Question 1110761
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Combine the tickets in groups in this way:


 40 advance tickets + 20 same day tickets = 

(20 advance tickets + 20 same say tickets) + 20 advance tickets.


Now the total cost is


1300 = 20*45 + 20a,  where "a" is the cost of the advance ticket.


It gives  20a = 1300-900 = 400.


Hence,  a = {{{400/20}}} = 20 dollars is the price of one advance ticket.


Then the price of one same day ticket is  45-20 = 25 dollars.
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Solved.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>The lesson to learn from this solution</U>:  &nbsp;&nbsp;You do not need to solve the system of equations to get the answer.