Question

1. \\(\frac{2x + 12}{3x - 9} cdot \frac{6 - 2x}{3x + 18}\\)

Explanation:

Step1: Factor each numerator and denominator

Factor \(2x + 12\) as \(2(x + 6)\), \(3x - 9\) as \(3(x - 3)\), \(6 - 2x\) as \(-2(x - 3)\), and \(3x + 18\) as \(3(x + 6)\). So the expression becomes \(\frac{2(x + 6)}{3(x - 3)} \cdot \frac{-2(x - 3)}{3(x + 6)}\).

Step2: Multiply the numerators and denominators

Multiply the numerators: \(2(x + 6) \cdot -2(x - 3)= -4(x + 6)(x - 3)\). Multiply the denominators: \(3(x - 3) \cdot 3(x + 6)=9(x + 6)(x - 3)\). The expression is now \(\frac{-4(x + 6)(x - 3)}{9(x + 6)(x - 3)}\).

Step3: Cancel common factors

Cancel out the common factors \((x + 6)\) and \((x - 3)\) (assuming \(x
eq - 6\) and \(x
eq 3\)). We are left with \(\frac{-4}{9}\).

Answer:

\(-\frac{4}{9}\)