Solving The Equation 4(x-1)-5(3-x)=14x-2(5x-3) A Step-by-Step Guide
Introduction to Linear Equations
In the realm of mathematics, linear equations form the bedrock of many algebraic concepts. These equations, characterized by their simple structure and predictable solutions, are fundamental to understanding more complex mathematical models. A linear equation is essentially an algebraic equation in which each term is either a constant or the product of a constant and a single variable. The variable in a linear equation is raised to the power of one, and the equation, when graphed, represents a straight line. Mastering the art of solving linear equations is crucial for anyone delving into algebra, calculus, or any quantitative field. The beauty of linear equations lies in their straightforward nature; they adhere to well-defined rules and can be solved using a systematic approach. This makes them an ideal starting point for anyone looking to strengthen their mathematical foundation.
Solving linear equations is not just a mathematical exercise; it’s a skill that translates into real-world problem-solving scenarios. From calculating financial investments to determining optimal resource allocation, linear equations provide a framework for modeling and understanding a wide range of phenomena. In this article, we will dissect and solve the linear equation 4(x-1)-5(3-x)=14x-2(5x-3), providing a step-by-step guide that demystifies the process. We will explore the underlying principles of algebraic manipulation, emphasizing the importance of maintaining balance and order when simplifying and solving equations. By the end of this discussion, you will not only have the solution to this specific equation but also a robust understanding of the techniques required to tackle a variety of linear equations with confidence. Understanding these techniques will empower you to approach mathematical challenges with a clear and structured mindset.
Moreover, linear equations serve as a gateway to more advanced mathematical concepts. The principles learned while solving linear equations, such as isolating variables and applying the distributive property, are directly applicable to solving systems of equations, inequalities, and even more complex algebraic expressions. A firm grasp of linear equations equips you with the tools necessary to explore these advanced topics, making your mathematical journey smoother and more rewarding. The journey through mathematics often begins with linear equations, and a solid foundation here can make all the difference in your future mathematical endeavors. So, let’s embark on this journey together, unraveling the intricacies of linear equations and equipping ourselves with the skills to conquer mathematical challenges.
Deconstructing the Equation: 4(x-1)-5(3-x)=14x-2(5x-3)
Our primary objective in mathematics is to solve the linear equation 4(x-1)-5(3-x)=14x-2(5x-3). This equation, while seemingly complex at first glance, is a classic example of a linear equation that can be solved by systematically applying the rules of algebra. The equation involves the variable 'x', and our goal is to isolate 'x' on one side of the equation to determine its value. This process involves several key steps, including distributing coefficients, combining like terms, and applying inverse operations to maintain the equality. Each step is crucial in simplifying the equation and bringing us closer to the solution. Before we dive into the detailed solution, let's break down the equation and identify the key components that we will be working with.
The equation consists of two sides, separated by an equals sign (=). The left-hand side (LHS) is 4(x-1)-5(3-x), and the right-hand side (RHS) is 14x-2(5x-3). Both sides involve expressions with 'x' as the variable, as well as constant terms. The presence of parentheses indicates that we need to apply the distributive property, which is a fundamental principle in algebra. The distributive property states that a(b+c) = ab + ac, and we will use this property to eliminate the parentheses and simplify the equation. By applying this property, we can expand the terms and make the equation easier to manipulate. This is a critical step in simplifying the equation and setting the stage for combining like terms.
Furthermore, the equation involves both addition and subtraction operations, which need to be handled carefully. We must adhere to the order of operations (PEMDAS/BODMAS) to ensure that we simplify the equation correctly. This means that we address parentheses first, then exponents, then multiplication and division (from left to right), and finally, addition and subtraction (from left to right). By following the order of operations, we can avoid errors and maintain the integrity of the equation. Understanding the structure of the equation and the operations involved is essential for developing a strategy to solve it. Now that we have deconstructed the equation, we are ready to embark on the journey of solving it, step by step, until we arrive at the final solution for 'x'.
Step-by-Step Solution
To begin solving the equation 4(x-1)-5(3-x)=14x-2(5x-3), our initial focus is on simplifying both sides of the equation. This involves applying the distributive property to remove the parentheses. The distributive property, a cornerstone of algebra, allows us to multiply a term by each term within the parentheses. On the left-hand side (LHS), we distribute the 4 across (x-1) and the -5 across (3-x). This yields 4 * x - 4 * 1 - 5 * 3 + 5 * x. Simplifying this gives us 4x - 4 - 15 + 5x. Similarly, on the right-hand side (RHS), we distribute the -2 across (5x-3), resulting in 14x - 2 * 5x + 2 * 3. This simplifies to 14x - 10x + 6. By applying the distributive property, we have effectively eliminated the parentheses, making the equation more manageable.
Next, we combine like terms on both sides of the equation. Like terms are terms that have the same variable raised to the same power. On the LHS, we have 4x and 5x as like terms, and -4 and -15 as like terms. Combining these, we get 4x + 5x - 4 - 15, which simplifies to 9x - 19. On the RHS, we have 14x and -10x as like terms, and 6 is a constant term. Combining the like terms on the RHS gives us 14x - 10x + 6, which simplifies to 4x + 6. At this point, our equation looks significantly simpler: 9x - 19 = 4x + 6. This simplification is a crucial step in solving the equation, as it reduces the complexity and brings us closer to isolating the variable 'x'.
