$$ \frac{y^{2}-2 y-15}{y^{2}-6 y+9} \div \frac{y-5}{y-3}=\square $$ (Simplify your answer.)

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UpStudy AI · Accepted Answer

To simplify the expression

$$
\frac{y^{2}-2y-15}{y^{2}-6y+9} \div \frac{y-5}{y-3},
$$

we first rewrite the division as multiplication by the reciprocal:

$$
\frac{y^{2}-2y-15}{y^{2}-6y+9} \cdot \frac{y-3}{y-5}.
$$

Next, we will factor the polynomials in the first fraction.

1. **Factoring $$y^{2}-2y-15$$**:
   We need two numbers that multiply to $$-15$$ and add to $$-2$$. These numbers are $$-5$$ and $$3$$. Thus, we can factor it as:

$$
   y^{2}-2y-15 = (y-5)(y+3).
   $$

2. **Factoring $$y^{2}-6y+9$$**:
   This is a perfect square trinomial. It can be factored as:

$$
   y^{2}-6y+9 = (y-3)(y-3) = (y-3)^{2}.
   $$

Now we can substitute these factored forms back into the expression:

$$
\frac{(y-5)(y+3)}{(y-3)^{2}} \cdot \frac{y-3}{y-5}.
$$

Next, we can simplify the expression by canceling out common factors. The $$y-5$$ in the numerator of the first fraction cancels with the $$y-5$$ in the denominator of the second fraction, and one $$y-3$$ in the numerator of the second fraction cancels with one $$y-3$$ in the denominator of the first fraction:

$$
\frac{(y+3)}{(y-3)}.
$$

Thus, the simplified expression is:

$$
\frac{y+3}{y-3}.
$$

Therefore, the final answer is:

$$
\boxed{\frac{y+3}{y-3}}.
$$
$$\frac{y+3}{y-3}$$

(Simplify your answer.)

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