y-6 < 0$. If $y-6 \geq 0$, then $|y-6| = y-6$ and the equation becomes $y-6 + 2y = 9$. Simplifying this equation, I get $3y = 15$ and $y = 5$. But I need to check if this value of $y$ satisfies the original condition that $y-6 \geq 0$. Plugging in $y = 5$, I see that $5-6 = -1$ which is not greater than or equal to zero, so this is not a valid solution. Therefore, I only need to consider the other case where $y-6 < 0$. If $y-6 < 0$, then $|y-6| = -(y-6) = 6-y$ and the equation becomes $6-y + 2y = 9$. Simplifying this equation, I get $y = 3$. This time, I need to check if this value of $y$ satisfies the original condition that $y-6 < 0$. Plugging in $y = 3$, I see that $3-6 = -3$ which is indeed less than zero, so this is a valid solution. Therefore, the only solution to the equation is $y = 3$. 
Solve the equation $|y-6| + 2y = 9$ for $y$.