An auditorium earned $25,000 in sold−out concert ticket sales. Front section tickets cost $75 per seat and back section tickets cost $50 per seat. The number of front section seats is twice the number of back section seats. How many seats are in the front section?
@jim_thompson5910 250 right ?

Question

An auditorium earned $25,000 in sold−out concert ticket sales. Front section tickets cost $75 per seat and back section tickets cost $50 per seat. The number of front section seats is twice the number of back section seats. How many seats are in the front section?
 @jim_thompson5910 250 right ?

whpalmer4 · Answer

let F be # of front section tickets ($75 each) and B be # of back section tickets ($50 each).

F = 2B (twice as many front section seats as back section)

75F + 50B = 25000

Easy route here would be substitution, as we've already got F in terms of B:

75(2B) + 50B = 25000
150B + 50B = 25000
200B = 25000
B = 125
F = 2B = 2(125) = 250

whpalmer4 · Answer

Checking, 250*75 = 18750, 125*50 = 6250, 18750+6250 = 25000, so we've met the other constraint as well.

gabylovesyou · Answer

so i was right! 250.. thank you

whpalmer4 · Answer

Yes, good work!