Admission ticket to a motion picture theater were priced at $5 for adult and $3 for the student. 820 tickets were sold and the total receipts were $2860, how many of the each type of ticket were sold?

Solution:
There were 2 types of tickets and were sold at different prices. The total sale was given to be $2860 with only 820 tickets sold. Now the problem is how many of each type of ticket were sold.

Based on the problem statements, we can derive the following variable representation:

Let
        a = adult ticket
        s = student ticket

Hence,
        $ a + s = 820 $             Eq. 1 (since there were 820 tickets sold)
        $ 5a + 3s = 2860 $       Eq. 2 (since the sum of the sold tickets was 2853)

The resulting equations form a system of two equations in two unknowns, a and s. We will solve this system using addition elimination method like below

        $ 5a + 3s = 2860 $
        $-5a -5s = -4100 $       (Eq. 1 multiplied by -5 to eliminate a)
        $ 0a - 2s = -1240 $
        $ s = 620 $                         (dividing both sides by -2)

With this value of s, Eq. 1 would then become
        $ a + s = 820 $
        $ a + 620 = 820 $
        $ a = 820 - 620 $
        $ a = 200 $

Therefore, there were 200 adult tickets sold at 5 dollars and 620 student tickets sold at 3 dollars.

Check:
Total ticket sold was 820: $ 200 + 620 = 820 $
Total sale for 820 tickets was 2860: $ 200(5) + 620(3) = 2860 $
                                                 $ 1000 + 1860 = 2860 $
                                                 $ 2860 = 2860 $