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{,}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 = 250 \\ 15x + 10y + 8z = 2825 \\ -2x + y = 0 \end{cases}$$ Using elimination on the first two equations to eliminate $$z$$ yields $$7x + 2y = 825$$. Using this new equation alongside the third equation ($$-2x + y = 0$$) allows us to eliminate $$y$$, resulting in $$x = 75$$. Substituting this back into the equations gives $$y = 150$$ and $$z = 25$$. Thus, the department sold $$75$$ adult tickets, $$150$$ student tickets, and $$25$$ child tickets.
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Updated 2026-06-03

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