Solve and check. If there is no solution, so indicate. \( \frac{y+9}{y^{2}-12 y+27}=\frac{1}{y-3}-\frac{1}{y-9} \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( y=\square \) (Simplify your answer.) B. There is no solution.

Let's solve the equation step by step. We start with the equation: \[ \frac{y+9}{y^2 - 12y + 27} = \frac{1}{y-3} - \frac{1}{y-9} \] First, let's simplify the right-hand side by finding a common denominator, which is \((y-3)(y-9)\): \[ \frac{1}{y-3} - \frac{1}{y-9} = \frac{(y-9) - (y-3)}{(y-3)(y-9)} = \frac{-6}{(y-3)(y-9)} \] Now the equation looks like this: \[ \frac{y+9}{y^2 - 12y + 27} = \frac{-6}{(y-3)(y-9)} \] Next, we factor \(y^2 - 12y + 27\): \[ y^2 - 12y + 27 = (y-3)(y-9) \] Now we have: \[ \frac{y+9}{(y-3)(y-9)} = \frac{-6}{(y-3)(y-9)} \] Since the denominators are the same, we equate the numerators: \[ y + 9 = -6 \] Solving for \(y\), we get: \[ y = -6 - 9 \] \[ y = -15 \] Next, we must check if \(y = -15\) is a valid solution by ensuring that it does not cause any denominators to become zero. Since inserting \(y = -15\) into \(y-3\) or \(y-9\) gives neither zero: - \(y - 3 = -15 - 3 = -18\) - \(y - 9 = -15 - 9 = -24\) Both evaluations are valid (not zero). Thus, the solution is: A. \( y = -15 \)