Solving The Equation 11(n-1) + 35 = 3n For N

JU07/08/2025 08, 2025 by THE IDEN 45 views

Solving for n in the Equation 11(n-1) + 35 = 3n: A Comprehensive Guide

Introduction

In the realm of mathematics, solving equations is a fundamental skill. Algebraic equations, in particular, form the bedrock of many mathematical concepts and applications. One common type of equation encountered is a linear equation, where the highest power of the variable is 1. In this article, we will delve into the step-by-step process of solving a specific linear equation: 11(n-1) + 35 = 3n. This equation involves a single variable, n, and our goal is to isolate n on one side of the equation to determine its value. Mastering the technique of solving such equations is crucial for success in algebra and beyond. The process involves applying the principles of algebraic manipulation, including the distributive property, combining like terms, and using inverse operations to maintain the equality. This comprehensive guide will walk you through each step, ensuring a clear understanding of the underlying concepts and techniques. By the end of this article, you will be equipped with the knowledge and confidence to tackle similar equations with ease. This detailed explanation not only focuses on the solution but also emphasizes the importance of understanding the logic behind each step, which is crucial for building a strong foundation in algebra. So, let's embark on this journey of solving for n and unravel the mystery behind this equation.

Understanding the Equation: 11(n-1) + 35 = 3n

Before diving into the solution, it's essential to understand the anatomy of the equation 11(n-1) + 35 = 3n. This equation is a linear equation in one variable, n. A linear equation is characterized by the fact that the variable appears only to the first power. The equation consists of two sides: the left-hand side (LHS) and the right-hand side (RHS), separated by an equals sign (=). The equals sign signifies that the expressions on both sides of the equation have the same value. On the LHS, we have the expression 11(n-1) + 35, which involves a combination of multiplication, subtraction, and addition. The term 11(n-1) indicates that the quantity (n-1) is multiplied by 11. The distributive property will be crucial in simplifying this term. The constant 35 is added to the result of this multiplication. On the RHS, we have the simple expression 3n, which represents 3 times the variable n. Our objective is to find the value of n that makes the LHS equal to the RHS. This is achieved by performing a series of algebraic operations that maintain the equality of the equation. Understanding the structure of the equation and the operations involved is the first step towards solving it successfully. This initial analysis sets the stage for a systematic approach, ensuring that each step taken is logically sound and contributes to the final solution. By breaking down the equation into its components, we can appreciate the relationships between the terms and the operations that need to be performed. This understanding forms the basis for the subsequent steps in solving for n.

Step-by-Step Solution

1. Applying the Distributive Property

The first step in solving the equation 11(n-1) + 35 = 3n is to simplify the left-hand side (LHS) by applying the distributive property. The distributive property states that a(b + c) = ab + ac. In our equation, we have 11(n-1), which can be expanded using the distributive property. Multiplying 11 by both n and -1, we get: 11 * n - 11 * 1 = 11n - 11. Now, substitute this back into the original equation: 11n - 11 + 35 = 3n. This step is crucial because it removes the parentheses, making it easier to combine like terms in the next step. The distributive property is a fundamental concept in algebra, and its correct application is essential for simplifying expressions and solving equations. By distributing the 11, we have effectively transformed the equation into a more manageable form. This process ensures that each term within the parentheses is properly accounted for, setting the stage for further simplification and the eventual isolation of the variable n. The clear and precise application of the distributive property is a hallmark of sound algebraic technique.

2. Combining Like Terms

After applying the distributive property, our equation is now 11n - 11 + 35 = 3n. The next step is to combine like terms on the left-hand side (LHS). Like terms are terms that have the same variable raised to the same power. In this case, -11 and +35 are constants, which are like terms. Adding -11 and 35, we get: -11 + 35 = 24. So, the equation becomes: 11n + 24 = 3n. Combining like terms simplifies the equation further, making it easier to isolate the variable n. This step consolidates the constant terms on the LHS, reducing the number of terms and making the equation more concise. The ability to identify and combine like terms is a fundamental skill in algebra, as it streamlines the process of solving equations. By combining the constants, we have taken a significant step towards isolating the variable term on the LHS. This simplification is crucial for the subsequent steps, as it brings us closer to the ultimate goal of determining the value of n. The accuracy in combining like terms is paramount, as any error in this step will propagate through the rest of the solution.

