Problem 24 - Algebra 2

Question

All 28 members in Crestview High School's Ski Club went on a one-day ski trip. Members can rent skis for \(\$ 16\) per day or snowboards for \(\$ 19\) per day. The club paid a total of \(\$ 478\) for rental equipment. Write a system of equations that represents the number of members who rented the two types of equipment.

Step-by-Step Solution

Answer

18 members rented skis, and 10 members rented snowboards.

Step 1: Define the Variables

Let's define the variables for the problem. Let \( x \) represent the number of members who rented skis, and \( y \) represent the number of members who rented snowboards.

Step 2: Establish the First Equation

Since all 28 members rented either skis or snowboards, we can write the equation \( x + y = 28 \). This represents the total number of members who rented equipment.

Step 3: Establish the Second Equation

Consider the total cost of rental equipment. Since skis cost \( \\(16 \) each and snowboards cost \( \\)19 \) each, and the total spent was \( \$478 \), we set up the equation \( 16x + 19y = 478 \).

Step 4: Formulate the System of Equations

From the first two steps, we now have a system of equations: \[ \begin{align*} x + y &= 28 \ 16x + 19y &= 478 \end{align*} \]

Step 5: Solve the System of Equations Using Substitution

Solve the first equation for \( y \): \( y = 28 - x \). Substitute this expression into the second equation: \( 16x + 19(28-x) = 478 \). This simplifies to \( 16x + 532 - 19x = 478 \).

Step 6: Simplify and Solve for x

Rearrange and combine like terms in the equation: \(-3x + 532 = 478 \). Subtract \( 532 \) from both sides: \(-3x = 478 - 532 \), which results in \(-3x = -54 \). Divide by -3, \( x = 18 \).

Step 7: Solve for y

Now substitute \( x = 18 \) back into the first equation: \( x + y = 28 \). Therefore, \( 18 + y = 28 \), which means \( y = 28 - 18 = 10 \).

Step 8: Verify the Solution

Verify that the solution \( (x, y) = (18, 10) \) satisfies the original equations. Check: \( 16(18) + 19(10) = 288 + 190 = 478 \), which matches the given total rental cost.

Key Concepts

Linear Equations

Linear equations are a type of algebraic equation where each term is either a constant or the product of a constant and a single variable. They often appear in the form of a line when graphed on a coordinate plane and are used to model relationships between quantities that have a constant rate of change. In the context of the exercise, we stated the linear equation:

• \( x + y = 28 \)
• \( 16x + 19y = 478 \)

These represent the total number of club members renting equipment and the total cost of rentals, respectively. Linear equations are fundamental in finding relationships in data and solving real-world problems where quantities vary linearly.

Substitution Method

The substitution method is a technique used to solve systems of equations by solving one of the equations for one variable in terms of the others, and then substituting this expression into the other equations. This allows for the elimination of one of the variables, making it possible to solve for the remaining variables. For our exercise, we solved for \( y \) in the first equation, resulting in:

• \( y = 28 - x \)

We then substituted this into the second equation, thereby simplifying and reducing the number of unknowns. The substitution method is particularly useful when one equation is easy to solve for one of the variables, leading to a straightforward path to the solution.

Algebraic Variables

In algebra, variables serve as symbols or placeholders that represent numbers. They allow us to create expressions and equations that can represent various relationships or quantities that may not be initially defined. In our problem, we used:

• \( x \) for the number of ski rentals
• \( y \) for the number of snowboard rentals

These algebraic variables enable us to set up equations that model real-life situations, like determining how many members rented each type of equipment given the constraints provided. Understanding variables gives power to describe and solve complex problems concisely and systematically.

Problem-Solving in Algebra

Problem-solving in algebra involves using mathematical concepts and techniques to find unknown values and relationships. It often includes:

• Defining what is known and unknown
• Setting up equations
• Solving equations
• Verifying solutions

In our exercise, we took the scenario of renting ski equipment and translated it into mathematical equations. We then used algebraic techniques, like substitution, to find the values of \( x \) and \( y \). Finally, we verified our solutions to ensure they met the original conditions, encapsulating the problem-solving process in algebra. This approach provides a methodical way to tackle issues, ensuring reliable and accurate outcomes.