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In a baseball stadium, there are three types of seats available. Box seats are $ 10, reserved seats are $6, and lawn seats are $4. The stadium capacity is 3900. If all the seats are sold, the total revenue to the club is $26, 820. If one half of the box seats sold, one half of the reserved seats are sold, and all the lawn seats are sold, the total revenue is $16,190. How many of each kind of seat are there?
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4 Answers

        B, R, L
R1:   1, 1, 1, 3900
R2: 10, 6, 4, 26820
R3:   5, 3, 4, 16190

R2 > R2 - 2*R3

R1: 1, 1, 1, 3900
R2: 0, 0,-4, -5560
R3: 5, 3, 4, 16190

R3 > R3 + R2

R1: 1, 1, 1, 3900
R2: 0, 0,-4, -5560
R3: 5, 3, 0, 10630

R2 > R2/(-4)

R1: 1, 1, 1, 3900
R2: 0, 0, 1, 1390
R3: 5, 3, 0, 10630

R1 > R1 - R2

R1: 1, 1, 0, 2510
R2: 0, 0, 1, 1390
R3: 5, 3, 0, 10630

R3 > R3 - 5*R1

R1: 1, 1, 0, 2510
R2: 0, 0, 1, 1390
R3: 0,-2, 0, -1920

R3 > R3/(-2)

R1: 1, 1, 0, 2510
R2: 0, 0, 1, 1390
R3: 0, 1, 0, 960

R1 > R1 – R3

R1: 1, 0, 0, 1550
R2: 0, 0, 1, 1390
R3: 0, 1, 0, 960

(B,R,L) = (1550,960,1390)
Hi Paula,
suppose Number of Box seats are x.
             Number of reserved seats are y
             Number of lawn seats are z.
The stadium capacity is 3900
so x+y+z= 3900................... equation 1
 
The total revenue to the club is $26,820.
 
so 10x+6y+4z= 26820..............equation 2
 
If one half of the Box seats (1/2 x) sold,one half of reserved seats (1/2 y)are sold, all lawn seats(z) are sold.
 
10 (1/2 x) +6(1/2 y)+4z = 16190
5x+ 3y +4z=16190.........equation 3
 
from equation 1, taking value of z= 3900-x-y........equation 4
& substituting in equation 2 & 3,
we get,
 
10x+6y+4(3900-x-y)= 26820,          5x+3y+4(3900-x-y)=16190
 
10x+6y+15600-4x-4y=26820,          5x+3y+15600-4x-4y=16190
simplifying,
6x+2y=11220                                 x-y=590
divide by 2
3x+y = 5610
 
now adding equations,
3x+y = 5610
  x-y =   590  
 
we get, 4x= 6200
             x= 1550
 
x-y=590
1550-y=590
y= 960
 
substituting the value of x & y in equation 4, we get
z= 3900-1550-960=3900-2510=1390.
 
 Ans:  There are 1550 Box seats, 960 reserved seats, 1390 lawn seats.            
 
Check: 1550+960+1390= 3900 i.e the capacity of stadium.                          
 
 
 
 
 
 
Hi Paula;
Total box seats...2b
Total reserved seats...2r
Total lawn seats...l
FIRST EQUATION...2b+2r+l=3900
 
SECOND EQUATION...[2b($10)]+[2r($6)]+[l($4)]=$26,820
The units of dollars on both sides cancels...
................................[2b(10)]+[2r(6)]+[l(4)]=26,820
................................20b+12r+4l=26,820
Divide both sides by 4......5b+3r+l=6,705
 
THIRD EQUATION...[(b)(10)]+[(r)(6)]+[(l)(4)]=16,190
..............................10b+6r+4l=16,190
Divide both sides by 2.....5b+3r+2l=8,095
 
Let's subtract the SECOND from the THIRD...
THIRD.........5b+3r+2l=8,095
SECOND...-(5b+3r+l=6,705)
....................l=1390
 
Let's plug l=1390 into all three equations...
FIRST EQUATION...2b+2r+l=3900
............................2b+2r+1390=3900
............................2(b+r)=2510
............................b+r=1255
 
SECOND EQUATION...5b+3r+l=6705
................................5b+3r+1390=6705
................................5b+3r=5315
 
THIRD EQUATION...5b+3r+2l=8095
.............................5b+3r+[(2)(1390)]=8095
.............................5b+3r+2780=8095
.............................5b+3r=5315
 
Let's take the FIRST EQUATION and isolate one variable.  I randomly select b...
FIRST EQUATION...b+r=1255
...........................b=1255-r
Let's plug this into the THIRD EQUATION...
THIRD EQUATION...5b+3r=5315
...........................[(5)(1255-r)]+3r=5315
............................6275-5r+3r=5315
............................6275-2r=5315
............................-2r=-960
...........................r=480
 
FIRST EQUATION...2b+2r+l=3900
...........................2b+[(2)(480)]+1390=3900
...........................2b+960+1390=3900
...........................2b+2350=3900
............................2b=1550
.............................b=775
 
SECOND EQUATION...5b+3r+l=6705
................................[(5)(775)]+[(3)(480)]+1390=6705
...............................3875+1440+1390=6705
...............................6705=6705
 
THIRD EQUATION...5b+3r+2l=8095
............................[(5)(775)]+[(3)(480)]+[(2)(1390)]=8095
............................3875+1440+2780=8095
............................8095=8095
 
TOTAL BOX SEATS...2b...(2)(775)=1550
TOTAL RESERVE SEATS....2r...(2)(480)=960
TOTAL LAWN......l...1390

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