Problem 28 - Elementary Algebra

Question

The relationship of the distance driven, \(x\), and the cost of gasoline, \(y\), is a direct variation. For a trip of \(250 \mathrm{mi}\), the cost is \(\$ 90\). a. Find the constant of proportionality. Include the units of measurement. b. Write an equation that represents this relationship. c. Find the cost of gasoline to drive \(225 \mathrm{mi}\). d. What does \(k\) represent in this equation?

Step-by-Step Solution

Answer

The constant of proportionality is \(0.36 \, \text{dollars per mile}\), the equation is \( y = 0.36x \), and the cost to drive 225 miles is 81 dollars. The constant \( k \) represents the cost per mile.

Step 1 - Identify the relationship

The relationship between distance driven, \(x\), and the cost of gasoline, \(y\), is given by a direct variation. This means \( y = kx \) for some constant \( k \).

Step 2 - Find the constant of proportionality

We know that for a trip of \(250 \, \text{mi}\), the cost is \(90 \, \text{dollars}\). Using the formula \( y = kx \), substitute the given values to find \( k \). \[ 90 = k \times 250 \] Solving for \( k \): \[ k = \frac{90}{250} = 0.36 \, \text{dollars per mile} \]

Step 3 - Write the equation

Now that we have found \( k \), we can write the equation that represents this relationship: \( y = 0.36x \)

Step 4 - Find the cost of gasoline for 225 miles

To find the cost of driving \(225 \, \text{mi}\), use the equation from Step 3: \[ y = 0.36 \times 225 \] Calculate the value: \[ y = 81 \, \text{dollars} \]

Step 5 - Interpret the constant of proportionality

In the equation \( y = kx \), the constant of proportionality \(k = 0.36\) represents the cost of gasoline per mile driven.

Key Concepts

constant of proportionality

In mathematics, particularly in direct variation, the constant of proportionality is a key concept that determines the relationship between two variables. This constant, denoted as \( k \), connects the input variable \( x \) to the output variable \( y \) through the equation \( y = kx \).

Let’s look at our exercise: we know that driving 250 miles costs \(90. To find the constant of proportionality (\( k \)), we use the relationship formula \( y = kx \). Here, \( y = 90 \) dollars, and \( x = 250 \) miles. By plugging these values into the equation, we get:

• 
  90 = k \times 250 \
  
• 
  
  Then, solving for \( k \) gives:
  k = \frac{90}{250} = 0.36 \text{ dollars per mile} \
  
  

So, \( k\) is 0.36 dollars per mile. This means for every mile driven, it costs \)0.36 for gasoline. Understanding this constant lets us easily calculate the cost for any distance driven by simply multiplying it by this constant of proportionality.

distance and cost relationship

The distance and cost relationship, when described as a direct variation, is a straightforward concept. In our case, it means that the cost of driving is directly proportional to the distance driven. The more miles you drive, the more you will pay for gasoline.

The formula representing this relationship is \( y = kx \), where \( y \) is the cost of gasoline, and \( x \) is the distance driven. With our constant of proportionality \( k = 0.36 \) dollars per mile, the equation becomes

• y = 0.36x

Using this formula, we can compute the cost for any given distance. For example, to find the cost of driving 225 miles:

• y = 0.36 \times 225 \
  y = 81 \text{ dollars} \
  

This tells us that it costs $81 to drive 225 miles. Thanks to the direct variation formula, changes in distance produce stable, predictable changes in cost, which makes planning and budgeting for trips easier.

algebraic equations

Algebraic equations are mathematical statements that show the relationship between different quantities. In our problem, the equation \( y = 0.36x \) is an algebraic equation that indicates how the cost of gasoline (\( y \)) varies with distance driven (\( x \)).

Knowing how to manipulate and understand these equations is essential for solving direct variation problems. Here are some key points useful for recognizing and working with algebraic equations:

• Identify the variables: In our problem, \(x \) is the distance, and \( y \) is the cost.


• Determine the constant: We found the constant of proportionality \( k \) and made sure it's in the correct units, dollars per mile.


• Use the equation: Plug in the values to find unknown variables.

Whether you are finding the cost for a specific distance or understanding how costs add up over different trips, algebraic equations help translate real-world relationships into mathematical terms. This makes problem-solving more straightforward and more accurate. By practicing with these concepts, students can gain a deeper understanding of how math describes everyday phenomena.