3. Isolating the Variable Term

Now that our equation is 11n + 24 = 3n, the next goal is to isolate the variable term. To do this, we need to get all the terms involving n on one side of the equation. A common strategy is to subtract the smaller variable term from both sides. In this case, we will subtract 3n from both sides of the equation: 11n - 3n + 24 = 3n - 3n. This simplifies to: 8n + 24 = 0. Subtracting 3n from both sides ensures that the equation remains balanced, as we are performing the same operation on both sides. This step is crucial for isolating the variable term and moving closer to solving for n. By subtracting the smaller variable term, we have effectively reduced the number of terms involving n, making the equation simpler to solve. The strategic decision of which term to subtract is important for efficient problem-solving. In this case, subtracting 3n results in a positive coefficient for the n term on the LHS, which is generally preferred. This step demonstrates the importance of algebraic manipulation in simplifying equations and preparing them for the final solution.

4. Isolating the Variable

After isolating the variable term, our equation is now 8n + 24 = 0. To isolate the variable n, we need to get it by itself on one side of the equation. First, we subtract 24 from both sides: 8n + 24 - 24 = 0 - 24. This simplifies to: 8n = -24. Now, to solve for n, we divide both sides by 8: (8n) / 8 = (-24) / 8. This gives us: n = -3. Dividing both sides by the coefficient of n is the final step in isolating the variable. This operation effectively undoes the multiplication, leaving n by itself on one side of the equation. The result, n = -3, is the solution to the original equation. This value of n satisfies the equation, meaning that when -3 is substituted for n in the original equation, both sides will be equal. The systematic approach of isolating the variable, by performing inverse operations, is a fundamental technique in algebra. This step demonstrates the power of algebraic manipulation in solving equations and finding the value of the unknown variable.

Verifying the Solution

To ensure that our solution, n = -3, is correct, we need to substitute it back into the original equation: 11(n-1) + 35 = 3n. Substituting n = -3, we get: 11(-3 - 1) + 35 = 3(-3). Simplify the left-hand side (LHS): 11(-4) + 35 = -44 + 35 = -9. Simplify the right-hand side (RHS): 3(-3) = -9. Since the LHS equals the RHS (-9 = -9), our solution is correct. Verifying the solution is a crucial step in the problem-solving process. It provides a check on the accuracy of our calculations and ensures that the value we have found for n indeed satisfies the original equation. By substituting the solution back into the equation, we can confirm that both sides are equal, which validates our answer. This step is particularly important in algebra, where errors can easily occur during the manipulation of equations. The verification process reinforces the understanding of the equation and the solution, building confidence in the result. It also highlights the importance of precision and accuracy in algebraic problem-solving. This final check ensures that we have not only found a solution but also a correct one.

Conclusion

In this comprehensive guide, we have successfully solved the equation 11(n-1) + 35 = 3n for n. We followed a step-by-step approach, starting with understanding the equation, applying the distributive property, combining like terms, isolating the variable term, and finally, isolating the variable to find the solution, n = -3. We then verified our solution by substituting it back into the original equation, confirming its accuracy. Solving algebraic equations is a fundamental skill in mathematics, and mastering this skill opens doors to more advanced concepts and applications. The techniques we have used in this article, such as the distributive property, combining like terms, and using inverse operations, are widely applicable in various mathematical contexts. By understanding the logic behind each step and practicing these techniques, you can confidently tackle similar equations and build a strong foundation in algebra. This process not only provides the solution to a specific problem but also enhances problem-solving skills and analytical thinking. The ability to solve equations is essential for success in mathematics and in many fields that rely on mathematical reasoning. So, continue to practice and explore different types of equations to further strengthen your algebraic skills.