Problem 36 - Prealgebra

Question

Solve each equation by finding a number to replace \(n\) that will make the equation a true statement. $$3 \cdot n=27$$

Step-by-Step Solution

Answer

The value of \(n\) is 9.

Step 1: Identify the Problem

The equation given is \(3 \cdot n = 27\). Our goal is to find the value of \(n\) that will satisfy this equation.

Step 2: Understand the Equation

This equation is a multiplication equation where \(n\) is being multiplied by 3 to result in 27. To find the value of \(n\), we need to isolate \(n\) on one side of the equation.

Step 3: Solve for \(n\)

To isolate \(n\), divide both sides of the equation by 3. This gives us: \(n = \frac{27}{3}\).

Step 4: Perform the Division

Divide 27 by 3: \(n = 9\).

Step 5: Verify the Solution

Substitute \(n = 9\) back into the original equation to verify: \(3 \cdot 9 = 27\). Both sides of the equation equal 27, confirming that our solution is correct.

Key Concepts

Prealgebra

Prealgebra serves as the foundation of so many mathematical concepts. It introduces simple operations that form the backbone of more complex mathematics. One of its key principles is understanding how numbers interact through different operations, like addition, subtraction, multiplication, and division.

When faced with an equation like \(3 \cdot n = 27\), we aim to uncover what number the variable \(n\) represents. Prealgebra helps us see this by relying on our understanding of basic arithmetic operations. We're looking for a number that fits perfectly into the equation when multiplied by 3, producing a product of 27.

• Multiplication tells us how many times we add a number to itself. For example, \(3 \cdot 4 = 12\) means we've added 4 three times (\(4+4+4\)).
• Division is often the key to undoing multiplication, allowing us to find unknown variables like \(n\).

By building on these prealgebra concepts, we pave the way to deeper mathematical learning and understanding.

Basic Algebra Concepts

Algebra introduces the concept of variables, which are simply symbols used to represent unknown numbers. This marks a shift from working only with numbers to working with equations. An equation like \(3 \cdot n = 27\) is known as a linear equation, which can be solved to find the value of \(n\).

Understanding this fundamental concept in algebra involves grasping a few crucial points:

• Variables: Symbols like \(n\) stand in for numbers we are trying to find.
• Equations: Statements that show two expressions are equal, such as \(3 \cdot n = 27\).
• Operations: The actions we take to maintain equilibrium in an equation. When one side changes in an operation like division, the other side must undergo the same change.

Algebra uses these concepts to resolve questions about unknown numbers or quantities. By practicing basic algebra, such as solving for \(n\), students develop skills to interpret and solve more complex issues. It builds confidence in attacking persistent problems through logical reasoning.

Isolation of Variables

Isolating variables is a pivotal skill in solving equations because it allows us to easily identify the values we are looking to find. In our exercise, \(3 \cdot n = 27\), the goal is to isolate \(n\). This means getting \(n\) by itself on one side of the equation.

To achieve this, we can use the inverse operation to 'undo' what the multiplication has done. Since \(n\) has been multiplied by 3, we need to divide both sides by 3 to isolate \(n\). Here’s how this process works:

• Identify the operation involved with the variable—in this case, multiplication by 3.
• Apply the inverse operation (division) to both sides: \(\frac{3 \cdot n}{3} = \frac{27}{3}\).
• This simplifies to \(n = 9\).

By isolating variables, students can solve for unknowns systematically, verifying their solutions by substituting back into the original equation. This technique simplifies problem-solving and is a core method employed through all levels of algebra.