Question

question 1 (1 point)
when fully simplified, excluding restrictions, \\(\frac{3x + 1}{x^2 - 4} \times \frac{x - 2}{3x + 1}\\) is equal to

a) \\(\frac{1}{x + 2}\\)

b) \\(\frac{x - 2}{x^2 - 4}\\)

c) \\(\frac{1}{x - 2}\\)

d) \\(x + 2\\)

Explanation:

Factor the denominators

$$ x^2 - 4 = (x - 2)(x + 2) $$

Write the product as a single expression

$$ \frac{3x + 1}{(x - 2)(x + 2)} \times \frac{x - 2}{3x + 1} = \frac{(3x + 1)(x - 2)}{(x - 2)(x + 2)(3x + 1)} $$

Cancel common factors

$$ \frac{\cancel{(3x + 1)}\cancel{(x - 2)}}{\cancel{(x - 2)}(x + 2)\cancel{(3x + 1)}} = \frac{1}{x + 2} $$

Answer:

• (A) \(\frac{1}{x+2}\) (Correct answer)
• (B) \(\frac{x-2}{x^2-4}\)
• (C) \(\frac{1}{x-2}\)
• (D) \(x+2\)