- Solve.
- Bob works for car dealer A which offers him a base rate of $800 and $15 for every car he sells. Tom works for car dealer B which gives him a base rate of $700 and $20 for each car he sells. If they sold the same number of cars last month, and Tom earned more than Bob. What is the minimum number of cars Tom sold last month?
- The cost of a charity dinner is $300 plus $18 per guest. The ticket to the dinner is $30 per person. How many tickets must be sold to cover the costs?

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## Introduction to Multi-Step Linear Inequalities

Multi-step linear inequalities are an essential concept in algebra that builds upon basic inequality principles. This lesson begins with an introduction video, which is crucial for grasping the fundamental ideas and techniques involved. The video provides a visual and auditory explanation of multi-step linear inequalities, setting the foundation for the rest of the lesson. As we progress, we'll delve into solving inequalities using various methods, including addition and subtraction in inequalities. We'll also explore graphing solutions on a number line, a vital skill for visualizing and interpreting inequality results. Additionally, we'll examine how these mathematical concepts apply to real-world problems, demonstrating their practical relevance. By mastering multi-step linear inequalities, students will enhance their problem-solving abilities and gain a deeper understanding of algebraic relationships. This knowledge is invaluable for more advanced mathematical topics and many real-life applications.

Solving inequalities using various methods, including addition and subtraction in inequalities, is a fundamental skill. We'll also delve into multiplication and division in inequalities, which are crucial techniques for solving more complex problems. Understanding these methods will allow students to tackle a wide range of mathematical challenges. Furthermore, we'll revisit graphing solutions on a number line to reinforce the visual aspect of interpreting inequalities. This comprehensive approach ensures that students are well-prepared for advanced topics in algebra and beyond.

Bob works for car dealer A which offers him a base rate of $800 and $15 for every car he sells. Tom works for car dealer B which gives him a base rate of $700 and $20 for each car he sells. If they sold the same number of cars last month, and Tom earned more than Bob. What is the minimum number of cars Tom sold last month? Express the problem in an inequality.

#### Step 1: Identify the Variables

First, we need to identify the variables involved in the problem. In this case, the variable is the number of cars sold by both Bob and Tom. Let's denote this variable as **X**. So, let **X** be the number of cars sold by both Bob and Tom last month.

#### Step 2: Write the Earnings Expressions

Next, we need to write expressions for the earnings of both Bob and Tom based on the given information:

**Bob's Earnings:**Bob has a base rate of $800 and earns an additional $15 for each car he sells. Therefore, Bob's total earnings can be expressed as:**800 + 15X**.**Tom's Earnings:**Tom has a base rate of $700 and earns an additional $20 for each car he sells. Therefore, Tom's total earnings can be expressed as:**700 + 20X**.

#### Step 3: Set Up the Inequality

According to the problem, Tom earned more than Bob last month. This means that Tom's earnings are greater than Bob's earnings. We can express this relationship as an inequality:

**700 + 20X > 800 + 15X**

#### Step 4: Simplify the Inequality

To find the minimum number of cars Tom sold, we need to solve the inequality for **X**. First, we will simplify the inequality by isolating the variable **X** on one side:

- Subtract 15X from both sides:
**700 + 20X - 15X > 800** - Simplify the expression:
**700 + 5X > 800** - Subtract 700 from both sides:
**5X > 100** - Divide both sides by 5:
**X > 20**

#### Step 5: Interpret the Solution

The inequality **X > 20** means that Tom must have sold more than 20 cars to earn more than Bob. Therefore, the minimum number of cars Tom sold last month is 21.

**Q1: What is the difference between a linear equation and a linear inequality?**

A linear equation has a single solution, while a linear inequality has a range of solutions. For example, x + 2 = 5 has one solution (x = 3), but x + 2 > 5 has many solutions (any x greater than 3).

**Q2: How do you solve a multi-step linear inequality?**

To solve a multi-step linear inequality, follow these steps: 1) Simplify the inequality by combining like terms. 2) Isolate the variable term on one side of the inequality. 3) Perform the same operations on both sides to isolate the variable. 4) Remember to flip the inequality sign when multiplying or dividing by a negative number.

**Q3: Why do we flip the inequality sign when multiplying or dividing by a negative number?**

Flipping the inequality sign when multiplying or dividing by a negative number is necessary to maintain the correct relationship between the two sides. This is because negative numbers reverse the order of inequalities. For example, if x > 5, then -x < -5.

**Q4: How do you graph a linear inequality on a number line?**

To graph a linear inequality on a number line: 1) Identify the boundary point. 2) Use an open circle () for strict inequalities (< or >) or a closed circle () for non-strict inequalities ( or ) at the boundary point. 3) Shade the line to the left for "less than" or to the right for "greater than" inequalities.

**Q5: Can you provide an example of a real-world application of linear inequalities?**

A common real-world application of linear inequalities is budgeting. For example, if you have $100 to spend on groceries and each item costs $x, you could express this as a linear inequality: nx 100, where n is the number of items. This inequality helps determine how many items you can buy while staying within your budget.

Understanding multi-step linear inequalities requires a solid foundation in several key mathematical concepts. One of the most crucial prerequisites is solving linear equations using addition and subtraction. This skill forms the basis for manipulating inequalities, as the same principles apply when working with inequality symbols.

Another essential concept is determining the number of solutions to linear equations. This knowledge directly translates to inequalities, helping students understand the range of possible solutions and how to represent them. Similarly, graphing linear inequalities in two variables builds upon this understanding, allowing for visual representation of solution sets.

Proficiency in solving linear equations using multiplication and division is crucial when dealing with multi-step inequalities. These operations are frequently used to isolate variables and simplify expressions. Additionally, understanding the negative exponent rule and its application to inequalities is vital, especially when dealing with reciprocals or fractional inequalities.

The concept of dividing integers plays a significant role in solving inequalities, particularly when combining like terms or simplifying expressions. This skill is closely related to the application of integer operations, which is fundamental in manipulating inequality expressions.

For more complex problems, understanding solution sets of linear systems provides valuable insight into how multiple inequalities interact. This knowledge is particularly useful when dealing with systems of inequalities or more advanced applications.

Lastly, familiarity with compound inequalities is essential for tackling multi-step problems. These often involve combining multiple inequality statements, requiring a deep understanding of how to interpret and manipulate complex inequality expressions.

By mastering these prerequisite topics, students will be well-equipped to tackle the challenges of solving multi-step linear inequalities. Each concept builds upon the others, creating a comprehensive framework for understanding and solving complex inequality problems. As students progress, they'll find that these foundational skills are not only crucial for solving inequalities but also form the basis for more advanced mathematical concepts in algebra and beyond.