Problem solver fun?

Question

Tickets for a train excursion were $120 for a sleeping
room, $80 for a berth, and $50 for a coach seat. The total ticket sales were $8600. If there were 20 more berth tickets sold than sleeping room tickets and 3 times as many coach tickets as sleeping room tickets, how many of each type of ticket were
sold?

Answer 1

b = # berth rooms
s = # sleeping rooms
c = # coach rooms

b = 20 + s
c = 3s
80b + 120s + 50c = 8600

Substitute the first two into the last one:
80(20 + s) + 120s + 50(3s) = 8600
1600 + 80s + 120s + 150s = 8600
1600 + 350s = 8600
350s = 7000
s = 20

Now use that to solve for b and c.
b = 20 + s = 20 + 20 = 40
c = 3s = 3(20) = 60

So... 20 sleeping, 40 berth, 60 coach.

Answer 2

This is a system of three variables in three equations. Let us say that x many sleeping rooms were sold, y many berths, and z many coach seats. With total tickets sales of $8600 and the given prices, we have 120x + 80y + 50z = 8600. If 20 more berths were sold than sleeping rooms, we have y = x + 20. If three times as many coach seats were sold as sleeping rooms, we have z = 3x. Now, solve the system:

120x + 80y + 50z = 8600
y = x + 20
z = 3x

This is somewhat easier than a typical system, because y and z are already expressed as functions of x only. Just substitute those expressions for y and z into the first equation and x will be the only unknown. After you solve for x, use it in the formulae for y and z to find them as well.

Answer 3

b = s + 20
c = 3s

8600 = 120s + 80b + 50c

8600 = 120s + 80(s+ 20) + 50(3s)

8600 = 350s + 1600

s = 20 (edit: I fixed an algebra error after seeing it from the above answer.) ;-)