Solving For B In 3a + 6b = 4 A Step By Step Guide

In the realm of algebra, the ability to solve for a specific variable within an equation is a fundamental skill. This skill is crucial for various applications, ranging from simplifying complex formulas in physics and engineering to optimizing financial models in economics. In this comprehensive guide, we will delve into the process of solving for the variable b in the equation 3a + 6b = 4. We will break down the steps involved, providing a clear and concise explanation for each, ensuring that you grasp the underlying principles and can confidently apply them to similar problems. By the end of this guide, you will not only be able to solve this particular equation but also possess the knowledge and skills to tackle a wide range of algebraic manipulations. Let's embark on this journey of algebraic discovery and unlock the secrets of variable isolation!

Understanding the Basics of Algebraic Manipulation

Before we dive into the specifics of solving for b, it's essential to establish a solid foundation in the basic principles of algebraic manipulation. At its core, solving for a variable involves isolating that variable on one side of the equation, leaving all other terms and constants on the other side. This isolation is achieved by performing a series of operations on both sides of the equation, ensuring that the equality is maintained. The key principle here is the golden rule of algebra: whatever operation you perform on one side of the equation, you must perform the same operation on the other side. This rule guarantees that the equation remains balanced and the solution remains valid. For instance, if you subtract a term from one side, you must subtract the same term from the other side. Similarly, if you multiply one side by a constant, you must multiply the other side by the same constant. These operations can include addition, subtraction, multiplication, and division, as well as more advanced techniques like factoring and exponentiation. The choice of operation depends on the specific equation and the variable you are trying to isolate. The goal is to strategically apply these operations to gradually peel away the terms surrounding the target variable until it stands alone. Understanding this fundamental concept is crucial for mastering algebraic problem-solving and will serve as the bedrock for our exploration of solving for b in the given equation.

Step-by-Step Solution for Isolating 'b'

Now, let's tackle the equation 3a + 6b = 4 and systematically isolate b. Our primary objective is to get b by itself on one side of the equation. To achieve this, we'll follow a series of algebraic manipulations, ensuring we maintain the equality throughout the process. The first step involves addressing the term 3a, which is added to 6b. To eliminate this term from the left side of the equation, we apply the inverse operation: subtraction. We subtract 3a from both sides of the equation. This gives us: 3a + 6b - 3a = 4 - 3a. On the left side, the 3a and -3a terms cancel each other out, leaving us with 6b = 4 - 3a. Now, b is closer to being isolated, but it's still being multiplied by 6. To undo this multiplication, we perform the inverse operation: division. We divide both sides of the equation by 6, resulting in (6b)/6 = (4 - 3a)/6. On the left side, the 6 in the numerator and the 6 in the denominator cancel each other out, leaving us with b = (4 - 3a)/6. This is our solution for b. We have successfully isolated b on one side of the equation, expressing it in terms of a. This step-by-step process demonstrates the power of algebraic manipulation in solving for specific variables. Each step is carefully chosen to move us closer to the goal of isolation, while adhering to the fundamental principle of maintaining equality.

Simplifying the Solution

While we have successfully solved for b, the solution b = (4 - 3a)/6 can be further simplified. Simplification in algebra is crucial for expressing solutions in their most concise and easily interpretable form. It often involves identifying common factors and reducing fractions to their lowest terms. In our case, we can examine the fraction (4 - 3a)/6 to see if any simplification is possible. We notice that both the numerator and the denominator have factors that might be simplified. The numerator, 4 - 3a, doesn't have any immediately obvious common factors. However, we can consider whether the entire numerator shares a common factor with the denominator, 6. The factors of 6 are 1, 2, 3, and 6. We can try to see if we can divide both terms in the numerator (4 and -3a) by any of these factors. While 4 is divisible by 2, -3a is not. Similarly, neither 4 nor -3a is divisible by 3. However, we can rewrite the fraction by distributing the division: b = 4/6 - (3a)/6. Now, we can simplify each fraction separately. 4/6 can be reduced by dividing both the numerator and denominator by their greatest common divisor, which is 2. This gives us 2/3. Similarly, (3a)/6 can be simplified by dividing both the numerator and denominator by 3, resulting in a/2. Therefore, our simplified solution becomes b = 2/3 - a/2. This simplified form is equivalent to the original solution, b = (4 - 3a)/6, but it is often preferred as it is more concise and easier to work with in further calculations or analysis. The process of simplification highlights the importance of not only finding a solution but also expressing it in its most elegant and manageable form.

