For a class play, student tickets cost $2 and adult tickets cost $5. A total os 30 tickets were sold. If the total sales must exceed $90, what is the miniuim number of adult tickets that must be sold?

Let s = # student tickets and a = # adult

s = 30 - a

2s + 5a > 90

Substitute 30-a for s in second equation and solve for a. Insert that value into the first equation and solve for s. Check by inserting both values into the second equation.

Let's assume the number of student tickets sold is x, and the number of adult tickets sold is y.

According to the given information, the cost of a student ticket is $2, and the cost of an adult ticket is $5.

We also know that the total number of tickets sold is 30. Therefore, we have the equation:

x + y = 30 (Equation 1)

We are also given that the total sales must exceed $90. In terms of the number of tickets sold, the total sales can be expressed as:

2x + 5y > 90 (Equation 2)

To find the minimum number of adult tickets that must be sold, we need to minimize the value of y.

Let's solve the equations:

From Equation 1, we can express x in terms of y:

x = 30 - y

Substituting this value in Equation 2, we get:

2(30 - y) + 5y > 90

Simplifying further:

60 - 2y + 5y > 90

3y > 30

y > 10

Therefore, the minimum number of adult tickets that must be sold is 11, as we need y to be greater than 10 to satisfy the conditions.

To find the minimum number of adult tickets that must be sold, let's work through the problem step by step.

Let's assume x represents the number of student tickets sold.
Therefore, the number of adult tickets sold can be represented as 30 - x, since the total number of tickets sold is 30.

The cost of a student ticket is $2, and the cost of an adult ticket is $5. To find the total sales, we can multiply the number of student tickets by $2 and the number of adult tickets by $5, and then sum these two values.

The equation for the total sales is:
2x + 5(30 - x) > 90

Simplifying the equation:
2x + 150 - 5x > 90
-3x > 90 - 150
-3x > -60

Dividing both sides by -3:
x < (-60)/(-3)
x < 20

Since x represents the number of student tickets, it must be a positive whole number. The maximum value for x can be 19 (since it must be less than 20).

Now, let's find the minimum number of adult tickets. Since the total number of tickets sold is 30 and the number of student tickets is 19, we subtract 19 from 30 to find the number of adult tickets:
30 - 19 = 11

Therefore, the minimum number of adult tickets that must be sold is 11.