SOLUTION: A Broadway theater has 400 ​seats, divided into​ orchestra, main, and balcony seating. Orchestra seats sell for $70, main seats for $55, and balcony seats for $35. If

The two key phrases in the condition are:


    "If all the seats are​ sold, the gross revenue to the theater is $20,400"

and 

    "If all the main and balcony seats are​ sold, but only half the orchestra seats are​ sold, the gross revenue is $17,600"


From these two statements, it follows immediately that half of the orchestra seats cost 20400 - 17600 = 2800 dollars;

hence,  all the orchestra seats cost twice of it, i.e. 5600 dollars.

Since the price of each single orchestra seat is $70, the number of the orchestra seats is   = 80.


In this way, we just reduced the problem from 3 unknown to only 2.


Let x be the number of the main seats and y be the number of the balcony seats.

Then you have this system of 2 equations


  x +   y = 400 - 80
55x + 35y = 20400 - 5600


or, equivalently

  x +   y =   320,     (1)
55x + 35y = 14800.     (2)


Use the elimination method. For it, multiply eq(1) by 55 (both sides), keeping eq(2) as is. You will get


55x + 55y = 55*320,     (1')
55x + 35y = 14800.      (2')


Now subtract eq(2') from eq(1') to get

      20y = 55*320 - 14800  ====>  y =  = 140.


Then from eq(1),  x = 320 - 140 = 180.


ANSWER.  180 main seats;  140 balcony seats;  and  80 orchestra seats.


CHECK.   70*80 + 180*55 + 140*35 = 20400 dollars.    ! Correct !

    From the first glance, the problem is about 3 equations in 3 unknowns.

    But carefully reading, the student can find the hidden way to reduce it to the system of only 2 equations in 2 unknowns,

    which is much easier to solve.