Question

must find $30 as well A basketball team sells tickets that cost ( $ 10, $ 20 ), or, for VIP seats, ( $ 30 ). The team has sold 3317 tickets overall. It has sold 126 more ( $ 20 ) tickets than ( $ 10 ) tickets.

Answer 1

The question is about solving a system of linear equations for ticket sales. The provided information allows us to set up two equations with three unknowns, representing the number of tickets sold at different price points. Additional information is required to solve for the exact numbers of each type of ticket.

Step-by-step explanation:

The subject question involves solving a system of linear equations related to ticket sales by a basketball team. There are three types of tickets being sold: at $10, $20, and $30. The total number of tickets sold is 3317, and there are 126 more $20 tickets sold than $10 tickets. Let's denote the number of $10 tickets as x, $20 tickets as y, and $30 tickets as z. We can form the following system of equations:

1. x + y + z = 3317 (Total tickets equation)
2. y = x + 126 (126 more $20 tickets than $10 tickets equation)

We are given two equations, but we have three unknowns; additional information is needed to solve for individual values of x, y, and z. Without further information, such as the total revenue generated or another relationship, we cannot uniquely determine the number of tickets sold at each price point. However, the question may require solving the system of equations given more data.