SOLUTION: A theater owner want to divide an 1800 seat theatre in 3 sections. With tickets costing 20, 35, and 50. He wants to have twice as many 20 dollar tickets as the sum of the other tic

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Question 345374: A theater owner want to divide an 1800 seat theatre in 3 sections. With tickets costing 20, 35, and 50. He wants to have twice as many 20 dollar tickets as the sum of the other tickets and he wants to earn 48,000 from a full house. To find out how many seats in each section, solve the equations.
x+y+z=1800
x=2(y+z)
20x+35y+50z=48,000
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Substitute x=2%28y%2Bz%29 into eq. 1 and eq. 2,
1.x%2By%2Bz=1800
1.2y%2B2z%2By%2Bz=1800
1.3y%2B3z=1800
1.y%2Bz=600
.
.
.
2.20x%2B35y%2B50z=48000
2.40y%2B40z%2B35y%2B50z=48000
2.75y%2B90z=48000
2.5y%2B6z=3200
From eq. 1,
y=600-z
Substitute into eq. 2,
5%28600-z%29%2B6z=3200
3000-5z%2B6z=3200
highlight%28z=200%29
Then,
y=600-200
highlight%28y=400%29
And finally,
x=2%28y%2Bz%29
x=2%28400%2B200%29
highlight%28x=1200%29