Question

Tickets to a school play cost $3 for students and $8 for adults. On opening night, $1000 was collected and 150 tickets sold. How many of each kind of ticket were sold? Write a system of equations and use substitution to solve.

Answer 1

To solve this problem, set up a system of equations representing the number of student and adult tickets sold. Solve the system using substitution to find the numbers of each type of ticket sold.

Step-by-step explanation:

To solve this problem, we can set up a system of equations to represent the number of student tickets and adult tickets sold.

Let x be the number of student tickets sold and y be the number of adult tickets sold.

We know that the total number of tickets sold is 150, so we have the equation:

x + y = 150

We also know that the total amount collected is $1000, so we have the equation:

3x + 8y = 1000

We can solve this system of equations using substitution.

From the first equation, we can solve for x in terms of y as:

x = 150 - y

Substituting this into the second equation, we have:

3(150 - y) + 8y = 1000

Simplifying, we get:

450 - 3y + 8y = 1000

Combining like terms, we get:

5y = 550

Dividing both sides by 5, we get:

y = 110

Substituting this value back into the first equation, we have:

x + 110 = 150

Simplifying, we get:

x = 40

Therefore, 40 student tickets and 110 adult tickets were sold.

Answer 2

Let the number of students who bought tickets be s,
Let the number of adults who bought tickets be a,

Equation 1:


`3s \: + \: 8a \: = \: 1000`

Equation 2:


`s \: + \: a \: = \: 150 \\ s \: = \: 150 \: - \: a`

Substitute ( 2 ) into ( 1 ),


`3(150 \: - \: a) \: + \: 8a \: = \: 1000 \\ 450 \: - \: 3a \: + \: 8a \: = \: 1000 \\ 5a \: = \: 550 \\ a \: = \: 110`

Substitute a = 110 into ( 2 ),


`s \: + \: 110 \: = \: 150 \\ s \: = \: 40`

Ans: 40 Student Tickets, 110 Adult Tickets