Tickets for a harlem globetrotter show cost $28 general admission, $43 courtside, or $173 bench seats. Nine times as many general admission tickets were sold as bench tickets, & the number of general admission tickets sold was 55 more than the sum of the number of courtside tickets & bench tickets. Sales of all three kinds of ticks totaled $97,605. How many of each kind of ticket were sold algebra

general admission tickets  1170

courtside tickets  985

tickets bench seats.  130

Step-by-step explanation:

To solve the problem, it is necessary to generate a system of equations with the information provided by the statement.

First be

x = # general admission tickets

y = # courtside tickets

z = # tickets bench seats.

The first equation would be that they sold nine times more general admission tickets than bench seats tickets.

That is: x = 9 * z (1)

And the second equation is that the number of general admission tickets sold was 55 more than the sum of the number of courtside tickets and bench seats tickets.

That is: x = 55 + y + z (2)

Now the third equation would be the money raised.

28 * x + 43 * y + 173 * z = 97605 (3)

Now if I replace I rearrange (1):

z = x / 9 (4) and replacement in 2, I have:

x = 55 + y + x / 9, rearranging:

y = (8/9) * x - 55 (5)

Now replacing (4) and (5) in 3, we have:

28 * x + 43 * ((8/9) * x - 55) + 173 * (x / 9) = 97605

Solving the above:

x = 1170, therefore

y = (8/9) * 1170 55 = 985

z = 1170/9 = 130

We can confirm this with equation (3)

28 * x + 43 * y + 173 * z = 97605

28 * 1170 + 43 * 985 + 173 * 130 = 97605

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.