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 $, student tickets for $, and child tickets for $. They sold a total of tickets and brought in $2{,}825. Furthermore, the number of student tickets sold was twice the number of adult tickets. Let , , and 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.0
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Solve an Application using a System of Three Linear Equations (Example 4.36)
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Solve an Application using a System of Three Linear Equations (Try It 4.71)
Solve an Application using a System of Three Linear Equations (Try It 4.72)
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