Verifying the Solution

After solving for a variable, it is always a prudent practice to verify the solution. This step is crucial to ensure that no errors were made during the algebraic manipulations and that the obtained solution is indeed correct. Verification provides confidence in the accuracy of the result and helps prevent the propagation of errors in subsequent calculations. The most common method for verifying a solution is to substitute the obtained value back into the original equation. If the equation holds true after the substitution, then the solution is verified. In our case, we have solved for b and obtained b = (4 - 3a)/6. To verify this solution, we substitute it back into the original equation, 3a + 6b = 4. Replacing b with (4 - 3a)/6, we get: 3a + 6*((4 - 3a)/6) = 4. Now, we simplify the equation. The 6 in the numerator and the 6 in the denominator cancel each other out, leaving us with: 3a + (4 - 3a) = 4. Next, we combine like terms. 3a and -3a cancel each other out, leaving us with: 4 = 4. This is a true statement, which confirms that our solution for b is correct. The verification process not only confirms the correctness of the solution but also reinforces the understanding of the algebraic manipulations performed. It provides a valuable check and balance in the problem-solving process, ensuring accuracy and building confidence.

Common Mistakes to Avoid

When solving for variables in algebraic equations, there are several common mistakes that students often make. Being aware of these pitfalls can help you avoid them and improve your accuracy in problem-solving. One frequent mistake is failing to apply the same operation to both sides of the equation. As we emphasized earlier, the golden rule of algebra dictates that any operation performed on one side must be performed on the other side to maintain equality. For example, if you subtract a term from the left side, you must subtract the same term from the right side. Another common mistake is incorrectly distributing a term. When multiplying a term across a parenthesis, you must multiply it by each term inside the parenthesis. For instance, if you have 2(x + 3), you must multiply 2 by both x and 3, resulting in 2x + 6. Neglecting to do so can lead to incorrect results. Sign errors are also a common source of mistakes. Pay close attention to the signs of the terms, especially when adding or subtracting negative numbers. For example, subtracting a negative number is the same as adding a positive number. Another pitfall is incorrectly simplifying fractions. When simplifying a fraction, you must divide both the numerator and the denominator by their greatest common divisor. Failing to do so can result in a fraction that is not in its simplest form. Finally, not verifying the solution is a mistake that can lead to overlooking errors. As we discussed earlier, verifying the solution by substituting it back into the original equation is a crucial step in ensuring accuracy. By being mindful of these common mistakes and taking steps to avoid them, you can significantly improve your algebraic problem-solving skills.

Conclusion: Mastering the Art of Solving for Variables

In conclusion, solving for a variable in an equation is a fundamental skill in algebra, and mastering this skill opens doors to a wide range of mathematical and scientific applications. In this comprehensive guide, we have meticulously explored the process of solving for b in the equation 3a + 6b = 4. We began by laying the groundwork with the basic principles of algebraic manipulation, emphasizing the importance of maintaining equality by performing the same operations on both sides of the equation. We then proceeded with a step-by-step solution, demonstrating how to isolate b by strategically applying inverse operations. We further refined our solution by simplifying the expression, highlighting the importance of expressing answers in their most concise and manageable form. The crucial step of verifying the solution was also discussed, underscoring the need to ensure accuracy and build confidence in the result. Finally, we addressed common mistakes to avoid, providing valuable insights into potential pitfalls and how to steer clear of them. By diligently following the principles and techniques outlined in this guide, you can confidently tackle a wide array of algebraic equations and master the art of solving for variables. This skill will not only serve you well in your mathematical pursuits but also empower you to solve real-world problems across various disciplines. Remember, practice is key to mastery. The more you practice, the more proficient you will become in algebraic manipulation. So, embrace the challenge, hone your skills, and unlock the power of algebra!