Example

Solving a Fine Arts Ticket Sales Application using a System of Three Linear Equations (Try It 4.71)

To solve a fine arts ticket sales application using a system of three linear equations, consider a fine arts department that sold adult tickets for $20, student tickets for $12, and child tickets for $10. They sold a total of 350 tickets and brought in $4,650. The number of child tickets sold was the same as the number of adult tickets. Let xx be the number of adult tickets, yy be the number of student tickets, and zz be the number of child tickets. The system of equations modeling this application is: {x+y+z=35020x+12y+10z=4650xz=0\begin{cases} x + y + z = 350 \\ 20x + 12y + 10z = 4650 \\ x - z = 0 \end{cases} To solve, substitute z=xz = x into the first two equations to get 2x+y=3502x + y = 350 and 30x+12y=465030x + 12y = 4650. From the first equation, y=3502xy = 350 - 2x. Substituting this into the second yields 30x+12(3502x)=465030x + 12(350 - 2x) = 4650, which simplifies to 6x=4506x = 450, meaning x=75x = 75. Since z=xz = x, z=75z = 75. Then y=3502(75)=200y = 350 - 2(75) = 200. The department sold 75 adult tickets, 200 student tickets, and 75 child tickets.

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Updated 2026-06-26

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