Suppose that there are two types of tickets to a show: advance and same-day. the combined cost of one advance ticket and one same-day ticket is 35 . for one performance, 15 advance tickets and 40 same-day tickets were sold. the total amount paid for the tickets was 900 . what was the price of each kind of ticket?

Let "a" and "s" represent the costs of advance and same-day tickets, respectively. Your problem statement gives you two relations.
.. a + s = 35 . . . . . the combined cost of one of each is 35
.. 15a +40s = 900 . . total paid for this combination of tickets was 900

There are many ways to solve these equations. You've probably been introduced to "substitution" and "elimination" (or "addition"). Using substitution for "a", we have
.. a = 35 -s
.. 15(35 -s) +40s = 900 . . substitute for "a"
.. 25s +525 = 900 . . . . . . . simplify
.. 25s = 375 . . . . . . . . . . . .subtract 525
.. s = 15 . . . . . . . . . . . . . . .divide by 25
Then
.. a = 35 -15 = 20

The price of an advance ticket was 20.
The price of a same-day ticket was 15.

Then just do it (lol)