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# Solve the word problem

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?

### 4 Answers by Expert Tutors

Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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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)
Ben T. | Experienced Math Tutor (From Algebra to Calculus)Experienced Math Tutor (From Algebra to ...
5.0 5.0 (28 lesson ratings) (28)
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x + y + z = 3900
10x + 6y + 4z = 26,820
5x + 3y + 4z = 16,190

x = 1550
y = 960
z = 1390
Meena S. | Patient & knowledgeable Mathematics & Statistics (SPSS) TutorPatient & knowledgeable Mathematics & St...
4.5 4.5 (57 lesson ratings) (57)
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Hi Paula,
suppose Number of Box seats are x.
Number of reserved seats are y
Number of lawn seats are z.
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

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.

Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
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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