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To solve the given expression \((3y - 5)(5y - 7)\), we will use the distributive property, often referred to as the FOIL method (First, Outer, Inner, Last):

\[
(3y - 5)(5y - 7)
\]

1. **First terms:**  
   Multiply the first terms from each binomial:  
   \[
   3y \cdot 5y = 15y^2
   \]

2. **Outer terms:**  
   Multiply the outer terms:  
   \[
   3y \cdot (-7) = -21y
   \]

3. **Inner terms:**  
   Multiply the inner terms:  
   \[
   (-5) \cdot 5y = -25y
   \]

4. **Last terms:**  
   Multiply the last terms:  
   \[
   (-5) \cdot (-7) = 35
   \]

Now, combine all the terms:

\[
15y^2 - 21y - 25y + 35
\]

Next, combine the like terms:

\[
15y^2 - 46y + 35
\]

So, the simplified expression is:

\[
\boxed{15y^2 - 46y + 35}
\]

Let me know if you need more details or have any questions!

Here are five related questions for you to explore:
1. How would you apply the FOIL method to \((2x + 3)(4x - 5)\)?
2. What happens when multiplying two binomials like \((y + 4)(y - 4)\)?
3. How do you factor a quadratic like \(15y^2 - 46y + 35\) back into binomials?
4. Can you expand and simplify \((4y + 2)(3y - 6)\)?
5. How do you solve a quadratic equation after simplifying it?

**Tip:** Always remember to combine like terms (terms with the same variable and exponent) after expanding!

Learn how to multiply and simplify the expression (3y - 5)(5y - 7) using the distributive property and FOIL method. This solution simplifies the quadratic expression step-by-step, suitable for students in Grades 8-10.

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