Solve the equation $|y-6| + 2y = 9$ for $y$.
We consider two cases, $y\ge 6$ and $y < 6$.

Case 1: $y \ge 6:$  If $y \ge 6$, then $|y-6| = y-6$ and our equation is $y-6+2y=9$.  So, we have $3y = 15$, or $y=5$. However, $y=5$ does not satisfy $y\ge 6$.  Testing $y=5$, we have $|5-6| + 2\cdot 5 =11$, not 9, and we see that $y=5$ is not a solution.

Case 2: $y < 6:$   If $y<6$, then $|y-6| = -(y-6) = -y+6$, so our equation is $-y+6+2y = 9$, from which we have $y=\boxed{3}$. This is a valid solution, since $y=3$ satisfies the restriction $y<6$.
3
Level 4
Algebra
Problem: Solve the equation $|y-6| + 2y = 9$ for $y$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation, first isolate the absolute value expression. Rewrite the equation as |y - 6| = 9 - 2y. The absolute value can yield two cases: 

Case 1: y - 6 = 9 - 2y. 
Rearranging gives 3y = 15, leading to y = 5.

Case 2: y - 6 = -(9 - 2y). 
This simplifies to y - 6 = -9 + 2y, which rearranges to y = -3.

Next, we need to check which solutions are valid by substituting back into the original equation. For y = 5, we have |5 - 6| + 2(5) = 1 + 10 = 11, which does not equal 9. 

For y = -3, we calculate |-3 - 6| + 2(-3) = 9 - 6 = 3, which also does not equal 9. 

Since neither solution satisfies the original equation, we must re-evaluate our approach. 

Instead, we can simplify the equation directly. Start with |y - 6| + 2y = 9. 

Consider the case where y - 6 is non-negative (y ≥ 6). Then |y - 6| = y - 6, leading to y - 6 + 2y = 9, which simplifies to 3y - 6 = 9. Solving gives 3y = 15, or y = 5. However, this contradicts our assumption that y ≥ 6.

Now consider the case where y - 6 is negative (y < 6). Then |y - 6| = -(y - 6) = 6 - y. The equation becomes 6 - y + 2y = 9, simplifying to y + 6 = 9. Solving gives y = 3.

Since y = 3 is less than 6, it is a valid solution. 

Thus, the only solution that satisfies the original equation is y = 3.<|END_OF_LATENT|> The final answer is $3$.