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In many real-world situations, we often have to deal with multiple relationships occurring simultaneously. This leads to a "System of Equations", where multiple equations work together to describe a scenario. In the problem where a group charters a bus, we have two main equations:

• The original equation for total cost: \( 900 = x \times y \) where \( x \) is the number of people, and \( y \) is the cost per person.
• The equation after some people opt out: \( 900 = (x - 5) \times (y + 2) \).

These two equations form a system because they both represent the same total charter cost under different circumstances. The goal here is to find a solution that satisfies both equations simultaneously. Solving systems of equations helps us find unknown values that fit within multiple conditions. Typically, one can solve these by substitution, elimination, or graphically, but in this problem, substitution is used for simplicity.

One way to solve quadratic equations is by using the quadratic formula. A vital part of this process is the discriminant. The discriminant is the part under the square root in the quadratic formula: \( b^2 - 4ac \). It tells us a lot about the nature of the solutions:

• If it’s positive, you get two real solutions.
• If it's zero, you get exactly one real solution.
• If negative, there are no real solutions, only complex ones.

In the bus chartering problem, substituting into the quadratic equation involves calculating the discriminant:\[ (-800)^2 - 4 \times 2 \times (-4500) = 640000 + 36000 = 676000 \]Since the discriminant is positive, we can expect two real solutions for \( x \), though logically, only one makes sense in the context of the problem (i.e., a positive number of people).

Cost analysis helps us understand and manage expenses effectively. In this exercise, analyzing costs is vital to understanding how changes in group size affect individual expenses. Originally, the cost is \( \\(900 \) spread across \( x \) participants, so each pays \( y \). When five individuals drop out, the remaining group faces a new cost: \( y + 2 \), because the fixed overall charter cost \( \\)900 \) is now divided among fewer people.

• Original cost per person: \( y = \frac{900}{x} \)
• New cost per person after dropout: \( y + 2 \)

This change in the cost distribution is a classic example of cost analysis, where the same total cost gets allocated differently based on circumstances, leading to a practical shift in per-person expense.

Variables are fundamental in algebra, representing unknown or changing quantities in problems. In the problem about chartering a bus, two variables are used:

• \( x \): the original number of people intending to go on the trip.
• \( y \): the cost per person initially planned.

Why use variables? They allow us to set up relationships and equations based on the problem context. Here, they help to express the total cost equation \( 900 = x \times y \). As we solve, these variables let us track and adjust as conditions change, such as when people decide not to go, leading to new variables \( x - 5 \) and \( y + 2 \). By carefully defining and solving for these variables, we can uncover the most accurate information and answer practical questions.