Let's solve this step-by-step.

We are tasked with writing and solving an inequality to determine $x$, the number of 40-kilogram crates that can be loaded into the container.

Key Information:

1. Maximum weight the container can hold: $23500 \, \text{kg}$.
2. Weight of shipments already loaded: $12100 \, \text{kg}$.
3. Weight of each crate: $40 \, \text{kg}$.

Step 1: Setting up the inequality

The total weight in the container is the sum of:

• The weight already loaded: $12100 \, \text{kg}$,
• The weight of $x$ crates, each weighing $40 \, \text{kg}$: $40x \, \text{kg}$.

This total weight must not exceed the maximum capacity of the container $23500 \, \text{kg}$. Hence, the inequality is: $12100 + 40x \leq 23500$

Step 2: Solving the inequality

1. Subtract $12100$ from both sides: $40x \leq 23500 - 12100$

2. Simplify the right-hand side: $40x \leq 11400$

3. Divide both sides by $40$: $x \leq \frac{11400}{40}$

4. Simplify: $x \leq 285$

Final Answer:

The inequality is: $12100 + 40x \leq 23500$

And the solution is: $x \leq 285$

Thus, at most 285 crates can be loaded into the container.

Would you like me to explain this in more detail or provide further clarification?

Related Questions:

1. How would the inequality change if the weight of each crate was increased to 50 kg?
2. What happens if the maximum weight capacity is reduced to 20000 kg?
3. If 150 crates have already been added, how much weight is left for additional crates?
4. How would you express the situation if the number of crates had to be at least 100?
5. What if the weight already loaded increased to 15000 kg? How would the solution change?

Tip:

When working with inequalities, always ensure you maintain the direction of the inequality while performing operations like addition, subtraction, multiplication, or division!