Tickets to a school play cost $3 for students and $8 for adults. On opening night, $1,000 was collected and 150 tickets sold. Use substitution to solve a system of equations to find how many of each kind of ticket were sold
nvm it 40 and 110
i need a answer
To solve this problem using substitution, we need to set up a system of equations based on the given information.
Let's define:
Let x be the number of student tickets sold.
Let y be the number of adult tickets sold.
From the given information:
The total revenue collected from selling student tickets is $3 multiplied by the number of student tickets sold, so this can be represented as 3x.
The total revenue collected from selling adult tickets is $8 multiplied by the number of adult tickets sold, so this can be represented as 8y.
We also know that the total revenue collected on opening night is $1000, so we can set up the first equation:
3x + 8y = 1000 Equation 1
We are also told that a total of 150 tickets were sold on opening night, which gives us the second equation:
x + y = 150 Equation 2
Now, we have a system of two equations:
3x + 8y = 1000 Equation 1
x + y = 150 Equation 2
To solve this system of equations using substitution, we solve one equation for one variable and substitute the expression into the other equation. Let's solve Equation 2 for x:
x = 150 - y
Now, substitute this expression for x in Equation 1:
3(150 - y) + 8y = 1000
Simplify:
450 - 3y + 8y = 1000
Combine like terms:
5y = 550
Solve for y:
y = 550/5
y = 110
Now that we have the value of y, we can substitute it back into Equation 2 to solve for x:
x + 110 = 150
x = 150 - 110
x = 40
Therefore, 40 student tickets and 110 adult tickets were sold on opening night.