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