School play tickets are $5.00 for none students and $3.00 for students. 937 ticket were sold for $3943.00

How many student tickets were sold and how many none student tickets were sold ?

Sold X none student tickets.

Sold Y student tickets.

Eq1: x + y = 937.
Eq2: 5x + 3y = 3943.
Multiply Eq1 by -3 and use Elimination
Method.

To determine the number of student and non-student tickets sold, we can set up a system of equations based on the given information.

Let's assume the number of student tickets sold as "S" and the number of non-student tickets sold as "NS".

The price of a student ticket is $3.00, so the total revenue from student tickets would be 3S.
The price of a non-student ticket is $5.00, so the total revenue from non-student tickets would be 5NS.

From the information given, we know that a total of 937 tickets were sold, so we can express this as:
S + NS = 937 -- Equation (1)

We also know that the total revenue from ticket sales was $3943.00, so we can express this as:
3S + 5NS = 3943 -- Equation (2)

We now have a system of linear equations with two variables (S and NS). To solve this system, we can use the method of substitution or elimination.

Since Equation (1) expresses S in terms of NS, we can substitute the value of S from Equation (1) into Equation (2):

3(937 - NS) + 5NS = 3943

Expanding the equation:

2811 - 3NS + 5NS = 3943

Combining like terms:

2NS = 1132

Dividing both sides by 2:

NS = 566

Now, substituting the value of NS back into Equation (1) to find S:

S + 566 = 937
S = 937 - 566
S = 371

Therefore, there were 371 student tickets sold and 566 non-student tickets sold.