Solving $x - \frac{x-1}{2} = 1 - \frac{x-2}{3}$ A Step-by-Step Guide

Introduction

This article provides a detailed walkthrough on solving the linear equation $x - \frac{x-1}{2} = 1 - \frac{x-2}{3}$. Linear equations are fundamental in algebra, and mastering their solution is crucial for more advanced mathematical concepts. We will break down each step, explaining the underlying principles and techniques involved. By the end of this guide, you will have a solid understanding of how to approach and solve similar equations. This comprehensive guide aims to not only provide the solution but also to enhance your understanding of the underlying mathematical principles. Let's embark on this mathematical journey to conquer this linear equation and fortify your algebraic skills. Understanding the techniques involved in solving this equation will enable you to confidently tackle more complex problems in mathematics and other related fields. This introduction serves as a roadmap for our exploration, ensuring you grasp the significance of each step and its contribution to the final solution.

Problem Statement: $x - \frac{x-1}{2} = 1 - \frac{x-2}{3}$

The given equation is a linear equation in one variable, $x$. Our goal is to find the value of $x$ that satisfies this equation. To do this, we will systematically manipulate the equation using algebraic operations until we isolate $x$ on one side. This process involves several key steps, including clearing fractions, combining like terms, and isolating the variable. Each step is essential to arriving at the correct solution. The equation presents a common challenge in algebra – dealing with fractions. To simplify the equation and make it easier to solve, we will first eliminate these fractions by multiplying both sides by the least common multiple (LCM) of the denominators. This technique is a cornerstone of solving linear equations involving fractions. It transforms the equation into a more manageable form, paving the way for subsequent algebraic manipulations. Throughout this article, we will highlight and explain the rationale behind each step, ensuring that you understand not only how to solve this particular equation but also the general principles that apply to a wide range of algebraic problems. This comprehensive approach is designed to build your confidence and competence in solving linear equations.

Step 1: Clearing the Fractions

To eliminate the fractions, we need to find the least common multiple (LCM) of the denominators, which are 2 and 3. The LCM of 2 and 3 is 6. We will multiply both sides of the equation by 6:

$6 \left(x - \frac{x-1}{2}\right) = 6 \left(1 - \frac{x-2}{3}\right)$

This step is crucial because it simplifies the equation by removing the fractions, making it easier to work with. The principle behind this operation is that multiplying both sides of an equation by the same non-zero number preserves the equality. By choosing the LCM as the multiplier, we ensure that each fraction will simplify to an integer, thus eliminating the fractions altogether. The choice of the LCM is deliberate and strategic. It is the smallest number that is divisible by both 2 and 3, which guarantees that the fractions will cancel out completely without introducing new fractions. This technique is not only applicable to this specific equation but is a general method for clearing fractions in any algebraic equation. Understanding and mastering this step is fundamental for solving equations involving fractions. The next step will involve distributing the 6 on both sides of the equation, which will further simplify the expression and bring us closer to isolating the variable $x$.

Step 2: Distributing the Multiplication

Now, distribute the 6 on both sides of the equation:

$6x - 6\left(\frac{x-1}{2}\right) = 6 - 6\left(\frac{x-2}{3}\right)$

Simplify the terms:

$6x - 3(x-1) = 6 - 2(x-2)$

This step involves the distributive property, a fundamental concept in algebra. The distributive property states that $a(b+c) = ab + ac$. Applying this property correctly is essential for simplifying algebraic expressions and equations. In this case, we are distributing the 6 on the left side and the 6 on the right side of the equation. It is crucial to pay attention to the signs when distributing, as an incorrect sign can lead to an incorrect solution. The simplification of the terms after distribution is equally important. Here, we divide 6 by 2 to get 3 and 6 by 3 to get 2. These simplified coefficients make the subsequent steps easier to manage. This process of distribution and simplification is a key skill in algebra, allowing us to transform complex expressions into simpler, more manageable forms. The next step will involve further simplification by distributing the -3 and -2 into their respective parentheses, which will bring us closer to combining like terms and isolating the variable $x$.

Step 3: Further Simplification

Distribute the -3 and -2:

$6x - 3x + 3 = 6 - 2x + 4$

This step continues the simplification process by removing the parentheses. It is essential to correctly apply the distributive property, paying close attention to the signs. For example, -3 multiplied by -1 becomes +3, and -2 multiplied by -2 becomes +4. Incorrectly handling the signs is a common source of errors in algebra, so it is vital to be meticulous. This step prepares the equation for combining like terms, which is the next logical step in solving for $x$. Removing the parentheses allows us to group terms with the same variable and constant terms, making the equation more manageable. The accuracy of this step directly impacts the correctness of the final solution. Therefore, it is crucial to double-check the distribution and the signs before proceeding to the next step. Each step in solving an equation builds upon the previous one, so any error in an earlier step will propagate through the rest of the solution.

