nikkicooper27
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A stadium has 55,000 seats. Seats sell for ​$28 in Section​ A, ​$16 in Section​ B, and ​$12 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 ​$1,158,000 from each​ sold-out event. How many seats does each section​ hold?

Verified Answer :

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

We can suggest this system of equations:

A+B+C=55,000
A=B+C                                      ⇒A-B-C=0
28A+16B+12C=1,158,000

We solve this system of equations by Gauss Method.

1             1              1               55,000
1            -1            -1                         0
28         16           12          1,158,000 


1             1              1               55,000
0            -2             -2             -55,000                                    (R₂-R₁)
0           12             16           382,000                                    (28R₁-R₂) 


1            1                1              55,000
0           -2              -2             -55,000
0            0                4               52,000                                     (6R₂+R₃)

Therefore:

4C=52,000
C=52,000/4
C=13,000


-2B-2(13,000)=-55,000
-2B-26,000=-55,000
-2B=-55,000+26,000
-2B=-29,000
B=-29,000 / -2
B=14,500.

A + 14,500+13,000=55,000
A+27,500=55,000
A=55,000-27,500
A=27,500.

Answer: there are 27,500 seats in section A, 14,500 seats in section B and 13,000 seats in section C.

MrRoyal

The number of seats is an illustration of simultaneous equations.

The number of seats in each section is: \(\mathbf{A = 27500}\)    \(\mathbf{B = 14500}\)      \(\mathbf{C = 13000}\)

From the question, we have:

The take in from the sold-out event

\(\mathbf{28A + 16B + 12C = 1158000}\)

The total number of seats is:

\(\mathbf{A + B + C = 55000}\)

The relationship between seats in sections A, B and C

\(\mathbf{A = B + C }\)

Substitute \(\mathbf{A = B + C }\) in \(\mathbf{A + B + C = 55000}\) and \(\mathbf{28A + 16B + 12C = 1158000}\)

\(\mathbf{A + B + C = 55000}\)

\(\mathbf{B + C + B + C = 55000}\)

\(\mathbf{2B + 2C = 55000}\)

Divide through by 2

\(\mathbf{B + C =27500}\)

Make B the subject

\(\mathbf{B =27500 - C}\)

Substitute \(\mathbf{A = B + C }\) in \(\mathbf{28A + 16B + 12C = 1158000}\)

\(\mathbf{28A + 16B + 12C = 1158000}\)

\(\mathbf{28(B + C) + 16B + 12C = 1158000}\)

\(\mathbf{28B + 28C+ 16B + 12C = 1158000}\)

\(\mathbf{28B + 16B+ 28C + 12C = 1158000}\)

\(\mathbf{44B+ 40C = 1158000}\)

Substitute \(\mathbf{B =27500 - C}\)

\(\mathbf{44(27500- C)+ 40C = 1158000}\)

\(\mathbf{1210000- 44C+ 40C = 1158000}\)

\(\mathbf{1210000- 4C = 1158000}\)

Collect like terms

\(\mathbf{ 4C = 1210000-1158000}\)

\(\mathbf{ 4C = 52000}\\\)

Divide both sides by 4

\(\mathbf{C = 13000}\)

Substitute \(\mathbf{C = 13000}\) in \(\mathbf{B =27500 - C}\)

\(\mathbf{B = 27500 - 13000}\)

\(\mathbf{B = 14500}\)

Recall that: \(\mathbf{A = B + C }\)

\(\mathbf{A = 14500 + 13000}\)

\(\mathbf{A = 27500}\)

Hence, the number of seats in each section is:

\(\mathbf{A = 27500}\)  

\(\mathbf{B = 14500}\)

\(\mathbf{C = 13000}\)

Read more about simultaneous equations at:

https://brainly.com/question/16763389

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