Question

which expression is equivalent to the following complex fraction?\\(\frac{\frac{x}{x - 3}}{\frac{x^2}{x^2 - 9}}\\)\\(\bigcirc\\ \frac{x - 3}{x}\\)\\(\bigcirc\\ \frac{x + 3}{1}\\)\\(\bigcirc\\ \frac{x + 3}{x}\\)\\(\bigcirc\\ \frac{x}{x + 3}\\)

Explanation:

Step1: Rewrite division as multiplication

$\frac{\frac{x}{x-3}}{\frac{x^2}{x^2-9}} = \frac{x}{x-3} \times \frac{x^2-9}{x^2}$

Step2: Factor the difference of squares

$x^2-9=(x+3)(x-3)$, so substitute:
$\frac{x}{x-3} \times \frac{(x+3)(x-3)}{x^2}$

Step3: Cancel common factors

Cancel $x$ and $(x-3)$:
$\frac{1}{1} \times \frac{x+3}{x} = \frac{x+3}{x}$

Answer:

C. $\frac{x+3}{x}$