Question

one adult ticket for a school play costs 10 dollars and one student ticket costs 6 dollars. one evening,750 people attended the play and the total receipts were 6860. how many of each type of ticket were sold? A full algebraic solution is required.

Answer 1

The algebraic solution to the situation is as follows:

x + y = 750

10x + 6y = 6860

The number of adult and student are 590 and 160 respectively.

How to find and solve an algebraic equation?

One adult ticket for a school play costs 10 dollars and one student ticket costs 6 dollars.

One evening,750 people attended the play and the total receipts were 6860.

Therefore,

let

x = number of adult ticket

y = number of student tickets

Therefore,

x + y = 750

10x + 6y = 6860

10x + 10y = 7500

10x + 6y = 6860

4y = 640

y = 640 / 4

y = 160

Hence,

x  = 750 - y

x = 750 - 160

x = 590

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