Question 1100563
.A stadium has 50,000 seats. Seats sell for $25 in section A, $20 in section B, $15 in section C. 
The number of seats in section A equals the total number of seats in sections B and C. 
Suppose the stadium takes in 1,074,500 from each sold-out event. How many seats does each section holds?
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Let x be the number of seats in section B, and y be the number of seats in section C.


Then the number of seats in section A is x+y, the sum of seats in sections B and C, according to the condition.


The condition says

seatsA + seatsB + seatsC = 50000,   or

(x+y)  + x      + y      = 50000,   which is the same as

2(x+y) = 50000,                     which implies

x + y = {{{50000/2}}} = 25000.


Thus we found that the sum of seats in sections B and C is 25000,  

and, therefore, the number of seats in section A is 25000.


Thus we have now the problem  <U>FOR TWO UNKNOWNS only</U>, since we just excluded A.

Now our system of two equations is

  x +   y = 25000,                  (1)
20x + 15y = 1074500 - 25*25000.     (2)


Equation (2) express the money payed for seats in sections B and C together.


Simplify (1),(2) step by step:

  x +   y = 25000,                  (3)
20x + 15y = 449500.                 (4)


Multiply eq(1) by 20 (both sides). Then subtract eq(2) from it:

20x + 20y = 500000,                 (3)
20x + 15y = 449500.                 (4)
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====> 5y = 500000-449500 = 50500  ====>  y = {{{50500/5}}} = 10100.


Then  x = 25000 - y = 25000 - 10100 = 14900.


<U>Answer</U>.  There are 25000 sears in section A,  14900 in section B  and  10100 in section C.
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