1. **State the problem:** Solve the inequality $$|2x - 7| < 1$$ and graph the solution. 2. **Recall the definition of absolute value inequality:** For $$|A| < B$$ where $$B > 0$$, the inequality is equivalent to $$-B < A < B$$. 3. **Apply this to our problem:** $$|2x - 7| < 1 \implies -1 < 2x - 7 < 1$$ 4. **Solve the compound inequality:** Add 7 to all parts: $$-1 + 7 < 2x - 7 + 7 < 1 + 7$$ $$6 < 2x < 8$$ 5. **Divide all parts by 2:** $$\frac{6}{2} < \frac{2x}{2} < \frac{8}{2}$$ Show cancellation: $$\frac{\cancel{6}}{\cancel{2}} < x < \frac{\cancel{8}}{\cancel{2}}$$ Simplify: $$3 < x < 4$$ 6. **Interpretation:** The solution is all $$x$$ values strictly between 3 and 4. 7. **Graphing:** This corresponds to an open interval (3,4) on the number line, meaning open circles at 3 and 4 and shading between them. **Final answer:** $$3 < x < 4$$