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In this problem, Mr. Clarke wants to buy 3 security cameras and spend less than $450 in total. We need to determine how much he can spend on each camera and express this situation as an inequality.

The variable $ x $ represents the amount Mr. Clarke wants to spend on each camera. The inequality that describes the situation is:

\[
3x < 450
\]

### To solve the inequality:
1. Divide both sides of the inequality by 3:
   \[
   x < \frac{450}{3}
   \]
   \[
   x < 150
   \]

This means Mr. Clarke wants to spend less than **$150** on each camera.

### The completed sentence:
Mr. Clarke wants to spend less than **$150** on each camera.

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Would you like more details or have any further questions on this?

Here are 5 related questions to expand your understanding:
1. What would happen if Mr. Clarke could spend at most $450 instead of less than $450?
2. How would the inequality change if Mr. Clarke wanted to buy 5 cameras instead of 3?
3. What if Mr. Clarke wanted to buy a different number of cameras? How would you set up the inequality?
4. How could you write this as a word problem where Mr. Clarke spends more than $450?
5. Can you think of a real-life scenario where you might need to solve a similar inequality?

**Tip:** When solving inequalities, remember that dividing or multiplying by a negative number reverses the inequality symbol.

Mr. Clarke plans to buy 3 security cameras to install outside his donut shop. He wants to spend less than $450 on the cameras in all. Let x represent how much Mr. Clarke wants to spend on each camera. Which inequality describes the problem?

Learn how to solve an inequality involving Mr. Clarke's purchase of 3 security cameras for his shop, where he wants to spend less than $450 in total. This algebra problem covers solving basic inequalities and understanding linear equations. Suitable for students in Grades 6-8.

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