Perform the indicated operation. \( \frac{y^{2}-2 y-15}{y^{2}-6 y+9} \div \frac{y-5}{y-3} \)

To solve the expression \( \frac{y^{2}-2y-15}{y^{2}-6y+9} \div \frac{y-5}{y-3} \), we first need to simplify the fractions involved. Start by factoring the quadratic expressions: 1. The numerator \( y^2 - 2y - 15 \) factors as \( (y - 5)(y + 3) \). 2. The denominator \( y^2 - 6y + 9 \) can be factored as \( (y - 3)(y - 3) \) or \( (y - 3)^2 \). Thus, we can rewrite the original expression as: \[ \frac{(y - 5)(y + 3)}{(y - 3)^2} \div \frac{y - 5}{y - 3} \] Next, we can rewrite the division as multiplication by the reciprocal: \[ \frac{(y - 5)(y + 3)}{(y - 3)^2} \cdot \frac{y - 3}{y - 5} \] Now, cancel the common factors: The \( (y - 5) \) terms in the numerator and the denominator cancel out, and we can also cancel one \( (y - 3) \) from the numerator and denominator: \[ \frac{(y + 3)}{(y - 3)} \] Thus, the simplified expression is: \[ \frac{y + 3}{y - 3} \] So, the final result is: \[ \frac{y + 3}{y - 3} \]