Absolute value inequality range

Using the property $|a| + |b| = |a - b|$ when $(a - 20/3)(a - 40/3) \le 0$: $|3x - 20| + |40 - 3x| = |(3x - 20) + (40 - 3x)| = 20$

This holds if and only if: $3x - 20 \ge 0 \quad \text{and} \quad 40 - 3x \ge 0$ $\frac{20}{3} \le x \le \frac{40}{3}$ $6.66 \le x \le 13.33$

We need to find which interval is entirely contained inside $[6.66, 13.33]$:

• Option A: $6 < x < 11$ (starts at 6, not contained)
• Option B: $7 < x < 12$ ($7 > 6.66$ and $12 < 13.33$, fully contained)
• Option C: $10 < x < 15$ (ends at 15, not contained)
• Option D: $9 < x < 14$ (ends at 14, not contained)

Thus, Option B necessarily holds.