1. **State the problem:** We need to find the possible amounts of food and supplies $f$ (in load units) that the catering van can carry, given the compound inequality: $$24,000 \leq 3,000f + 24,000 \leq 60,000$$ 2. **Understand the inequality:** This inequality shows the total weight of the van with food and supplies. The empty van weighs 24,000 pounds, and each load unit of food and supplies weighs 3,000 pounds. 3. **Isolate $f$ in the inequality:** Subtract 24,000 from all parts of the inequality: $$24,000 - 24,000 \leq 3,000f + 24,000 - 24,000 \leq 60,000 - 24,000$$ which simplifies to: $$0 \leq 3,000f \leq 36,000$$ 4. **Divide all parts by 3,000 to solve for $f$:** $$\frac{0}{3,000} \leq \frac{3,000f}{3,000} \leq \frac{36,000}{3,000}$$ Using cancellation notation: $$0 \leq \cancel{3,000}f / \cancel{3,000} \leq 12$$ which simplifies to: $$0 \leq f \leq 12$$ 5. **Interpretation:** The catering van can carry between 0 and 12 load units of food and supplies. **Final answer:** $$0 \leq f \leq 12$$