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Admission tickets to a local circus show were priced at USD4 for adult and USD3 for the student.

Admission tickets to a local circus show were priced at 4 dollars for adult and 3 dollars for the student. For the first 3 days of the show, 810 tickets were sold and the total receipts were $2853. How many of the each type of ticket were sold?

 
Solution:
 
Based on the details given in the problem, we can have this variable representation:
 
x = adult tickets
y = student tickets
 
Given the conditions in the problem, we can derive the following equations
 
x + y                = 810               [Eq. 1]
[The sum of the adult and student tickets is 810 ]
 
4x + 3y            = 2853             [Eq. 2]
[The total sales of the adult and the student tickets is 2853]
 
Solving Eq. 1 for x gives:
 
            x + y                = 810
            x                      = 810 – y         [Eq. 3]
 
Substituting Eq. 3 in Eq. 2 gives:
 
4x + 3y                        = 2853             [Eq. 2]
4(810 – y) + 3y            = 2853
3240 – 4y + 3y            = 2853
3240 – y                      = 2853
3240 – 2853                = y
                                    387      = y
 
Using Eq. 1 to solve for x gives:
 
x + y                = 810
x + 387            = 810
x                      = 810 – 387
x                      = 423
 
Therefore, there were 423 adult tickets sold and 387 student tickets sold.
 
Check:
 
x + y                = 810
423 + 387        = 810
            810      = 810
 
4x + 3y                        = 2853
4(423) + 3(387)          = 2853
4(423) + 3(387)          = 2853
1692 + 1161                = 2853
2853    = 2853