Problem 25 - 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. How many members rented skis and how many rented snowboards?

Step-by-Step Solution

Answer

18 members rented skis, and 10 rented snowboards.

Step 1: Define Variables

Let \( x \) be the number of members who rented skis, and \( y \) be the number of members who rented snowboards. We know that the total number of members is 28, so we can write the equation \( x + y = 28 \).

Step 2: Set Up Cost Equation

Each ski rental costs \\(16, and each snowboard rental costs \\)19. The total rental cost is \$478. Therefore, the cost equation is \( 16x + 19y = 478 \).

Step 3: Solve the System of Equations

We have the system of equations: 1. \( x + y = 28 \)2. \( 16x + 19y = 478 \)First, solve equation (1) for \( y \): \( y = 28 - x \).Substitute \( y \) in equation (2): \( 16x + 19(28 - x) = 478 \)

Step 4: Distribute and Simplify

Distribute the \( 19 \) in the equation: \( 16x + 532 - 19x = 478 \).Combine like terms: \( -3x + 532 = 478 \).

Step 5: Solve for x

Subtract 532 from both sides of the equation to isolate the term with \( x \):\( -3x = 478 - 532 \).Simplify the right side: \( -3x = -54 \).Divide both sides by \(-3\) to solve for \( x \):\( x = 18 \).

Step 6: Find y Using x

Substitute \( x = 18 \) back into the equation \( y = 28 - x \):\( y = 28 - 18 = 10 \).

Step 7: Conclusion

Therefore, 18 members rented skis, and 10 members rented snowboards.

Key Concepts

Algebraic Expressions

In order to solve problems involving systems of equations, we often use algebraic expressions. An algebraic expression is a combination of numbers, variables, and operators (like addition and subtraction). These expressions are used to represent real-world situations in mathematical form. They allow us to manipulate and solve equations to find unknown values.

In this ski club problem, we use algebraic expressions to translate the scenario's costs and quantities into mathematical equations. We define variables where:

• \(x\) represents the number of members who rented skis.
• \(y\) represents the number of members who rented snowboards.

The equations that we form from these expressions help us to respect the conditions given in the problem. This includes the total number of members and the total rental costs. These expressions set the ground for forming equations that can then be solved using various algebraic techniques.

Variable Substitution

Variable substitution is a crucial technique used in solving systems of equations. It involves replacing one variable with an equivalent expression derived from another equation, making it simpler to solve. This method is particularly useful when solving linear equations, which often involve two or more variables.

In our ski club example, we start with two equations:

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

To solve these equations, we first solve one equation for one variable. Here, we solved \(x + y = 28\) for \(y\), yielding \(y = 28 - x\). This expression is then substituted into the second equation \(16x + 19y = 478\). By replacing \(y\) with \(28 - x\), the equation becomes a one-variable equation, \(16x + 19(28 - x) = 478\), which is easier to solve.

This technique is a powerful tool in simplifying complex problems and finding solutions in steps, making it more manageable to determine unknown values like how many members rented skis or snowboards.

Linear Equations

Linear equations are fundamental in algebra. They are equations where the highest power of the variable is one, forming a straight line when graphed. They can describe relationships with constant rates of change, and are often part of solving real-life problems like budgeting, rates, or memberships.

In problems with scenarios like the ski rental case, linear equations capture the constraints given in the situation. For instance, the equation \(x + y = 28\) is a linear equation representing the total number of ski club members. Another linear equation, \(16x + 19y = 478\), models the total rental cost.

The solution to a system of linear equations involves finding values for the variables that satisfy all equations involved. In this case, solving these linear equations allowed us to determine exactly how many club members rented skis and snowboards. Understanding linear equations helps to bridge the gap between realistic scenarios and their mathematical representations, making them a pivotal component of algebra.