Solving Linear Equations in Ticket Sales
In this article, we will walk through a problem that involves solving a system of linear equations to determine how many adult and student tickets were sold at a museum. The system of equations approach allows us to efficiently solve for the unknowns and provides a clear method for understanding similar types of problems.
The Problem
The cost of an adult ticket to a museum is $4, while a student ticket costs $3. A total of 105 tickets were sold, and the total revenue from ticket sales was $380. We need to determine how many adult and student tickets were sold.
Setting Up the Equations
Let nx represent the number of adult tickets sold, and ny represent the number of student tickets sold. We can set up the following equations based on the information provided:
nx ny 105 4nx 3ny 380Step-by-Step Solution
1. **Express y in terms of x**:
From the first equation: ny 105 - nx
2. **Substitute y in the second equation**:
4nx 3(105 - nx) 380
3. **Simplify and solve for x**:
Distributing the 3 gives: 4nx 3105 - 3nx 380
Combining like terms: nx 315 380
Isolating x**:
nx 380 - 315
nx 65
4. **Find y**:
Substituting x back into the equation for y**:
ny 105 - 65
ny 40
Conclusion
Therefore, the number of adult tickets sold is 65, and the number of student tickets sold is 40.
Alternative Methods
Here are a few alternative methods to solve the same problem:
5. Eliminate s**: a s 105 4a 3s 380 Eliminate s by multiplying the first equation by 3… eq 3a 3s 315 Subtract the second equation minus this new equation… a 65 Substitute into the first equation… 65 s 105 s 40
6. Substitution method**: Let A be the number of adults and S be the number of students. A S 105 4A 3S 380 Multiply the first equation by 3… 3A 3S 315 Subtract the second equation minus this new equation… A 65 Substitute into the first equation… S 105 - 65 40
7. Elimination method**: x y 105 4x 3y 380 Multiply the first equation by 3… eq 3x 3y 315 Subtract the two equations… x 65 Substitute into the first equation… y 105 - 65 40