Problem 61 - Prealgebra

Question

A stationary store has staplers on sale for \(1.50 apiece. A business purchases an unknown number of these and the total cost of their purchase is \)36.00. How many were purchased?

Step-by-Step Solution

Answer

The business purchased 24 staplers.

Step 1: Define the variables

Let the number of staplers purchased be represented by \( x \). The cost of each stapler is given as \( 1.50 \). We need to find the value of \( x \).

Step 2: Set up the equation

The total cost of the staplers purchased is given as \( 36.00 \). The equation representing the total cost is \( 1.50x = 36.00 \), where \( x \) is the number of staplers.

Step 3: Solve for x

To find \( x \), divide both sides of the equation \( 1.50x = 36.00 \) by \( 1.50 \): \[ x = \frac{36.00}{1.50} \] Calculating this gives \( x = 24 \).

Step 4: Verify the solution

Multiply the number of staplers \( x = 24 \) by the cost per stapler \( 1.50 \), and ensure it equals the total cost:\[ 1.50 \times 24 = 36.00 \] The calculation is correct.

Key Concepts

Understanding Variable Definition in Solving Equations

When solving equations, the first step often involves defining a variable. A variable is essentially a placeholder for an unknown value that you need to find. In algebra, variables are typically represented by letter symbols such as \( x \), \( y \), or \( z \). In this exercise, we define the variable \( x \) to represent the unknown number of staplers being purchased.

Defining a variable helps us translate a real-world situation into a mathematical one. This step is crucial as it allows us to form an equation that we can solve to find the unknown value. However, choosing the right variable and clearly defining what it represents is important. In our example, stating "Let \( x \) be the number of staplers purchased" gives us a starting point for our problem-solving process. It turns the problem into an equation: \( 1.50x = 36.00 \).

The concept of defining variables is foundational in mathematics and many scientific fields, as it allows for precise communication and solution discovery.

Exploring Linear Equations

Linear equations are equations of the first degree, meaning they involve terms where the variable is raised to the power of one. The general form of a linear equation is \( ax + b = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. In the exercise at hand, we have the linear equation \( 1.50x = 36.00 \).

Linear equations have several characteristics that make them relatively simple to solve:

• They graph as straight lines on the coordinate plane.
• They have a single solution for the variable. If an equation has two variables, it solves for a line that represents numerous solutions.
• They involve operations that are reversible, such as addition and multiplication.

To solve a linear equation like \( 1.50x = 36.00 \), we perform basic operations to isolate the variable. We divided both sides by \( 1.50 \) to find \( x = 24 \). This demonstrates the straightforward nature of solving linear equations. By understanding these characteristics, students can approach such problems with greater confidence.

Basic Arithmetic in Equation Solving

Basic arithmetic lies at the heart of solving equations and involves fundamental operations such as addition, subtraction, multiplication, and division. In our example, we used multiplication and division to work through the equation \( 1.50x = 36.00 \).

Here's how basic arithmetic helps in solving the equation:

• **Multiplication:** We initially dealt with the multiplication of \( 1.50 \) with \( x \), which gave us the total cost of the staplers.
• **Division:** To solve for \( x \), we needed to divide both sides by \( 1.50 \), a step that simplified the equation to just the variable \( x \).

These operations are powerful tools in manipulating and rearranging equations. Performing arithmetic correctly ensures that a solution is accurate.

In more complex problems, a strong grasp of basic arithmetic is essential as it forms the basis of tackling any algebraic operation. Practice with these skills enhances fluency in solving not just linear equations but more challenging mathematical tasks as well.