Problem 6 - 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? Write a verbal model for the problem.

Step-by-Step Solution

Answer

The cost per gallon of regular gasoline is $1.19 and the cost per gallon of premium gasoline is $1.39.

Step 1: Formulate the equations

Let \( R \) be the cost of regular gasoline and \( P \) be the cost of premium gasoline. From the information, we can create two equations. Equation 1: \( 10R + 15P = 32.75 \), which states the total cost for regular and premium gasoline. Equation 2: \( P = R + 0.20 \), which represents that premium gasoline costs 20 cents more per gallon than regular gasoline.

Step 2: Solve the system of equations

We will substitute the second equation into the first in order to solve for R. Replacing \( P \) in equation 1 we get: \( 10R + 15(R+.20) = 32.75 \). Simplify to obtain \( 25R + 3 = 32.75 \). Solving for R we get \( R = 1.19 \) which represents the cost per gallon of regular gasoline.

Step 3: Find the cost of premium gasoline

Substitute \( R = 1.19 \) into equation 2, we have \( P = 1.19 + 0.20 = 1.39 \). This represents the cost per gallon of premium gasoline.

Key Concepts

Cost Calculation

Cost calculation is a fundamental element when dealing with financial problems. In this exercise, we are tasked with determining the individual prices of two types of gasoline. We know that the total cost for purchasing 10 gallons of regular gasoline and 15 gallons of premium gasoline is $32.75. When faced with such a problem, the goal is to break down total expenses into the cost components.
To do this, we assign variables to the costs of each type of gasoline:

• Let the cost per gallon of regular gasoline be represented by \( R \).
• Similarly, let the cost per gallon of premium gasoline be represented by \( P \).

This is crucial because it helps us set up algebraic equations to model the relationship between the components. For instance, the equation for total cost can be expressed as \( 10R + 15P = 32.75 \). Breaking down costs into such simple terms helps us easily solve for unknowns. Always remember to translate word problems into mathematical expressions for easier computation.

Substitution Method

The substitution method is an effective way to solve systems of linear equations. It involves replacing one variable in terms of another variable within the equations. This is especially handy when given an additional relationship between the variables.
For the gasoline cost problem, we are provided with this relationship: Premium gasoline costs $0.20 more per gallon than regular gasoline. It is formulated as \( P = R + 0.20 \). By using this relationship in conjunction with our cost equation, \( 10R + 15P = 32.75 \), we can substitute \( P \) in the cost equation.

• Replace \( P \) with \( R + 0.20 \).
• The new equation becomes \( 10R + 15(R + 0.20) = 32.75 \).

By performing this substitution, we now have a single equation in terms of \( R \). Solving this equation will give us the cost per gallon of regular gasoline, \( R = 1.19 \). Substituting \( R \) back to find \( P \) gives \( P = 1.39 \), the cost of premium gasoline. The substitution method simplifies solving processes by reducing the number of variables and is a key technique taught in algebra.

Algebraic Expressions

Algebraic expressions form the basis of creating relationships in problems involving costs and quantities. They are concise mathematical statements used to represent situations described in word problems, translating them into mathematical computations.
In this exercise, we encountered equations such as \( 10R + 15P = 32.75 \) and \( P = R + 0.20 \). These are algebraic expressions that capture essential information from the word problem.

• They help simplify complex relationships by expressing them in terms of mathematical operations like addition and multiplication.
• Understanding how to manipulate these expressions through techniques such as substitution or elimination enables us to find unknown values effectively.

Algebraic expressions are like the language of mathematics that allow us to model real-world problems with accuracy. Grasping their construction and manipulation is crucial for anyone working with systems of equations and looking to solve for unknowns.