Problem 24 - Beginning and Intermediate Algebra

Question

Write a system of equations and solve. At Sparkle Car Wash, two deluxe car washes and three regular car washes would cost \(\$ 26.00 .\) Four regular washes and one deluxe wash would cost \(\$ 23.00 .\) What is the cost of a deluxe car wash and of a regular wash? (THE IMAGES CANNOT COPY)

Step-by-Step Solution

Answer

The cost of a deluxe car wash (x) is \(\$7.00\) and the cost of a regular car wash (y) is \(\$4.00\).

Step 1: Define variables

Let the cost of one deluxe car wash be denoted by \(x\), and the cost of one regular car wash be denoted by \(y\).

Step 2: Convert information into equations

Write two equations based on the given information: 1. Two deluxe car washes and three regular car washes cost $26.00: \(2x + 3y = 26\) 2. Four regular washes and one deluxe wash cost $23.00: \(x + 4y = 23\) Now we have a system of two linear equations: \[2x + 3y = 26\] \[x + 4y = 23\]

Step 3: Solve the system of equations

To solve the system, we can use either the substitution method, the elimination method, or the matrix method. In this case, let's use the elimination method: Multiply the second equation by -2 to eliminate \(x\): \[-2(x + 4y = 23) \Rightarrow -2x - 8y = -46\] Now, add the newly obtained equation to the first equation: \[(2x + 3y) + (-2x - 8y) = 26 + (-46) \Rightarrow -5y = -20\] Divide both sides by -5 to find the value of \(y\): \[-5y = -20 \Rightarrow y = 4\] Now that we have found the value of \(y\), substitute its value in one of the equations to find the value of \(x\): \[x + 4(4) = 23 \Rightarrow x + 16 = 23 \Rightarrow x = 7\]

Step 4: State the Answer

The cost of a deluxe car wash (x) is \(\$7.00\) and the cost of a regular car wash (y) is \(\$4.00\).

Key Concepts

Linear Equations

Linear equations are algebraic expressions representing straight lines when graphed on a coordinate plane. They are comprised of variables which are typically denoted by letters like \(x\) or \(y\). These expressions can be simplified by performing arithmetic operations on both sides of the equation to maintain equality. In this exercise, we are given two linear equations:

• \(2x + 3y = 26\)
• \(x + 4y = 23\)

These equations have what we call a constant term on the right-hand side, and a variable term on the left-hand side. Our goal is to find out the values of these variables, specifically the cost of the cars washes in this scenario. This is achieved by manipulating these linear equations, which form a 'system of linear equations.'

Elimination Method

The elimination method is a strategy to solve a system of linear equations. It involves combining the equations in such a manner that one variable is removed, allowing us to solve for the other. Here’s how it works:First, we make the coefficients of one of the variables in both equations identical. In our exercise, the elimination method was used on variable \(x\). The second equation \(x + 4y = 23\) was multiplied by \(-2\):

• \(-2x - 8y = -46\)

Now, we add the transformed equation to the first:

• \(2x + 3y + (-2x - 8y) = 26 + (-46)\)

This leads to:

• \(-5y = -20\)

Dividing both sides by -5 yields the value for \(y\). The result simplifies one of the equations, as only one variable is left, making it easier for us to find the value of the other variable, \(x\). This systematic removal of variables is why the elimination method is an efficient approach.

Variables in Algebra

In algebra, variables are symbols representing quantities that can change or vary. They allow for mathematical expressions and equations to be generalized and manipulated. Variables are usually denoted by letters like \(x\) and \(y\). In the given exercise:

• \(x\) represents the cost of one deluxe car wash.
• \(y\) represents the cost of one regular car wash.

Algebraic equations involving variables can express complex relationships and can be solved to find specific values that meet all conditions of the system. Variables serve as placeholders for these unknown values until they are determined through solving. They provide the flexible foundation for mathematical modeling in equations.

Solving Algebra Problems

Solving algebra problems involves a systematic approach to finding the values of variables that satisfy given equations. The key steps often include:

• Identifying the variables and what they represent.
• Writing equations based on the problem statement.
• Choosing an appropriate method to solve the equations, such as elimination, substitution, or graphing.

In the exercise, after setting up the system of equations, we used the elimination method. Once one equation is simplified to find one variable, we substituted back to find the other. This two-step approach of elimination followed by substitution ensures all equations in the system contribute to finding a solution. Algebra problem-solving requires careful handling of variables and equations to arrive at accurate results that reflect the real-world situation described.