Question

\\\frac{3}{x - 3} + \frac{x}{x + 3} =\\

(a) \\\frac{1}{x - 3}\\
(b) \\\frac{x + 3}{x - 3}\\
(c) \\\frac{3 + x}{2x}\\
(d) \\\frac{x^2}{x^2 - 9}\\
(e) \\\frac{x^2 + 9}{x^2 - 9}\\

Explanation:

Find a common denominator

$$ \text{LCD} = (x - 3)(x + 3) = x^2 - 9 $$

Rewrite each fraction with the common denominator

$$ \frac{3(x + 3)}{(x - 3)(x + 3)} + \frac{x(x - 3)}{(x - 3)(x + 3)} $$

Combine and simplify the numerators

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

Answer:

• (A) \(\frac{1}{x - 3}\)
• (B) \(\frac{x + 3}{x - 3}\)
• (C) \(\frac{3 + x}{2x}\)
• (D) \(\frac{x^2}{x^2 - 9}\)
• (E) \(\frac{x^2 + 9}{x^2 - 9}\) (Correct answer)