A drama club earns $1040 from production. A total of 64 adult tickets and 132 student tickets are sold. An adult ticket cost twice as much as a student ticket. Write a system of linear equations that represents this situation. What is the price of each type of ticket?

To solve this problem, we can set up a system of linear equations. Let's represent the cost of an adult ticket as "x" and the cost of a student ticket as "y".

From the problem, we know that:

1) The drama club earns $1040 from production, so the total revenue can be represented as:
1040 = (64 * x) + (132 * y)

2) An adult ticket costs twice as much as a student ticket, so we can write:
x = 2y

These two equations form the system of linear equations representing the situation.

To find the price of each type of ticket, we can substitute the value of x from equation 2 into equation 1:

1040 = (64 * (2y)) + (132 * y)

Now we simplify this equation:

1040 = 128y + 132y

Combine like terms:

1040 = 260y

Now we solve for y:

y = 1040 / 260

Simplifying this expression, we find:

y = 4

So, the price of a student ticket is $4.

Substituting this value back into equation 2, we can find x:

x = 2 * (4)

Simplifying this, we find:

x = 8

Therefore, the price of an adult ticket is $8.

Let's call the price of an adult ticket "A" and the price of a student ticket "S".

We know that the drama club earns $1040, which can be represented by the equation:
A(64) + S(132) = 1040

We are also given that an adult ticket costs twice as much as a student ticket, which can be represented by the equation:
A = 2S

Now we have a system of linear equations:
Equation 1: A(64) + S(132) = 1040
Equation 2: A = 2S

To solve this system of equations, we can substitute Equation 2 into Equation 1:
(2S)(64) + S(132) = 1040
128S + 132S = 1040
260S = 1040
S = 1040/260
S = 4

Now we can find the price of an adult ticket by substituting the value of S into Equation 2:
A = 2(4)
A = 8

Therefore, the price of an adult ticket is $8 and the price of a student ticket is $4.

a = 2 s

64 a + 132 s = 1040

substituting ... 64 (2 s) + 132 s = 1040 ... 128 s + 132 s = 260 s = 1040

solve for s , substitute back to find a