Formulate a system of equations for the situation below and solve.

A theater has a seating capacity of 660 and charges $2 for children, $4 for students, and $6 for adults. At a certain screening with full attendance, there were half as many adults as children and students combined. The receipts totaled $2640. How many children attended the show?

A system of equations has no solution if y=8x + 7 is one of the equations, which could be the other equation

To solve this problem, let's start by representing the unknown quantities with variables.

Let:
C = number of children attending the show
S = number of students attending the show
A = number of adults attending the show

Using the given information, we can create a system of equations:

Equation 1: A + C + S = 660 (The total number of attendees should equal the seating capacity)
Equation 2: A = (1/2)(C + S) (The number of adults is half the sum of children and students)
Equation 3: 2C + 4S + 6A = 2640 (The total receipts from ticket sales)

Now, we can solve the system of equations.

First, substitute Equation 2 into Equation 1:
(1/2)(C + S) + C + S = 660
(1/2)C + (1/2)S + C + S = 660
(3/2)C + (3/2)S = 660

Multiply the equation by 2 to eliminate the fractions:
3C + 3S = 1320 (Equation 4)

Next, substitute Equation 2 and Equation 4 into Equation 3:
2C + 4S + 6((1/2)(C + S)) = 2640
2C + 4S + 3C + 3S = 2640
5C + 7S = 2640 (Equation 5)

Now, we have a system of two equations, Equation 4 and Equation 5.

Solve the system of equations using any method you prefer, such as substitution or elimination.

Let's solve by elimination:

Multiply Equation 4 by 5 and Equation 5 by 3 to eliminate the coefficient of C:

15C + 15S = 6600 (Equation 6)
15C + 21S = 7920 (Equation 7)

Subtract Equation 6 from Equation 5:

15C + 21S - 15C - 15S = 7920 - 6600
6S = 1320
S = 220

Now substitute the value of S = 220 back into Equation 5:

5C + 7(220) = 2640
5C + 1540 = 2640
5C = 1100
C = 220

Therefore, the number of children attending the show is 220.