Problem 115 - College Algebra

Question

A gym membership with two personal training sessions costs \(\$ 125,\) while gym membership with five personal training sessions costs \(\$ 260 .\) What is cost per session?

Step-by-Step Solution

Answer

The cost per personal training session is $45.

Step 1: Define Variables

First, define the variables. Let \( x \) be the cost of the gym membership and \( y \) be the cost per personal training session.

Step 2: Set Up Equations

Based on the problem description, we have two equations. One for the gym membership with two sessions: \( x + 2y = 125 \), and another for the gym membership with five sessions: \( x + 5y = 260 \).

Step 3: Subtract the Equations

To eliminate \( x \) and find \( y \), subtract the first equation from the second equation: \((x + 5y) - (x + 2y) = 260 - 125\). This simplifies to \(3y = 135\).

Step 4: Solve for y

Solve the equation \(3y = 135\) for \(y\). Divide both sides by 3 to isolate \(y\): \(y = 45\). So, the cost per personal training session is \(\$45\).

Key Concepts

Variables in Equations

In the world of mathematics, variables act as placeholders that represent unknown values. Using letters like \( x \) or \( y \), we can symbolize these unknowns in an equation.
In our gym membership exercise, the variable \( x \) is the cost of the gym membership, while \( y \) stands for the cost of a single personal training session.
By turning unknown values into variables, it becomes easier to structure and solve problems.

• Variables can simplify information and make it manageable.
• They allow us to form equations that describe relationships between different quantities.

Think of an equation as a balance, where variables help us figure out the "weight" of missing pieces. In this exercise, understanding what each variable symbolizes is the first step towards solving the problem.

Equation Systems

When we have more than one equation that needs to be solved together, we deal with a system of equations. These systems help us find multiple unknowns at once. In this example, two equations describe the same gym membership scenario.
The first equation, \( x + 2y = 125 \), relates to a membership with two personal training sessions. The second, \( x + 5y = 260 \), involves five sessions.
Together, these equations form a system which we can solve to find the values of both \( x \) and \( y \).

• Two or more equations with the same variables form an equation system.
• The goal is to find the values of variables that satisfy all equations simultaneously.

Recognizing these systems allows us to address complex relationships using straightforward mathematical operations.

Solving Linear Equations

Solving linear equations involves finding the values of variables that make each equation true. In our gym membership problem, the combination of equations can be manipulated to find \( y \).
We used the subtraction method here. By subtracting the first equation \((x + 2y = 125)\) from the second \((x + 5y = 260)\), we eliminated \( x \). This simplifies everything to one equation: \( 3y = 135 \).
This technique makes it easier to handle multiple variables simultaneously.

• Eliminating a variable helps simplify the equation system into a single equation.
• Divide or multiply as needed to isolate variables.
• The goal is to have each variable represented by a number.

In the solution, dividing by 3 gives us \( y = 45 \), meaning each personal training session costs \$45. Through proper step-by-step approaches, linear equations reveal values hiding within the problem.