Example

Solve an Application using a System of Three Linear Equations (Example 4.36)

To solve a real-world application involving a system of three linear equations, such as determining ticket sales, we translate the problem's conditions into an algebraic system.

Consider a theater that sold adult tickets for $1515, student tickets for $1010, and child tickets for $88. They sold a total of 250250 tickets and brought in $2,8252{,}825. Furthermore, the number of student tickets sold was twice the number of adult tickets. Let xx, yy, and zz represent the number of adult, student, and child tickets, respectively. This gives the system:

{x+y+z=25015x+10y+8z=28252x+y=0\begin{cases} x + y + z = 250 \\\\ 15x + 10y + 8z = 2825 \\\\ -2x + y = 0 \end{cases}

Using elimination on the first two equations to eliminate zz yields 7x+2y=8257x + 2y = 825. Using this new equation alongside the third equation (2x+y=0-2x + y = 0) allows us to eliminate yy, resulting in x=75x = 75. Substituting this back into the equations gives y=150y = 150 and z=25z = 25. Thus, the department sold 7575 adult tickets, 150150 student tickets, and 2525 child tickets.

Image 0

0

1

Updated 2026-05-25

Contributors are:

Who are from:

Tags

OpenStax

Intermediate Algebra @ OpenStax

Ch.4 Systems of Linear Equations - Intermediate Algebra @ OpenStax

Algebra

Related