Question

Tickets for the Harlem Globetrotters show at Michigan State University in 2010 cost $16, $23, or, for VIP seats, $40. If 9 times as many $16 tickets were sold as VIP tickets, and the number of $16 tickets sold was 55 more than the sum of the number of $23 tickets and VIP tickets, sales of all three kinds of tickets would total $46,575. How many of each kind of ticket would have been sold?

Answer 1

Can you please explain how to do this one without using the matrix method?

Answer 2

I will just start from the point after the equations have been derived.

x = 9z ----(1)
x - y - z = 55 ----(2)
16x + 23y + 40z = 46575 ----(3)

Substitute (1) into (2) and (3), you get
8z - y = 55 ---- (4) [derived from (2)]and
184z + 23y = 46575 ----(5) [derived from (3)]

With that, manipulating equation (4) yields
y = 8z - 55 -----(6), which, when substituted into (5), yields
184z + 23(8z - 55) = 46575
368z = 47840
z = 130

Substitute z = 130 into (6) and (1)
y = 8(130) - 55 = 985
x = 9(130) = 1170

Answer 3

So I have broken the solution into parts. If you don't understand how I get my equations, please take a look at the question analysis table in the pdf document. ^^

Answer 4

I don't understand where the 184 comes from in the fifth equation.

Answer 5

x = 9z ----(1)
16x + 23y + 40z = 46575 ----(3)

Substitute (1) into (3):
16(9z) + 23y + 40z = 46575
144z + 23y + 40z = 46575
184z + 23y = 46575 (QED)

Answer 6

ok I see how you did that, thank you.

Answer 7

No problems ^^