Let x be the number of $35 tickets and

let y be the number of $25 tickets.


Then you have these two equations

  x +   y =   350,           (1)     (counting tickets)
35x + 25y = 10250  dollars   (2)     (counting dollars)


These are your basic equations, and as soon as you got this system, the setup is completed.


There are different methods of solving such system (Substitution, Elimination, using determinants). 

I will use the Substitution method here. 
From equation (1) express y = 350-x and substitute it into equation (2). You will get a single equation for the unknown x

35x + 25(350-x) = 10250.      (3)


Simplify and solve for x:

35x + 25*350 - 25x = 10250  ====>  10x = 10250 - 25*350 = 1500  ====>  x =  = 150.


Thus, 150  $35 tickets were sold.


Hence the number of $25 tickets was 350 - 150 = 200.


Check.  150*35 + 200*25 =  = 10250 dollars.   ! Correct. !


Answer.  150  of  the  $35 tickets  and  200  of  $25  tickets.

Let x be the number of the $35 tickets.

Then the number of the $25 tickets is (350-x),  according to the condition.


The revenue equation is

35x + 25(350-x) = 10250.     (4)


It is your basic equation in the frame of this approach,
and as soon as you got this equation, the setup is completed.


Notice that this equation coincides exactly with the equation (3) of the solution 1 above.

Solve equation (4) by the same method as it was done in the Solution 1.


Surely, you will get the same answer.

Let assume for a minute that all 350 tickets were at 35 dollars.

Then the total revenue would be 35*350 = 12250 dollars.

It is by 12250 - 10250 = 2000 dollars more than the given revenue value.


Why did we get the difference ?  - But of course, because we accounted the $25 tickets as $35 tickets.

So, this difference is due to the difference between the $35 and the $25 tickets price, which (the difference) is $35 - $25 = $10.

Then it is clear that the number of $25 tickets was   = 200,  exactly as in the solutions 1 and 2 above.


and you got the same answer.