SOLUTION: A stadium has 48000 seats. Seats sell for ​$35 in Section​ A, ​$30 in Section​ B, and ​$25 in Section C. The number of seats in Section A equals the total number of seats

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Question 1189692: A stadium has 48000 seats. Seats sell for ​$35 in Section​ A, ​$30 in Section​ B, and ​$25 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 ​$1510500 from each​ sold-out event. How many seats does each section​ hold
Found 2 solutions by math_tutor2020, MathTherapy:
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Let
A = number of seats in section A
B = number of seats in section B
C = number of seats in section C

The stadium has 48000 seats total
A+B+C = 48000

"The number of seats in Section A equals the total number of seats in Sections B and C."
So,
A = B+C

The previous equation
A+B+C = 48000
updates to
B+C+B+C = 48000
2B+2C = 48000
2(B+C) = 48000
after replacing A with B+C

Let's solve for C
2(B+C) = 48000
B+C = 48000/2
B+C = 24000
C = 24000-B

Now use these two facts
"Seats sell for ​$35 in Section​ A, ​$30 in Section​ B, and ​$25 in Section C. "
"he stadium takes in ​$1510500 "
to form this equation
35A+30B+25C = 1510500

Plug in A = B+C
35A+30B+25C = 1510500
35(B+C)+30B+25C = 1510500
35B+35C+30B+25C = 1510500
65B+60C = 1510500

Now plug in C = 24000-B and solve for B
65B+60C = 1510500
65B+60(24000-B) = 1510500
65B+1440000-60B = 1510500
5B+1440000 = 1510500
5B = 1510500-1440000
5B = 70500
B = 70500/5
B = 14100

We can use this value to find C
C = 24000-B
C = 24000-14100
C = 9900

Now we can find A
A = B+C
A = 14100+9900
A = 24000

===================================================================================================

To summarize

A = 24000
B = 14100
C = 9900

which represents the number of seats from sections A, B, and C in that order.

As a check,
35A+30B+25C = 1510500
35*24000+30*14100+25*9900 = 1510500
840000+423000+247500 = 1510500
1510500 = 1510500

Also,
A+B+C = 24000+14100+9900 = 48000
The answer is confirmed.
Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

A stadium has 48000 seats. Seats sell for ​$35 in Section​ A, ​$30 in Section​ B, and ​$25 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 ​$1510500 from each​ sold-out event. How many seats does each section​ hold
With a 48,000-seat capacity, and with Section A's total capacity being that of both B and C, it's clear that 
Section A's capacity, as well as Sections B and C, combined, is:
Section A's capacity is 24,000, and with each seat selling for $35, proceeds from Section A = 35(24,000) = $840,000
As proceeds from Section A totals $840,000, proceeds from Sections B and C, combined = $670,500 ($1,510,500 - $840,000)

Let seating capacity in Section B, be B
As both Sections B and C have a capacity of 24,000, seating capacity in Section C = 24,000 - B
We then get: 30B + 25(24,000 - B) = 670,500
6B + 5(24,000 - B) = 134,100 ------ Factoring out GCF, 5
6B + 5(24,000) - 5B = 134,100
6B - 5B = 134,100 - 5(24,000)
Seating capacity in Section B, or B = 134,100 - 120,000 = 14,100
Seating capacity in Section C: 24,000 - 14,100 = 9,900
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