Solve the word problem

Question

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?

Answer 1

        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)

Answer 2

x + y + z = 3900

10x + 6y + 4z = 26,820

5x + 3y + 4z = 16,190

 

 

x = 1550

y = 960

z = 1390

Answer 3

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.                          

 

 

 

 

 

 

Answer 4

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