1. State the problem: Solve the inequality $|2n-18|\le 2$.\n\n2. Use the absolute value rule (two cases): Since $|A|\le b$ with $b\ge 0$, we can write it as an AND of two inequalities: $-b\le A\le b$.\n\n3. Substitute $A=2n-18$ and $b=2$: $-2\le 2n-18\le 2$.\n\n4. Split into two separate inequalities (AND): $-2\le 2n-18$ and $2n-18\le 2$.\n\n5. Solve the first inequality $-2\le 2n-18$: add $18$ to all three parts to isolate the $2n$.\n\n6. Compute: $$-2+18\le 2n-18+18$$\n\n7. Simplify with a canceled shift (show the intermediate step): $$\cancel{(-18)}-2+18\le 2n\,.$$\n\n8. Simplify: $16\le 2n$.\n\n9. Divide both sides by $2$ (positive, so the inequality direction stays the same): $$\frac{16}{2}\le \frac{2n}{2}$$\n\n10. Cancel and simplify: $$\cancel{2}n\ge \cancel{2} \cdot 8\;\Rightarrow\; n\ge 8.$$\n\n11. Solve the second inequality $2n-18\le 2$: add $18$ to all three parts.\n\n12. Compute: $$2n-18+18\le 2+18$$\n\n13. Simplify: $2n\le 20$.\n\n14. Divide both sides by $2$ (positive): $$\frac{2n}{2}\le \frac{20}{2}$$\n\n15. Cancel and simplify: $$n\le 10.$$\n\n16. Combine the two results (because it was AND): $8\le n\le 10$.\n\n17. Write the solution set in set-builder form: $\{n\mid 8\le n\le 10\}$.\n\n18. Final answer: $8\le n\le 10$.\n