Now, our goal is to isolate 'x' on one side of the equation. To achieve this, we will move all terms involving 'x' to one side and all constant terms to the other side. A common strategy is to subtract 4x from both sides to eliminate the 'x' term on the RHS. This gives us 9x - 19 - 4x = 4x + 6 - 4x, which simplifies to 5x - 19 = 6. Next, we add 19 to both sides to isolate the 'x' term further. This gives us 5x - 19 + 19 = 6 + 19, which simplifies to 5x = 25. Finally, we divide both sides by 5 to solve for 'x'. This gives us 5x / 5 = 25 / 5, which simplifies to x = 5. Therefore, the solution to the equation 4(x-1)-5(3-x)=14x-2(5x-3) is x = 5. This step-by-step solution demonstrates the systematic approach to solving linear equations, emphasizing the importance of applying algebraic principles and maintaining balance throughout the process.
Verification of the Solution
After arriving at the solution x = 5 for the equation 4(x-1)-5(3-x)=14x-2(5x-3), it is crucial to verify the solution. Verification is a vital step in the problem-solving process, as it ensures that the calculated solution is correct and satisfies the original equation. By substituting the solution back into the equation, we can confirm whether the left-hand side (LHS) is equal to the right-hand side (RHS). This process not only validates our answer but also reinforces our understanding of the equation and the steps involved in solving it. Let's proceed with the verification process by substituting x = 5 into the original equation.
Substituting x = 5 into the original equation, we get 4(5-1)-5(3-5)=14(5)-2(5*5-3). Now, we simplify both sides of the equation following the order of operations (PEMDAS/BODMAS). On the LHS, we have 4(4)-5(-2). This simplifies to 16 + 10, which equals 26. On the RHS, we have 14(5)-2(25-3). This simplifies to 70 - 2(22), which further simplifies to 70 - 44, which also equals 26. Comparing both sides, we find that the LHS (26) is equal to the RHS (26), which confirms that our solution x = 5 is indeed correct. This verification step is essential, as it provides confidence in our solution and validates the steps we took to solve the equation.
Moreover, the verification process highlights the importance of accuracy and attention to detail in mathematics. Even a small error in the calculation can lead to an incorrect solution. By verifying our solution, we are essentially double-checking our work and ensuring that we have not made any mistakes. This practice is invaluable in developing a strong mathematical foundation and problem-solving skills. The satisfaction of verifying a solution and confirming its correctness is a rewarding experience, as it solidifies our understanding and mastery of the mathematical concepts involved. Therefore, always remember to verify your solutions, as it is the final step in the journey of solving an equation and a testament to your mathematical proficiency.
Common Mistakes and How to Avoid Them
When solving linear equations, especially those with multiple steps like 4(x-1)-5(3-x)=14x-2(5x-3), it's common to encounter certain pitfalls. Recognizing these common mistakes is crucial for improving accuracy and efficiency in problem-solving. One of the most frequent errors is related to the distributive property. Students often forget to distribute the negative sign correctly, leading to incorrect signs within the equation. For example, in the term -5(3-x), the -5 needs to be multiplied by both 3 and -x. A common mistake is to only multiply -5 by 3, resulting in -15 instead of -15 + 5x. To avoid this, always double-check that you have distributed the coefficient correctly, paying close attention to the signs. A careful and methodical approach is essential to prevent this common error.
Another common mistake is combining like terms incorrectly. Like terms must have the same variable raised to the same power. For instance, in the equation we solved, terms with 'x' can only be combined with other terms with 'x', and constant terms can only be combined with other constant terms. A mistake could be made by incorrectly adding a term with 'x' to a constant term, which is not mathematically valid. To avoid this, underline or circle like terms before combining them to ensure that you are only adding or subtracting terms that are compatible. Organization and attention to detail are key in this step. Keeping track of terms and their signs will help prevent errors in combining like terms.
Furthermore, errors can occur when performing inverse operations. When isolating the variable, it's essential to perform the same operation on both sides of the equation to maintain balance. For example, if you subtract a term from one side, you must subtract the same term from the other side. A common mistake is to forget to apply the operation to all terms on both sides, leading to an unbalanced equation and an incorrect solution. To avoid this, make sure to write out each step clearly and explicitly, ensuring that the same operation is applied to every term on both sides of the equation. This practice will help maintain the integrity of the equation and lead to a correct solution. By being aware of these common mistakes and implementing strategies to avoid them, you can significantly improve your accuracy and confidence in solving linear equations.
Conclusion
In conclusion, solving the linear equation 4(x-1)-5(3-x)=14x-2(5x-3) has been a comprehensive exercise in applying algebraic principles. We began by deconstructing the equation, identifying the need for the distributive property to eliminate parentheses. We then meticulously simplified the equation by combining like terms on both sides. Through the strategic application of inverse operations, we isolated the variable 'x' and arrived at the solution x = 5. To ensure the accuracy of our solution, we performed a crucial verification step, substituting x = 5 back into the original equation and confirming that both sides were equal. This step-by-step process underscores the importance of a systematic approach in solving linear equations.
Throughout this discussion, we also highlighted common mistakes that students often make, such as errors in distributing coefficients, combining unlike terms, and performing inverse operations incorrectly. By recognizing these pitfalls and implementing strategies to avoid them, we can enhance our problem-solving skills and increase our confidence in handling algebraic equations. The ability to solve linear equations is a foundational skill in mathematics, and mastering this skill opens doors to more advanced mathematical concepts and applications. Linear equations are not just abstract mathematical constructs; they are powerful tools for modeling and solving real-world problems.
Ultimately, the journey of solving the equation 4(x-1)-5(3-x)=14x-2(5x-3) is more than just finding a numerical answer. It's about developing a logical and structured approach to problem-solving, understanding the underlying principles of algebra, and cultivating a mindset of accuracy and verification. The skills and insights gained through this exercise will undoubtedly serve as a valuable foundation for future mathematical endeavors. So, embrace the challenge of solving linear equations, learn from your mistakes, and celebrate the satisfaction of arriving at a correct solution. The world of mathematics awaits, and with a solid understanding of linear equations, you are well-equipped to explore its vast and fascinating landscape.
