Question

which expression is equivalent to the following complex fraction?

\frac{\frac{x}{x-3}}{\frac{x^2}{x^2-9}}

\circ \frac{x-3}{x}
\circ \frac{x+3}{1}
\circ \frac{x+3}{x}
\circ \frac{x}{x+3}

Explanation:

Rewrite division as multiplication by the reciprocal

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

Factor the difference of squares

$$ x^2-9 = (x-3)(x+3) $$

Simplify the expression by canceling common factors

$$ \frac{x}{x-3} \cdot \frac{(x-3)(x+3)}{x^2} = \frac{x(x-3)(x+3)}{x^2(x-3)} = \frac{x+3}{x} $$

Answer:

• (A) \(\frac{x-3}{x}\)
• (B) \(\frac{x+3}{1}\)
• (C) \(\frac{x+3}{x}\) (Correct answer)
• (D) \(\frac{x}{x+3}\)