1. **State the problem:** Solve the inequality $$\left|\frac{x+9}{x-9}\right| \le 2.$$\n\n2. **Rewrite the absolute value inequality:** This means $-2 \le \frac{x+9}{x-9} \le 2$.\n\n3. **Solve the left inequality:** \n$$-2 \le \frac{x+9}{x-9}$$\nMultiply both sides by $x-9$, but consider the sign of $x-9$ because it affects the inequality direction.\n\n- For $x > 9$, multiply without changing inequality:\n$$-2(x-9) \le x+9$$\n$$-2x + 18 \le x + 9$$\n$$18 - 9 \le x + 2x$$\n$$9 \le 3x$$\n$$x \ge 3$$\n\n- For $x < 9$, multiply and reverse inequality:\n$$-2(x-9) \ge x+9$$\n$$-2x + 18 \ge x + 9$$\n$$18 - 9 \ge x + 2x$$\n$$9 \ge 3x$$\n$$x \le 3$$\n\n4. **Solve the right inequality:** \n$$\frac{x+9}{x-9} \le 2$$\nMultiply both sides by $x-9$, again considering sign:\n\n- For $x > 9$:\n$$x+9 \le 2(x-9)$$\n$$x + 9 \le 2x - 18$$\n$$9 + 18 \le 2x - x$$\n$$27 \le x$$\n\n- For $x < 9$:\n$$x+9 \ge 2(x-9)$$\n$$x + 9 \ge 2x - 18$$\n$$9 + 18 \ge 2x - x$$\n$$27 \ge x$$\n\n5. **Combine intervals and consider domain:** \nThe expression is undefined at $x=9$.\n\n- For $x > 9$, from left inequality: $x \ge 3$ (always true), from right inequality: $x \ge 27$. So for $x > 9$, solution is $x \ge 27$.\n\n- For $x < 9$, from left inequality: $x \le 3$, from right inequality: $x \le 27$ (always true). So for $x < 9$, solution is $x \le 3$.\n\n6. **Final solution:** \n$$(-\infty, 3] \cup [27, \infty)$$\n\n7. **Summary:** The values of $x$ that satisfy the inequality are all real numbers less than or equal to 3, and all real numbers greater than or equal to 27, excluding $x=9$ where the expression is undefined.