Step 4: Combining Like Terms

Combine like terms on both sides of the equation:

$(6x - 3x) + 3 = (-2x) + (6 + 4)$

$3x + 3 = 10 - 2x$

Combining like terms is a fundamental algebraic technique that simplifies equations by grouping similar terms together. In this step, we combine the $x$ terms on the left side ($6x$ and $-3x$) and the constant terms on the right side (6 and 4). This process reduces the number of terms in the equation, making it easier to solve. The goal is to isolate the variable $x$, and combining like terms is a crucial step in achieving this. It's important to ensure that only terms with the same variable and exponent (like terms) are combined. For example, $3x$ and $-2x$ can be combined, but $3x$ and 3 cannot. The result of this step is a simplified equation that is closer to the standard form of a linear equation, which makes it easier to manipulate and solve for the unknown variable. This step is a prerequisite for the next step, which involves isolating the variable on one side of the equation.

Step 5: Isolating the Variable

To isolate $x$, we need to get all the $x$ terms on one side and the constants on the other side. Add $2x$ to both sides:

$3x + 2x + 3 = 10 - 2x + 2x$

$5x + 3 = 10$

Subtract 3 from both sides:

$5x + 3 - 3 = 10 - 3$

$5x = 7$

This step is the core of solving the equation. We aim to isolate the variable $x$ on one side of the equation. To achieve this, we perform the same operations on both sides of the equation, maintaining the balance. First, we add $2x$ to both sides to eliminate the $-2x$ term on the right side. Then, we subtract 3 from both sides to eliminate the constant term on the left side. These operations effectively move all the $x$ terms to one side and all the constant terms to the other side. The principle behind this is that performing the same operation on both sides of an equation preserves the equality. This technique is a cornerstone of algebraic manipulation and is used extensively in solving various types of equations. The result of this step is a simplified equation where the variable $x$ is almost isolated, with only a coefficient remaining. The final step will involve dividing both sides by this coefficient to obtain the value of $x$.

Step 6: Solving for x

Divide both sides by 5:

$\frac{5x}{5} = \frac{7}{5}$

$x = \frac{7}{5}$

This final step completes the solution process. We divide both sides of the equation by the coefficient of $x$, which is 5, to isolate $x$. This operation gives us the value of $x$ that satisfies the original equation. The principle behind this step is the same as in the previous step – performing the same operation on both sides of the equation preserves the equality. Dividing by the coefficient of $x$ is the standard method for isolating the variable in a linear equation. The solution $x = \frac{7}{5}$ represents the value that, when substituted back into the original equation, will make the equation true. This value is the answer to the problem. It is a common practice to check the solution by substituting it back into the original equation to verify its correctness. This step is a crucial part of the problem-solving process, ensuring that the solution is accurate and that no errors were made in the previous steps.

Step 7: Verification (Optional but Recommended)

To verify the solution, substitute $x = \frac{7}{5}$ back into the original equation:

$\frac{7}{5} - \frac{\frac{7}{5}-1}{2} = 1 - \frac{\frac{7}{5}-2}{3}$

Simplify:

$\frac{7}{5} - \frac{\frac{2}{5}}{2} = 1 - \frac{-\frac{3}{5}}{3}$

$\frac{7}{5} - \frac{1}{5} = 1 + \frac{1}{5}$

$\frac{6}{5} = \frac{6}{5}$

The left side equals the right side, so the solution is correct.

Verification is an essential step in problem-solving, especially in mathematics. It ensures that the solution obtained is accurate and satisfies the original equation. In this step, we substitute the value of $x$ that we found ($x = \frac{7}{5}$) back into the original equation. If the left side of the equation equals the right side after the substitution, then the solution is correct. This process helps to catch any errors that might have occurred during the solution process. Verification is not always required, but it is highly recommended, especially in situations where accuracy is critical. It provides confidence in the solution and ensures that the problem has been solved correctly. The process involves careful substitution and simplification, following the order of operations. Each step in the verification process should be checked to avoid errors. The successful verification confirms that the solution is correct and that the steps taken to solve the equation were valid.

Conclusion

We have successfully solved the equation $x - \frac{x-1}{2} = 1 - \frac{x-2}{3}$ and found that $x = \frac{7}{5}$. This process involved clearing fractions, distributing, combining like terms, isolating the variable, and solving for $x$. We also demonstrated the importance of verifying the solution to ensure accuracy. Mastering these techniques is essential for solving various algebraic problems. The step-by-step approach outlined in this article provides a clear and structured method for solving linear equations. Each step is crucial, and understanding the underlying principles is key to success. The techniques learned in this process are not only applicable to this specific equation but also to a wide range of algebraic problems. Solving linear equations is a fundamental skill in mathematics, and proficiency in this area is essential for further studies in mathematics and related fields. The ability to confidently solve equations allows for a deeper understanding of mathematical concepts and their applications. This article has aimed to provide not only the solution to a specific problem but also a broader understanding of the methods and principles involved in algebraic problem-solving. The systematic approach and the emphasis on verification are intended to promote accuracy and confidence in your mathematical abilities.