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A linear relationship involves two variables that create a straight-line pattern when graphed. In our scenario, the cost of renting a car for a given day is calculated based on a fixed daily rate and a variable rate depending on the miles driven. This is an example of a linear equation, where one variable, denoted as \(x\), is directly related to another, denoted as \(y\), through a constant rate of change.

In this particular exercise, the interesting aspect is how cost increases with each mile driven. The equation provided is \(y = 50 + 0.20x\), illustrating a clear relationship between miles driven, \(x\), and total cost, \(y\). The fixed rate of \(50 per day represents the y-intercept, while the \)0.20 per mile represents the slope. Every additional mile driven results in a 20-cent increase in the total cost.

Understanding this concept is key to solving problems involving linear equations, as it helps predict how changes in one variable impact the other.

Calculating costs in scenarios like this involves substituting known amounts into a linear equation. Let's break down each step to find out how much a person would pay when driving 100 miles:

• Base cost for one day: \(50
• Variable mile cost: \)0.20 per mile

To calculate, substitute \(x = 100\) miles into the formula \(y = 50 + 0.20x\):

\[y = 50 + 0.20(100) = 50 + 20 = 70\]
In this example, driving 100 miles for one day results in a total cost of $70.

Cost calculation follows a predictable pattern where anyone driving the same number of miles and renting for the same number of days will pay the same amount. This uniformity is due to the set structure of the linear cost model, ensuring that each variable change impacts the outcome identically, like a set price list.

Algebraic expressions are used to represent relationships and formulas in mathematics, particularly in cost calculations and situations involving variable components. The given expression \(y = 50 + 0.20x\) combines both constant and variable terms to illustrate how the rental costs vary with mileage.

In this case:

• \(50\) is the constant term, representing the base price regardless of distance driven.
• \(0.20x\) is the variable term, reflecting the additional cost incurred per mile driven.

The algebraic expression simplifies complex real-world problems into understandable mathematical statements.

It helps predict outcomes based on variable inputs, offering clarity and consistency in scenarios where costs or quantities might vary. Understanding how to manipulate these expressions through substitution or simplification is a fundamental skill in algebra, enabling solutions to diverse problems efficiently.