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? Solve the system and answer the question.

Short answer

The cost per gallon of regular gasoline is \(\$1.19\) and the cost per gallon of premium gasoline is \(\$1.39\)

Step-by-step solution

Establish the equations

Let's represent the cost per gallon of regular gasoline as \(x\) and premium gasoline as \(y\). From the given problem,\[\begin{align*}10x + 15y &= 32.75 \quad -(1)\y &= x + 0.20 \quad -(2)\end{align*}\]

Substitute Equation (2) into Equation (1)

Using Equation (2), we can replace \(y\) with \(x + 0.20\) in Equation (1):\[10x + 15(x + 0.20) = 32.75\]

Solve for x (Regular Gasoline Cost)

Solve Equation (3) for \(x\):\[10x + 15x + 3 = 32.75\]Combining like terms gives:\[25x = 29.75\]So \(x = 29.75/25 = 1.19\)

Find the cost of premium gasoline (y)

Using Equation (2), we substitute the value of \(x\) into it to find \(y\):\[y = 1.19+ 0.20 = 1.39\]

Key concepts

Algebraic Equations

Algebraic equations are the foundation of many mathematical problems and are used to describe the relationship between variables. An algebraic equation is a statement of equality between two algebraic expressions. It involves constants, variables, and arithmetic operators, such as addition, subtraction, multiplication, and division.

For instance, let's consider the equation \(x + 3 = 7\). This simple algebraic equation states that when 3 is added to a certain number \(x\), the result is 7. To solve it, we would isolate the variable \(x\) by subtracting 3 from both sides of the equation, obtaining \(x = 4\).

In the given problem, we have a practical application of algebraic equations where \(x\) represents the price per gallon of regular gasoline, and \(y\) stands for the price per gallon of premium gasoline. Two equations are formulated to solve for both variables, linking the cost of gasoline to quantities and the price difference between regular and premium gasoline.

Solving Linear Systems

Solving linear systems involves finding the values of the variables that make all equations in the system true simultaneously. A 'system' means that there are multiple equations that are all connected by shared variables. To solve a system of linear equations, various methods can be used, such as substitution, elimination, or graphing.

In the case of substitution, one equation is solved for one variable, and this expression is substituted into the other equation(s). This method proves particularly useful when one equation can be easily isolated for a variable, just like in our example where Equation (2), \(y = x + 0.20\), was easily isolated for \(y\) and was then substituted into Equation (1).

The process of substitution turned the system of equations into a single equation with one variable, \(x\), which was then solved to find the cost per gallon of regular gasoline. Understanding how to manipulate and solve such systems is invaluable for deciphering more complex problems or real-life situations that can be modeled by linear relationships.

Word Problems in Algebra

Word problems in algebra require translating a written narrative into mathematical equations. This skill is essential for students to learn as it bridges the gap between abstract mathematical concepts and real-world scenarios where measurements, costs, and various quantities interact.

The key to approaching these problems is to carefully read the narrative, extract the relevant information, and define what needs to be solved for. Typically, the unknown values are represented by variables, and the information given is used to establish equations.

In our exercise, the word problem translated into a system involving the total cost of gallons of gasoline and the cost difference between types of gasoline. Students are required to identify the key pieces of information - the number of gallons bought, the total cost, and the price difference - and use these to set up the equations. Once these equations are created and solved, as we did earlier in our steps, we successfully find the cost per gallon for regular and premium gasoline, thus finding a solution to the practical problem presented.