Question

1. \\(\frac{1}{k + 2} \cdot \frac{9k + 63}{k + 7}\\)

Explanation:

Step1: Factor the numerator

Factor \(9k + 63\) as \(9(k + 7)\). So the expression becomes \(\frac{1}{k + 2} \cdot \frac{9(k + 7)}{k + 7}\).

Step2: Cancel common factors

Cancel out the common factor \((k + 7)\) in the numerator and denominator. We get \(\frac{1}{k + 2} \cdot 9\).

Step3: Multiply the remaining terms

Multiply \(\frac{1}{k + 2}\) and \(9\) to get \(\frac{9}{k + 2}\).

Answer:

\(\frac{9}{k + 2}\)