x + 3\over x - 3 + x - 3\over x + 3 = \frac{5}{2}.

(x + 3)² + (x - 3)² \over (x + 3)(x - 3) = \frac{5}{2}.

⇒ (x + 3)² + (x - 3)² \over (x + 3)(x - 3) = \frac{5}{2}

⇒ (x² + 6x + 9) + (x² - 6x + 9) \over x² - 9 = \frac{5}{2}

⇒ 2x² + 18 \over x² - 9 = \frac{5}{2}

⇒ 2(2x² + 18) = 5(x² – 9)

⇒ 4x² + 36 = 5x² – 45

Subtract 4x² from both sides:

36 = x² – 45

⇒ x² = 36 + 45 = 81

Taking square roots, we get:

x = ± 9

Since x ≠ -3, 3, both solutions x = 9 and x = – 9 are acceptable.