A theater charges $50 per ticket for seats in section A, $30 per ticket in section B, and $20 per ticket in section C. For a certain performance, 4000 tickets were sold for a total of $120,000 in revenue. If 1000 more tickets were sold in section B than in section A, how many tickets were sold in each section?
A theater charges $50 per ticket for seats in section A, $30 per ticket in section B, and $20 per ticket in section C. For a certain performance, 4000 tickets were sold for a total of $120,000 in revenue. If 1000 more tickets were sold in section B than in section A, how many tickets were sold in each section?
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Problem: Theater Ticket Sales
A theater charges $50 per ticket for seats in Section A, $30 per ticket in Section B, and $20 per ticket in Section C. For a specific performance, 4000 tickets were sold, generating a total revenue of $120,000. If 1000 more tickets were sold in Section B than in Section A, how many tickets were sold in each section?
**Task:**
1. Formulate a system of equations to represent this scenario.
2. Solve the system to determine the number of tickets sold in each section.
Given these constraints and the total number of tickets sold and the revenue, you can derive the following system of equations:
1. Let \( x \) represent the number of tickets sold in Section A.
2. Let \( y \) represent the number of tickets sold in Section B.
3. Let \( z \) represent the number of tickets sold in Section C.
The conditions provided can be translated into the following equations:
1. **Total tickets sold:** \( x + y + z = 4000 \)
2. **Total revenue generated:** \( 50x + 30y + 20z = 120,000 \)
3. **Relation between Section A and Section B tickets:**
\[ y = x + 1000 \]
**Next Steps:**
1. Use these equations to solve for the values of \( x \), \( y \), and \( z \).
2. Substitute \( y = x + 1000 \) into the other equations to simplify and solve the system.
Proceed to solve this system in the next question.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F585970c1-8db0-450f-987b-10f4de58f48c%2Fd1f6332e-b075-442c-8748-6ee301c6d8c5%2Flpomzrm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem: Theater Ticket Sales
A theater charges $50 per ticket for seats in Section A, $30 per ticket in Section B, and $20 per ticket in Section C. For a specific performance, 4000 tickets were sold, generating a total revenue of $120,000. If 1000 more tickets were sold in Section B than in Section A, how many tickets were sold in each section?
**Task:**
1. Formulate a system of equations to represent this scenario.
2. Solve the system to determine the number of tickets sold in each section.
Given these constraints and the total number of tickets sold and the revenue, you can derive the following system of equations:
1. Let \( x \) represent the number of tickets sold in Section A.
2. Let \( y \) represent the number of tickets sold in Section B.
3. Let \( z \) represent the number of tickets sold in Section C.
The conditions provided can be translated into the following equations:
1. **Total tickets sold:** \( x + y + z = 4000 \)
2. **Total revenue generated:** \( 50x + 30y + 20z = 120,000 \)
3. **Relation between Section A and Section B tickets:**
\[ y = x + 1000 \]
**Next Steps:**
1. Use these equations to solve for the values of \( x \), \( y \), and \( z \).
2. Substitute \( y = x + 1000 \) into the other equations to simplify and solve the system.
Proceed to solve this system in the next question.
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