jmc

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 $y<6$.