Example

Solving a Zoo Ticket Mixture Application

Apply the system of equations method to solve a ticket mixture application.

Problem: The ticket office at the zoo sold 553553 tickets one day. The receipts totaled $3,936\$3{,}936. How many $9\$9 adult tickets and how many $6\$6 child tickets were sold?

  1. Name the unknowns. Let a=a = the number of adult tickets and c=c = the number of child tickets.
  2. Translate into a system of equations.
  • The total number of tickets sold is 553553: a+c=553a + c = 553
  • The total value of adult tickets is 9a9a and the total value of child tickets is 6c6c. The total receipts are $3,936\$3{,}936: 9a+6c=3,9369a + 6c = 3{,}936

The system of equations is: a+c=553a + c = 553 9a+6c=3,9369a + 6c = 3{,}936

  1. Solve the system using the elimination method. Multiply the first equation by 6-6: 6(a+c)=6(553)-6(a + c) = -6(553) 6a6c=3,318-6a - 6c = -3{,}318

Add this to the second equation: 9a+6c=3,9369a + 6c = 3{,}936 3a=6183a = 618 a=206a = 206

Substitute a=206a = 206 into the first equation: 206+c=553206 + c = 553 c=347c = 347

  1. Check the result. 206206 adult tickets at $9\$9 each is $1,854\$1{,}854. 347347 child tickets at $6\$6 each is $2,082\$2{,}082. 1,854+2,082=3,9361{,}854 + 2{,}082 = 3{,}936 \checkmark

  2. Answer the question. The zoo sold 206206 adult tickets and 347347 child tickets.

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Updated 2026-05-25

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