Let's go step-by-step to find the product of the given fractions:
[tex]\[
\frac{2y}{y-3} \cdot \frac{4y - 12}{2y + 6}.
\][/tex]
First, let's write down the two given fractions:
1. [tex]\(\frac{2y}{y-3}\)[/tex]
2. [tex]\(\frac{4y - 12}{2y + 6}\)[/tex]
Now, we'll multiply these fractions. According to the rules of fraction multiplication, we multiply the numerators together and the denominators together:
[tex]\[
\frac{2y}{y-3} \cdot \frac{4y - 12}{2y + 6} = \frac{(2y) \cdot (4y - 12)}{(y-3) \cdot (2y + 6)}.
\][/tex]
Next, we'll simplify the expression.
Consider the numerator first:
[tex]\[
2y \cdot (4y - 12) = 2y \cdot 4y - 2y \cdot 12 = 8y^2 - 24y.
\][/tex]
Now, consider the denominator:
[tex]\[
(y - 3) \cdot (2y + 6)
\][/tex]
Observe that [tex]\(2y + 6\)[/tex] can be factored:
[tex]\[
2y + 6 = 2(y + 3).
\][/tex]
Thus, the denominator factoring is:
[tex]\[
(y - 3) \cdot 2(y + 3).
\][/tex]
Putting it all together, we have:
[tex]\[
\frac{8y^2 - 24y}{2(y-3)(y+3)}.
\][/tex]
Now, notice that the numerator [tex]\(8y^2 - 24y\)[/tex] has a common factor of [tex]\(8y\)[/tex]:
[tex]\[
8y^2 - 24y = 8y(y - 3).
\][/tex]
So we have:
[tex]\[
\frac{8y(y - 3)}{2(y - 3)(y + 3)}.
\][/tex]
Next, we cancel out the common factor [tex]\((y - 3)\)[/tex]:
[tex]\[
\frac{8y}{2(y + 3)}.
\][/tex]
Now, simplify the fraction by dividing both the numerator and the denominator by 2:
[tex]\[
\frac{8y}{2(y + 3)} = \frac{4y}{y + 3}.
\][/tex]
Therefore, the product of the given fractions is:
[tex]\[
\frac{4y}{y + 3}.
\][/tex]
So, the correct answer is:
[tex]\[
\boxed{\frac{4 y}{y+3}}
\][/tex]
