Problem 32 - Elementary Algebra Within Reach

Question

In Exercises 29-34, use a system of linear equations to solve the problem. The total cost of 6 gallons of regular gasoline and 11 gallons of premium gasoline is \(\$ 68.33\). Premium gasoline costs \(\$ 0.20\) more per gallon than regular gasoline. Find the price per gallon for each type of gasoline.

Step-by-Step Solution

Answer

The price per gallon for regular gasoline is \$3.89 and for premium gasoline is \$4.09.

Step 1: Set up the equations

Let \( r \) represent the price of regular gasoline and \( p \) represent the price of premium gasoline. The information given can be written as: \[ 6r + 11p = 68.33 \] (total price of 6 gallons of regular and 11 of premium gasoline equals $68.33) and \[ p = r + 0.20 \] (premium gasoline costs $0.20 more per gallon than regular gasoline)

Step 2: Substitute the second equation into the first one

Replace \( p \) in the first equation with \( r+0.20 \) from the second one: \[ 6r + 11(r + 0.20) = 68.33 \]

Step 3: Solve for r

Distribute and combine like terms to solve for \( r \): \[ 6r + 11r + 2.20 = 68.33 \] \[ 17r = 66.13 \] \[ r = 3.89 \]

Step 4: Substitute r back into the second equation

Substitute \( r = 3.89 \) into the second equation to find \( p \): \[ p = 3.89 + 0.20 \] \[ p = 4.09 \]

Key Concepts

Understanding Linear Equations in a System

Linear equations are equations where the highest power of the variable is one. They depict relationships as straight lines when graphed. In the exercise provided, we work with a system of linear equations. This means there is more than one equation at play, all connected by similar variables. Each equation corresponds to different constraints or conditions given in the problem.

Systems of linear equations often enable us to find unknown values by using relationships between them. In our case, these equations help define the connection between the gallons of gasoline bought and their respective costs, using prices per gallon as variables.

The Substitution Method Explained

The substitution method is a straightforward way to find the solution to a system of linear equations. It involves solving one of the equations for a single variable, and then substituting that expression into the other equation.

Here's how it works: First, solve the second equation, \( p = r + 0.20 \), for one variable, which in this case is \( p \). Then substitute this expression into the first equation, \( 6r + 11p = 68.33 \). This reduces the system to a single equation with one variable: \( 6r + 11(r + 0.20) = 68.33 \).

By reducing the number of variables, it becomes simpler to solve for the remaining variable. Once that value is found, it can be substituted back to get the value of the other variable.

Price Calculation Demystified

Price calculation in this context involves determining the cost per unit, given certain totals and relationships between different items. We have to calculate the cost per gallon of regular and premium gasoline using the information that premium gasoline is \( \$0.20 \) more expensive per gallon than regular gasoline.

Start by assigning variables to the unknowns: \( r \) for the price of regular gasoline and \( p \) for the price of premium gasoline. Use the equations \( 6r + 11p = 68.33 \) and \( p = r + 0.20 \) to express the total cost condition and price difference condition, respectively. Solving these equations lets you find the exact prices for both types of gasoline.

Algebraic Problem Solving Techniques

Algebraic problem solving involves identifying unknowns and defining relationships between them with equations. It's much like solving a puzzle, where the information given needs to be pieced together with algebraic techniques.

The key steps include:

• Defining variables for the unknowns, which are the prices in this case.
• Formulating equations that represent the relationships described.
• Using methods such as substitution to manage and solve these equations.

With practice, solving these algebraic problems becomes easier. This particular problem involves defining price relationships between two types of gasoline and calculating their costs. It's a practical example of how algebra can be used in daily life for tasks like price negotiation and budgeting.