Question

what is the product?
\\(\frac{2y}{y - 3} cdot \frac{4y - 12}{2y + 6}\\)
\\(circ \frac{2}{3}\\)
\\(circ \frac{10}{9}\\)
\\(circ \frac{4y}{y - 3}\\)
\\(circ \frac{4y}{y + 3}\\)

Explanation:

Step1: Factor numerators and denominators

Factor \(4y - 12\) as \(4(y - 3)\) and \(2y + 6\) as \(2(y + 3)\). So the expression becomes \(\frac{2y}{y - 3} \cdot \frac{4(y - 3)}{2(y + 3)}\).

Step2: Cancel common factors

Cancel out the common factors \((y - 3)\) and \(2\) from the numerator and denominator. We have \(\frac{2y \cdot 4(y - 3)}{(y - 3) \cdot 2(y + 3)}=\frac{2y \cdot 2}{y + 3}=\frac{4y}{y + 3}\).

Answer:

\(\frac{4y}{y + 3}\) (the last option: \(\boldsymbol{\frac{4y}{y + 3}}\))