5. Solve the inequality \( |x-5|-|x-3| \geq 0 \) and represent the solution on the number line.

We need to solve the inequality   |x – 5| – |x – 3| ≥ 0. A good strategy is to break the number line into regions where the expressions inside the absolute values change sign. The critical values are x = 3 and x = 5. Step 1. Consider three cases: Case 1: x < 3   Here, both x – 5 and x – 3 are negative.   So, |x – 5| = 5 – x and |x – 3| = 3 – x.   Then the inequality becomes    (5 – x) – (3 – x) = 5 – x – 3 + x = 2.   Since 2 ≥ 0, the inequality holds for every x in (–∞, 3). Case 2: 3 ≤ x < 5   Now, x – 3 is nonnegative and x – 5 is negative.   So, |x – 3| = x – 3 and |x – 5| = 5 – x.   The inequality becomes:    (5 – x) – (x – 3) = 5 – x – x + 3 = 8 – 2x.   We require:    8 – 2x ≥ 0  ⟹ 8 ≥ 2x  ⟹ x ≤ 4.   Thus, in this case, the inequality holds for x between 3 and 4 (inclusive of 3 and 4). Case 3: x ≥ 5   Now, both x – 5 and x – 3 are nonnegative.   So, |x – 5| = x – 5 and |x – 3| = x – 3.   Then the inequality becomes:    (x – 5) – (x – 3) = x – 5 – x + 3 = –2.   Since –2 is not ≥ 0, the inequality fails for x ≥ 5. Step 2. Combine the results from all cases:   For x < 3, the inequality always holds.   For 3 ≤ x ≤ 4, the inequality holds.   For x > 4, the inequality fails. Thus, the solution set is:   x ∈ (–∞, 4]. Step 3. Representing the solution on the number line:   • Draw a horizontal line for the number line.   • Shade the entire line from –∞ up to 4.   • Place a closed (filled) circle at x = 4 to indicate that 4 is included.   • Do not shade any portion of the line for x > 4. Final Answer: The inequality |x – 5| – |x – 3| ≥ 0 is satisfied for all x in (–∞, 4].