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An arena that hosts sporting events and concerts has three sections for 3 levels of seating. For a basketball game seats in Section A cost $100, seats in section B cost $50 and seats in section C cost $35. The number of seats in section C equals the number of seats in sections A and B combined. The arena holds 12,000 seats and the game is sold out. If the total revenue from ticket sales is $560,000, determine the number of seats in each section.

Start by assigning your variables. Then write your system of equations. Solve using substitution or elimination. Show all work for this system of 3 equations.

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Answer:

There are 1000 seats in Section A, 5000 seats in section B and 6000 seats in Section C.

Step-by-step explanation:

Let the number of seats in Section A=x

Let the number of seats in Section B=y

Let the number of seats in Section C=z

The number of seats in section C equals the number of seats in sections A and B combined.

z=x+y......(I)

The arena holds 12,000.

x+y+z=12000......(II)

For a basketball game seats in Section A cost $100, seats in section B cost $50 and seats in section C cost $35.

Total revenue from ticket sales is $560,000

100x+50y+35z=560000.....(III)

The system of equation is given as:

z=x+y......(I)

x+y+z=12000......(II)

100x+50y+35z=560000.....(III)

Substitute z=x+y From (I) into Equation (II)

x+y+x+y=12000

2x+2y=12000

x+y=6000......(IV)

Substitute z=x+y From (I) into (III).

100x+50y+35(x+y)=560000

100x+35x+50y+35y=560000

135x+85y=560000.....(V)

From (IV), x=6000-y

135(6000-y)+85y=560000

810000-135y+85y=560000

50y=250000

y=5000

Recall From (IV), x=6000-y

x= 6000-5000=1000

From (I)

z=x+y=1000+5000=6000

There are 1000 seats in Section A, 5000 seats in section B and 6000 seats in Section C.

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Answer: section A has 1000 seats

Section B has 5000 seats and section C has 6000 seats

Step-by-step explanation:

Lety use "A" to represent the number of seats in section A, "B" to represent the number of seats in section B and "C" to represent the number of seats in section C.

Since the number of seats in section C equals the number of seats in section A and B combined, then:

A+B = C -------eqn 1

Again, the arena holds 12,000 seats.

So, A+B+C = 12,0000 --------eqn 2

Recall that seats in section A costs $100. Seat B costs $50 and seat C costs $35. If the game is sold out and the total revenue generated from ticket sales is $560,000.

Then:. (100×A)+(50×B)+(35×C)= $560,000

100A+50B+35C= 560,000 ---------eqn 3

Since A+B = C (in eqn 1), we substitute "A+B" for C in eqn 2

Therefore C + C = 12,000

2C = 12,000

C = 6,000 seats.

If C is equal to 6000, we then substitute C for 6000 seats in eqn 1

Therefore, A+B = 6,000

B= 6000 - A

We then substitute "B" for "6,000 - A" and "C" for 6000 in equation 3

100A + [50 × (6000 - A)]+ (35 × 6000) = 560,000

100A + (300,000 - 50A) + 210,000 = 560,000

50A + 510000 = 560000

50A = 50000

A = 1,000 seats

Substitute "A" for 1000 and "C" for 6000 in eqn 1

Therefore: 1000 + B = 6000

B = 6000 - 1000

B = 5000 seats

Summarily, there are 1,000 seats in section A; 5,000 seats in section B and 6000 seats in section C

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