A stadium has 55000 seats in total. Seats in section A sell for $30, $24 in section B, and $18 in section C. The total number of seats in section A equals the total number of seats in sections B and C. The stadium takes in $1,405,200 on a sold out night. How many seats does each section hold?

Answer: A = 27,500, B = 14,200, C = 13,300

Step-by-step explanation:

A = B + C

55,000 = A + B + C

30A + 24B + 18C = 1,405,200

Substitute #2 with A from #1

55,000 = B + C + B + C or 2B + 2C

Now substitute A for #3

30 (B + C) + 24B + 18C = 1,405,200

30B + 30C + 24B + 18C = 1,405,200

54B + 48C = 1,405,200

Now multiply are new #2 equation by - 24

(2B + 2C = 55,000) * -24

-48B + - 48C = - 1,320,000

Combine this equation to our new #3

[-48B + (-48C) = - 1,320,000] + [54B + 48C = 1,405,200]

6B = 85,200

B = 14,200

Now that we got B we have to continue ...

Substitute B in our new #2

2 (14,200) + 2C = 55,000

Solve.

28,400 + 2c = 55,000

-28,400

2c = 26,600

C = 13,300

Now we can use #1

A + 14,200 + 13,300 = 55,000

A + 27,500 = 55,000

-27,500

A = 27,500

(There's also a much simpler way ... Divide 55,000 in half to get A automatically.)

A = 27500 seats, B = 14200 seats, C = 13300 seats

Step-by-step explanation:

a=b+c

a+b+c=55000 seats

30a+24b+18c=1405200 dollars

substitue a for b+c

b+c+b+c=55000

30 (b+c) + 24b+18c=1405200

seats 2b+2c=55000 - -> divide by 2 on both sides b+c=27500 - -> c=27500-b

dollars 30b+30c+24b+18c=1405200 - -> 54b+48c=1405200 divide by 6 - -> 9b+8c=234200

substitute c for 13750-b

9b+8 (27500-b) = 234200

9b+220000-8b=234,200

b=14,200 seats

c=27500-b=27500-14200=13300 seats

a=55000-b-c=55000-27500=27500 seats