Question 151000
Because there are three unknowns (courtside, endzone, and balcony seats), there have to be a minimum of three equations to solve it. 
<br>You know the total number of seats, so you can set up the equation:
c + e + b = 15000
We'll call this equation A.
<br>You also know the total revenue when all the seats are sold out, plus the prices of each of the seats, so you can set up the equation:
9c + 7e + 4b = 81000
We'll call this equation B.
<br>The problem also tells you the revenue when only half the courtside and balcony seats are sold and when all the endzone seats are sold, giving the equation:
&frac12(9c) + 7e + &frac12(4b) = 47500
(9/2)c + 7e + 2b = 47500
We'll call this equation C. 
<br>When you have three equations, in order to solve them, try to add and subtract them from each other so that you end up with only two equations and two variables. Looking at the above equations, it looks like it would be easier to solve for c and b. Equation B minus equation C would eliminate e.
9c + 7e + 4b = 81000
-[(9/2)c + 7e + 2b = 47500]
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(9/2)c + 2b = 33500

<br>Equation B minus seven times equation A will also eliminate e while leaving b and c behind:
9c + 7e + 4b = 81000
- 7[c + e + b] = 7 * 15000
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2c - 3b = -24000
<br> Choose a variable in both equations and solve for it. We will solve for b.
(9/2)c + 2b = 33500
2b = 33500 - (9/2)c
b = 16750 - (9/4)c
<br> 2c - 3b = -24000
2c + 24000 = 3b
b = (2/3)c + 24000/3
b = (2/3)c + 8000
<br> Now you can combine the equations:
16750 - (9/4)c = b = (2/3)c + 8000
16750 - 8000 = (2/3)c + (9/4)c
8750 = (8/12)c + (27/12)c 
8750 = (35/12)c
c = 3000
<br> Plug this into one of the smaller equations:
b = (2/3)c + 8000
b = (2/3)(3000) + 8000
b = 2000 + 8000
b = 10000
<br> Now you can plug this into one of the original equations. Equation A looks easiest:
c + e + b = 15000
e = 15000 - c - b
e = 15000 - 3000 - 10000
e = 2000
<br>Therefore, the stadium has 3000 courtside seats, 2000 endzone seats, and 10000 balcony seats.