A stadium has 53,000 seats. Seats sell for $28 in Section A, $16 in Section B, and $12 in Section C. Th number of seats in Section A equals the total number of seast in Sections B and C. Suppose the stadium takes in $1,118,000 from each sold-out event. How many seats does each section hold?

Section A holds:
Section B holds:
Section C holds:

28a + 16b + 12c = 1,118,000

but a = b+c
28(b+c) + 16b + 12c = 1,118,000
44b + 40c = 1,118,000
11b+10c = 111,800

but we also know that a + b + c = 53000
b+c + b + c = 53000
b+c = 26500
c = 26500 - b

sub in that value of c into 11b+10c = 111,800
and you can find b
after that you can find c
and finally find a in a = b+c

A equals B plus C ... half of the seats are in A ... 53,000 / 2 = ?

value of A seats ... 28 * (53000 / 2) = 742,000

number of B and C seats ... B + C = 53,000 / 2 ... 12B + 12C = 6 * 53,000

value of B and C seats ... 16B + 12C = 1,118,000 - 742,000

subtract the two equations to eliminate C
... 4B = 1,118,000 - 742,000 - (6 * 53,000)

solve for B , then substitute back to find C

Let the number of seats in Section A be x.

According to the given information, the number of seats in Section B and Section C combined is also x.

The total number of seats in the stadium is 53,000, so we can write the equation: x + x + x = 53,000

Simplifying, we have 3x = 53,000

Dividing both sides by 3, we get: x = 53,000 / 3 = 17,666.67

Since the number of seats must be a whole number, we can round down to the nearest whole number to get x = 17,666.

Therefore, Section A holds 17,666 seats.

Since the total number of seats in Sections B and C combined is also x, the number of seats in each of these sections is also 17,666.

Therefore, Section B holds 17,666 seats and Section C holds 17,666 seats.

To solve this problem, we can set up a system of equations. Let's assume that the number of seats in Section B is 'x', and the number of seats in Section C is 'y'. Since the number of seats in Section A equals the total number of seats in Sections B and C, we can express it as 'x + y'.

According to the given information, the stadium has 53,000 seats, so we know that the sum of the seats in all three sections should equal 53,000:

x + y + (x + y) = 53,000

Simplifying the equation, we get:

2x + 2y = 53,000

Now, we need to express the revenue from each section. Section A sells for $28, Section B for $16, and Section C for $12.

The total revenue from each sold-out event is $1,118,000, so we can write another equation based on the revenue from each section:

28(x + y) + 16x + 12y = 1,118,000

Simplifying the equation, we get:

28x + 28y + 16x + 12y = 1,118,000

Combining like terms, we have:

44x + 40y = 1,118,000

Now we have a system of equations:

2x + 2y = 53,000
44x + 40y = 1,118,000

We can solve this system of equations to find the values of 'x' and 'y', which represent the number of seats in Sections B and C, respectively.

Using a method like substitution or elimination, we can find that x = 11,000 and y = 16,000.

Therefore, the number of seats in each section is as follows:

Section A: x + y = 11,000 + 16,000 = 27,000 seats
Section B: x = 11,000 seats
Section C: y = 16,000 seats

So, Section A holds 27,000 seats, Section B holds 11,000 seats, and Section C holds 16,000 seats.