Question

A sports stadium has 49,000 seats. Seats sell for $25 in Section A, $20 in Section B, and $15 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 $1,052,000 from each sold-out event. How many seats does each section have?

Answer 1

Where do you find difficulty ?

Answer 2

You there ?

Answer 3

I can't find the total amount of seats in each section

Answer 4

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

A = B + C
A + B + C = 49,000
25A + 20B + 15C = 1,052,000

Do you know how to solve system of 3 equations?

Answer 5

@dude @Vocaloid

Answer 6

@eviant, @mhchen helped you out. You have that A = B + C. Substitute B+C in place of A for the 2nd and 3rd equations.

Answer 7

@Hero sorry, I can't see @mchen's answer, did he reply directly to my post?

Answer 8

\(\color{#0cbb34}{\text{Originally Posted by}}\) @mhchen
Let A = number of seats in A
B = number of seats in B
C = number of seats in C

A = B + C
A + B + C = 49,000
25A + 20B + 15C = 1,052,000

Do you know how to solve system of 3 equations?
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 9

@mhchen @Hero I do not know how to solve them correctly

Answer 10

Have you tried solving a system of 2 equations using substitution? You can use the same method for a system of 3 equations.

We have A = B + C

We can use that to rewrite the 2nd equation as
(B + C) + B + C = 49,000
We can also rewrite the 3rd equation as
25(B+C) = 20B + 15C = 1,052,000

Now you have a system of 2 equations, so you can solve for B and C right?