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At a college production of a​ play, 490 tickets were sold. the ticket prices were​ $8, $10, and​ $12, and the total income from ticket sales was ​$4640. how many tickets of each type were sold if the combined number of​ $8 and​ $10 tickets sold was 6 times the number of​ $12 tickets​ sold?

Respuesta :

jcherry99

Answer:

  • number of 12 dollar tickets = 70
  • number of 10 dollar tickets = 220
  • number of 8 dollar tickets = 200

Explanation:

Givens

  • Let the 8 dollar ticket number   = x1
  • Let the 10 dollar ticket number = x2
  • Let the 12 dollar ticket number = x3

Equations

  • x1 + x2 + x3 = 490
  • (x1 + x2) = 6*x3
  • x1*8 + x2*10 * x3*12 = 4640

Solution

You can get x3 right away.

x1 + x2 + x3 = 490         Substitute 6*x3 for x1 + x2

6*x3 + x3 = 490             Combine

7*x3 = 490                     Divide by 7

7*x3/7 = 490/7               Do the division

x3 = 70                           Answer x3

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x1 + x2 + 70 = 490         Let x3 = 70    

x1 + x2+70-70=490-70  Subtract 70  

x1 + x2 = 420                  

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8x1 + 10x2 + 12x3 = 4640       Use the third equation to get an amount for x1,x2

8x1 + 10x2 + 12*70 = 4640

8x1 + 10x2 + 840 = 4640        Subtract 840  

8x1 + 10x2 = 4640 - 840         Combine  

8x1 + 10x2 = 3800  

====================

8x1 + 8x2 = 420*8

8x1 + 8x2 = 3360

====================

8x1 + 10x2 = 3800

8x1 + 8x2 =  3360                   Subtract

2*x2 = 440

2*x2/2 = 440/2

x2 = 220

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x1 + x2 + x3 = 490

x1 + 220 + 70 = 490

x1 + 290 = 490

x1 = 200