Problem 10 - Algebra 2

Question

Each week Carmen earns \(\$ 15\) plus \(\$ 0.17\) for every pamphlet that she delivers. Write an equation that can be used to find how much Carmen earns each week. How much will she earn the week she delivers 300 pamphlets?

Step-by-Step Solution

Answer

Carmen earns $66 for delivering 300 pamphlets.

Step 1: Understand the Problem

Carmen earns a fixed amount of $15 each week, plus an additional $0.17 for every pamphlet delivered. We need to create an equation to calculate her weekly earnings and determine her earnings if she delivers 300 pamphlets.

Step 2: Define Variables

Let \( x \) represent the number of pamphlets delivered in a week. Carmen's earnings each week can be represented as \( y \). Thus, \( x = 300 \) for the given problem.

Step 3: Formulate the Equation

Using the information provided, the earnings in a week can be expressed using the equation: \[ y = 15 + 0.17x \].

Step 4: Substitute and Solve for a Specific Case

To find out how much Carmen earns when delivering 300 pamphlets, substitute \( x = 300 \) into the equation: \[ y = 15 + 0.17(300) \]. Calculate this to find her total earnings.

Step 5: Calculate the Earnings

Compute \( 0.17 \times 300 = 51 \). Substitute this back into the equation to get: \[ y = 15 + 51 = 66 \].

Step 6: Conclusion

Therefore, Carmen earns $66 the week she delivers 300 pamphlets.

Key Concepts

Algebraic Expressions

An algebraic expression is a combination of variables, numbers, and arithmetic operations. It forms the basis of writing linear equations, which include terminology like terms and coefficients. In the exercise above, the expression "\(15 + 0.17x\)" is the key algebraic expression representing Carmen's weekly earnings. The fixed part of the earnings is represented by the constant \(15\). The variable part is \(0.17x\), where \(x\) indicates the number of pamphlets delivered in a week.

Understanding how to form and interpret algebraic expressions is crucial. It enables us to model real-life financial situations, like Carmen's earnings. A good grasp of these expressions helps in visualizing the incremental nature of earnings based on output, making such equations useful in various real-world applications, from budgeting to resource management.

Variable Substitution

Variable substitution involves replacing variables in an algebraic equation with specific values in order to solve the equation. For Carmen's earnings, we use the variable \(x\) to represent the number of pamphlets delivered. In this instance, knowing \(x = 300\) allows us to substitute directly into the equation \(y = 15 + 0.17x\) to find out how much she earns.

This process is crucial when solving problems with specific scenarios, like determining earnings for a concrete number of pamphlets. By substituting the value of 300 for \(x\), the calculation becomes straightforward: \(y = 15 + 0.17(300)\). This substitution is a foundational skill in algebra, as it transforms variable-dependent expressions into solvable numerical forms, aiding in direct computation or evaluation.

Arithmetic Operations

Arithmetic operations are the fundamental mathematical processes that include addition, subtraction, multiplication, and division. They are essential in solving algebraic equations. In our exercise, arithmetic operations are used to calculate Carmen's total earnings for a specific number of pamphlets.

To determine Carmen's earnings of delivering 300 pamphlets, we multiply \(0.17\) by \(300\), resulting in \(51\). Arithmetic multiplication here shows how incremental earnings are calculated per pamphlet delivered. Following the multiplication, we then add this figure to the fixed earning of \(15\), using addition to arrive at \(66\), the total earnings. Mastery of these operations allows one to handle both simple and complex calculations, translating abstract mathematical concepts into tangible results.