Problem 7 - Algebra 1

Question

Use the following problem. The total cost of 10 gallons of regular gasoline and 15 gallons of premium gasoline is \(\$ 32.75 .\) Premium costs \(\$ .20\) more per gallon than regular. What is the cost per gallon of each type of gasoline? Assign labels to the verbal model.

Step-by-Step Solution

Answer

The cost per gallon of regular gasoline is $1.35 and the cost per gallon of premium gasoline is $1.55.

Step 1: Create equation for total cost

Let's denote the cost per gallon of regular gasoline as \( R \), and the cost per gallon of premium gasoline as \( P \). According to the problem, the total cost of 10 gallons of regular gas and 15 gallons of premium gas is \$32.75. This can be expressed as the equation \( 10R + 15P = 32.75 \)

Step 2: Create equation for price difference

The problem states that premium gasoline costs $0.20 more per gallon than regular. This difference can be written as: \( P = R + 0.2 \)

Step 3: Substitute and solve

Now we can substitute equation 2 (price difference) into equation 1 (total cost). In place of \( P \) in equation 1 we substitute \( R + 0.2 \), giving us a new equation: \( 10R + 15(R + 0.2) = 32.75 \). Solving this equation yields \( R = 1.35 \). Then substitute \( R = 1.35 \) into equation 2 to get \( P = 1.35 + 0.2 = 1.55 \)

Key Concepts

Verbal Model

A verbal model is a powerful tool that helps us create a mathematical representation of a real-world situation using words and language. In essence, it's about translating the problem into everyday language before jumping into equations. This helps us clearly understand what the problem is asking and how different variables relate to each other.

• For our gasoline example, we can set up a verbal model by imagining a simple statement: "The cost of gasoline is a result of the total gallons multiplied by the cost per gallon."
• In this case, the verbal model tells us: "The total cost is the sum of the cost of regular and premium gasoline."

When you construct this model, you quickly see the relationships. Regular gas is one component, and premium gas is another, with their costs adding up to the total. By organizing our thoughts in this way, it makes the next steps—creating equations—much simpler and more intuitive.

Substitution Method

The substitution method is a technique used to solve systems of equations. It's particularly useful when you have one equation expressed in terms of a variable, which can then be replaced in another equation. This method can simplify your calculations and lead quickly to solutions.
Here's how it works in our gasoline problem:

• First, recognize you have two equations: one for total cost and one for the price difference.

We already know from the problem that the premium price is regular price plus 0.20 (i.e., \( P = R + 0.2 \)). We can substitute this expression into the cost equation \( 10R + 15P = 32.75 \), replacing \( P \) with \( R + 0.2 \).
This gives us a single equation with one variable, making it much easier to solve.

• Substitution reduces the complexity by turning a system of equations into a single equation with one unknown.

This approach exemplifies how breaking down complex problems into manageable parts can make them much easier to tackle.

Linear Equations

Linear equations are fundamental in algebra. They are equations of the first degree, meaning they involve variables raised only to the power of one. The form is usually expressed as \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants.
In our problem, we use linear equations to model the relationship between the amount of gasoline and its cost.

• The total cost equation \( 10R + 15P = 32.75 \) is linear because both variables \( R \) and \( P \) have a power of one.
• The price relationship equation \( P = R + 0.2 \) is also linear, showing a straightforward relationship between the two variables.

Solving these linear equations helps us find exact values for \( R \) and \( P \), giving us the costs per gallon for regular and premium gas, respectively.
Understanding linear equations is crucial because they provide a clear method to represent and solve problems involving proportional relationships, such as cost per unit in our gasoline example.