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.

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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