For the school play, one adult ticket costs $10 and one student ticket cost $6. Twice as many student tickets as adult tickets were sold. The total receipts were $3 300. How many of each kind of ticket were sold?

use the number of tickets, and the total income:

s = 2a
10a+6s = 3300

To solve this problem, we can set up a system of equations using the given information.

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

According to the problem, the cost of one adult ticket is $10, so the total cost of all the adult tickets sold would be 10 * a = 10a.

Similarly, the cost of one student ticket is $6, so the total cost of all the student tickets sold would be 6 * s = 6s.

It is also stated in the problem that twice as many student tickets were sold as adult tickets. So, s = 2a.

The total receipts from the ticket sales were $3,300, so we can write another equation:
10a + 6s = 3300.

Now, we have a system of two equations:

s = 2a ...(Equation 1)
10a + 6s = 3300 ...(Equation 2).

To solve this system, we can substitute the value of 's' from Equation 1 into Equation 2:

10a + 6(2a) = 3300
10a + 12a = 3300
22a = 3300
a = 3300 / 22
a ≈ 150.

Now, substitute the value of 'a' back into Equation 1 to find 's':

s = 2a
s = 2 * 150
s = 300.

Therefore, 150 adult tickets and 300 student tickets were sold for the school play.