a 6,000 seat theater has tickets for sale at $26 and $40 . How many tickets must be sold at each price for a sell out performance to generate a revenue of $178400?

Well, I have to say, this sounds like quite the "theater-ical" problem! Let's break it down and see what we can come up with.

Let's say we sell x tickets at $26 and y tickets at $40. Since there are 6,000 seats in total, we must have:

x + y = 6000

Now, given that the revenue is $178,400, we can set up an equation for that as well. The revenue from the $26 tickets would be 26x, and the revenue from the $40 tickets would be 40y. So, we have:

26x + 40y = 178400

Now, we can use these two equations to solve for x and y. I'd be happy to assist you further once you have the values of x and y. Trust me, it's going to be a "theater-rific" solution!

Let's assume x represents the number of tickets sold at $26 and y represents the number of tickets sold at $40.

We know that the total number of tickets sold must add up to 6,000, so we have the equation:

x + y = 6,000 (Equation 1)

We also know that the revenue from selling each ticket at $26 is $26x, and the revenue from selling each ticket at $40 is $40y. So, the total revenue is:

26x + 40y = 178,400 (Equation 2)

Now, we can solve this system of equations to find the values of x and y.

From Equation 1, we can isolate x:

x = 6,000 - y

Substituting this value of x into Equation 2, we have:

26(6,000 - y) + 40y = 178,400

Simplifying the equation:

156,000 - 26y + 40y = 178,400
14y = 22,400
y = 1,600

Substituting the value of y back into Equation 1, we find:

x + 1,600 = 6,000
x = 6,000 - 1,600
x = 4,400

Therefore, to generate a revenue of $178,400 for a sold-out performance, 4,400 tickets must be sold at $26 and 1,600 tickets must be sold at $40.

To find out how many tickets must be sold at each price, let's assign variables to represent the number of tickets sold at $26 and $40.

Let's say 'x' represents the number of tickets sold at $26, and 'y' represents the number of tickets sold at $40.

According to the given information, the total number of tickets sold should be sufficient to fill all 6,000 seats. Therefore, the equation for the total number of tickets sold is:

x + y = 6000 ----(1)

Now, let's calculate the revenue generated from each ticket price. The revenue from the $26 tickets would be 26x, and the revenue from the $40 tickets would be 40y. Since the total revenue required is $178,400, we can write:

26x + 40y = 178400 ----(2)

To solve this system of equations, we can use different methods such as substitution or elimination. Let's use the substitution method:

From equation (1), we can isolate the value of 'x' in terms of 'y':
x = 6000 - y

Now, substitute the value of 'x' in equation (2):
26(6000 - y) + 40y = 178400

Simplify and solve for 'y':
156000 - 26y + 40y = 178400
14y = 22400
y = 1600

Now substitute the value of 'y' back into equation (1) to find 'x':
x + 1600 = 6000
x = 6000 - 1600
x = 4400

Therefore, to generate a revenue of $178400, 4400 tickets must be sold at $26 each and 1600 tickets must be sold at $40 each.

work with seats:

If there are n seats at $26, then there are 6000-n seats at $40, if the place is sold out.

work with the money -- add up how much money is made for each ticket price.

26n + 40(6000-n) = 178400
26n + 240000 - 40n = 178400
14n = 61600
n = 4400

So, there were 4400 $26 seats and 1600 $40 seats