Question

Movie PRICES In Exercises 67 and \(68,\) a theater charges \(\$ 4\) per person before 6: 00 P.M. and \(\$ 7\) per person after 6: 00 P.M. The total ticket sales for Saturday were \(\$ 11,228\) Suppose no one attended the theater before 6: 00 P.M. How many people attended the theater after 6: 00 P.M.? Explain how you know.

Short answer

The number of people who attended the theater after 6:00 P.M. is 1,604.

Step-by-step solution

Understand the Problem

The problem gives the ticket price per person after 6:00 P.M. which is \$7. The total ticket sales for the day is \$11,228. It's also given that no one attended before 6:00 P.M. Therefore, all the ticket sales came from people who attended after 6:00 P.M.

Formulate the equation

Let's denote by \(X\) the number of people who attended the theater after 6:00 P.M. Since each of these persons has paid \$7 for the ticket, the total amount of money from these sales is \(7 \cdot X\), which equals to the total sales for the day, that is, \$11,228. We then get the equation: \(7 \cdot X = 11,228\)

Solve the equation

To find out the value of \(X\), we need to solve the equation. We do so by dividing both sides of the equation by 7. Therefore, \(X = 11,228/7\)

Calculate

After doing the division, we find out that \(X = 1,604\)

Key concepts

Solving Linear Equations

The process of solving linear equations is foundational in algebra and involves finding the value of the variable that makes the equation true. These equations typically include one or two variables and are set up as equalities. In our practical example, we’re given the equation \(7 \cdot X = 11,228\), where \(X\) represents the number of people. \(
\) \(
\) To find \(X\), we use inverse operations to isolate the variable. Since \(X\) is multiplied by 7, we perform the inverse operation—division. We divide both sides of the equation by 7, giving us \(X = 11,228 / 7\). Through this straightforward calculation, we arrive at the correct number of attendees,which is 1,604. \(
\) \(
\) It's important for students to understand the rationale behind each step; in doing so, they become equipped to apply these steps to various forms of linear equations, enhancing their problem-solving skills.

Word Problems

Word problems, like the theater ticket example provided, require translating a written scenario into mathematical terms—an essential skill in algebra. The key to tackling such problems is to carefully extract the relevant information and disregard any superfluous details. \(
\) \(
\) In this instance, we focused on the ticket price after 6:00 P.M. and the total sales, ignoring the time slot before 6:00 P.M. since no sales were made then. The next step involves defining a variable to represent the unknown; here, it's the number of attendees, dubbed \(X\). Once the proper equation is formulated, solving the word problem becomes a matter of solving this equation. \(
\) \(
\) Students should practice breaking down word problems into manageable parts and converting them into algebraic expressions or equations, which leads us to our final concept.

Algebraic Expressions

Algebraic expressions consist of constants, variables, and operators (such as addition, subtraction, multiplication, and division) combined to represent a specific value or set of values. They are fundamental blocks of algebra, often used to model real-life situations. \(
\) \(
\) In the theater ticket scenario, the expression \(7 \cdot X\) represents the total revenue from ticket sales after 6:00 P.M. The number 7 is the constant (price per ticket), and \(X\) is the variable representing the number of tickets sold. \(
\) \(
\) Learning to construct and manipulate algebraic expressions allows students to solve a myriad of problems efficiently. The practice helps in translating real-world situations into mathematical models, which can then be analyzed and solved.