Solving Linear Equations A Step-by-Step Guide For 2.25 - 11j - 7.75 + 1.5j = 0.5j - 1
Introduction
In the realm of mathematics, linear equations serve as the bedrock for countless applications, from simple problem-solving to complex modeling in various scientific and engineering disciplines. Mastering the art of solving linear equations is therefore a fundamental skill for anyone venturing into these fields. This article provides a comprehensive guide on how to solve the linear equation $2.25 - 11j - 7.75 + 1.5j = 0.5j - 1$. We will walk you through each step of the process, ensuring clarity and understanding along the way. This detailed explanation will not only help you solve this specific equation but also equip you with the skills to tackle similar problems with confidence. Our main focus is to break down the equation, isolate the variable 'j', and arrive at the solution using systematic algebraic manipulations. We will also highlight common pitfalls and strategies to avoid them, making this a valuable resource for both beginners and those looking to refresh their algebra skills. The journey of solving this equation will involve combining like terms, simplifying expressions, and applying the principles of equality to maintain balance on both sides of the equation. Through this meticulous approach, we aim to demystify the process and demonstrate the elegance and precision of mathematical problem-solving. So, let's embark on this mathematical journey and unravel the solution to the equation $2.25 - 11j - 7.75 + 1.5j = 0.5j - 1$. This will not only enhance your understanding of linear equations but also bolster your confidence in tackling more complex mathematical challenges.
Step 1: Combine Like Terms on the Left-Hand Side
The first crucial step in solving any linear equation involves simplifying both sides of the equation by combining like terms. In our equation, $2.25 - 11j - 7.75 + 1.5j = 0.5j - 1$, the left-hand side (LHS) contains both constant terms (2.25 and -7.75) and terms with the variable 'j' (-11j and 1.5j). To simplify, we group these like terms together. Combining the constant terms, we have $2.25 - 7.75$, which equals $-5.5$. Next, we combine the terms with 'j': $-11j + 1.5j$. This results in $-9.5j$. Therefore, by combining like terms on the LHS, the equation transforms into $-5.5 - 9.5j = 0.5j - 1$. This simplification is a critical step because it reduces the complexity of the equation and makes it easier to isolate the variable 'j'. It's important to pay close attention to the signs (positive or negative) of each term during this process to avoid errors. By accurately combining like terms, we set the stage for the subsequent steps in solving the equation. This process not only simplifies the equation but also highlights the importance of organization and attention to detail in mathematical problem-solving. Remember, a clear and methodical approach is key to success in mathematics, and combining like terms is a fundamental technique in achieving this clarity.
Step 2: Move Terms with 'j' to One Side
After simplifying the left-hand side of the equation, the next logical step is to gather all terms containing the variable 'j' on one side of the equation. This is achieved by performing the same operation on both sides, ensuring the equation remains balanced. In our case, the equation is $-5.5 - 9.5j = 0.5j - 1$. To move the term $0.5j$ from the right-hand side (RHS) to the LHS, we subtract $0.5j$ from both sides of the equation. This yields: $-5.5 - 9.5j - 0.5j = 0.5j - 1 - 0.5j$. On the RHS, $0.5j - 0.5j$ cancels out, leaving us with $-1$. On the LHS, we combine the 'j' terms: $-9.5j - 0.5j$, which equals $-10j$. Thus, the equation now becomes $-5.5 - 10j = -1$. This step is crucial because it isolates the variable 'j' terms, bringing us closer to solving for 'j'. The principle of performing the same operation on both sides is a fundamental concept in algebra, ensuring that the equation remains balanced and the solution remains valid. By carefully moving the 'j' terms to one side, we simplify the equation further and prepare it for the next stage of the solution process. This step demonstrates the power of algebraic manipulation in transforming equations into more manageable forms.
Step 3: Move Constant Terms to the Other Side
With the 'j' terms now isolated on the left-hand side, the next step involves moving all the constant terms to the right-hand side of the equation. This is achieved by adding or subtracting constants from both sides, maintaining the balance of the equation. Our equation currently stands as $-5.5 - 10j = -1$. To move the constant term $-5.5$ from the LHS to the RHS, we add $5.5$ to both sides of the equation. This gives us: $-5.5 - 10j + 5.5 = -1 + 5.5$. On the LHS, $-5.5 + 5.5$ cancels out, leaving us with $-10j$. On the RHS, $-1 + 5.5$ equals $4.5$. Therefore, the equation now simplifies to $-10j = 4.5$. This step is critical because it isolates the term containing 'j' on one side and the constant term on the other, paving the way for the final step of solving for 'j'. By carefully manipulating the equation and keeping it balanced, we are progressing towards the solution in a systematic manner. This process reinforces the importance of understanding the properties of equality in algebra. By isolating the variable term, we are one step closer to uncovering the value of 'j' that satisfies the original equation.
Step 4: Solve for 'j'
Having isolated the term with 'j' on one side and the constant term on the other, the final step is to solve for 'j'. Our equation is currently $-10j = 4.5$. To isolate 'j', we need to divide both sides of the equation by the coefficient of 'j', which is $-10$. Dividing both sides by $-10$ gives us: $ rac{-10j}{-10} = rac{4.5}{-10}$. On the LHS, $-10j$ divided by $-10$ simplifies to 'j'. On the RHS, $4.5$ divided by $-10$ equals $-0.45$. Therefore, the solution to the equation is $j = -0.45$. This step represents the culmination of our efforts, where we successfully determine the value of 'j' that satisfies the original equation. By applying the principles of algebra and performing each step with precision, we have navigated through the equation and arrived at the solution. This process highlights the power of algebraic manipulation in solving linear equations. The solution $j = -0.45$ is the value that, when substituted back into the original equation, will make the equation true. This final step underscores the importance of accuracy and attention to detail in mathematical problem-solving. With 'j' now determined, we have successfully solved the linear equation.
Conclusion
In conclusion, we have successfully navigated the process of solving the linear equation $2.25 - 11j - 7.75 + 1.5j = 0.5j - 1$. By systematically applying algebraic principles, we arrived at the solution $j = -0.45$. This journey involved several key steps: combining like terms, moving terms with 'j' to one side, moving constant terms to the other side, and finally, isolating 'j' to find its value. Each step was crucial in simplifying the equation and bringing us closer to the solution. This exercise not only demonstrates the mechanics of solving linear equations but also highlights the importance of a methodical and organized approach to mathematical problem-solving. Understanding these principles and techniques is fundamental for success in algebra and beyond. The ability to manipulate equations, isolate variables, and solve for unknowns is a skill that extends far beyond the classroom, finding applications in various fields of science, engineering, economics, and more. By mastering these skills, individuals can confidently tackle complex problems and make informed decisions. The solution $j = -0.45$ represents a specific value that satisfies the given equation, and this process can be applied to a wide range of linear equations. With practice and a solid understanding of algebraic principles, anyone can become proficient in solving these types of problems. This comprehensive guide serves as a valuable resource for students, educators, and anyone looking to enhance their mathematical skills. The journey of solving this equation reinforces the power and elegance of mathematics as a tool for problem-solving and critical thinking.
