Solving 1/((x-1)(x-2)) + 1/((x-2)(x-3)) = 2/3 A Step-by-Step Guide
Introduction
In this comprehensive article, we will delve into the step-by-step process of solving the given equation: 1/((x-1)(x-2)) + 1/((x-2)(x-3)) = 2/3, where x ≠1, 2, 3. This equation falls under the category of rational equations, which often require careful manipulation and a solid understanding of algebraic principles to arrive at the solution. Our discussion will cover the initial setup, simplification techniques, and the eventual resolution of the quadratic equation that arises. We will also address the crucial step of verifying the solutions to ensure they are valid within the given domain, excluding x = 1, 2, and 3. By exploring this specific equation, we aim to enhance your skills in handling similar mathematical challenges and reinforce fundamental concepts in algebra.
Solving rational equations involves a series of strategic steps, beginning with identifying the restrictions on the variable, which in our case are x ≠1, 2, 3. These restrictions are critical because they prevent division by zero, an undefined operation in mathematics. Next, we will simplify the equation by finding a common denominator, allowing us to combine the fractions. This process transforms the equation into a more manageable form, typically involving polynomials. After simplification, we often encounter a quadratic equation, which can be solved using various methods such as factoring, completing the square, or applying the quadratic formula. Each of these methods has its advantages, depending on the specific characteristics of the equation. Once we obtain the potential solutions, it is imperative to check them against the initial restrictions to eliminate any extraneous solutions. This verification step ensures that the final answers are mathematically sound and applicable within the original context of the problem. The journey through this equation will not only provide a solution but also reinforce a systematic approach to problem-solving in algebra.
Throughout this article, we will emphasize the importance of each step, providing clear explanations and justifications for the methods used. By doing so, we aim to make the process accessible and understandable, even for those who may find algebra challenging. Our goal is to equip you with the tools and knowledge necessary to tackle similar equations with confidence and precision. We will also highlight common pitfalls and mistakes that students often make when dealing with rational equations, helping you avoid these errors in your own problem-solving endeavors. Ultimately, this article serves as a comprehensive guide to not only solve this specific equation but also to cultivate a deeper understanding of algebraic principles and techniques. So, let's embark on this mathematical journey together, unraveling the complexities of the equation and discovering the solutions that lie within.
Step-by-Step Solution
1. Finding a Common Denominator
The first key step in solving our rational equation is to find a common denominator for the fractions on the left-hand side of the equation. The equation we are dealing with is 1/((x-1)(x-2)) + 1/((x-2)(x-3)) = 2/3. To combine these fractions, we need to identify a denominator that is a multiple of both (x-1)(x-2) and (x-2)(x-3). The least common denominator (LCD) in this case is (x-1)(x-2)(x-3). This is because it includes all the unique factors present in the original denominators. Once we have the LCD, we can rewrite each fraction with this new denominator, making it possible to add them together.
To rewrite the first fraction, 1/((x-1)(x-2)), we need to multiply both the numerator and the denominator by the factor (x-3). This gives us (x-3)/((x-1)(x-2)(x-3)). Similarly, for the second fraction, 1/((x-2)(x-3)), we multiply both the numerator and the denominator by (x-1), resulting in (x-1)/((x-1)(x-2)(x-3)). By performing these operations, we ensure that both fractions now have the same denominator, allowing us to combine them. The equation now looks like this: (x-3)/((x-1)(x-2)(x-3)) + (x-1)/((x-1)(x-2)(x-3)) = 2/3. This is a crucial step because it sets the stage for simplifying the equation and ultimately solving for x.
This process of finding a common denominator is a fundamental technique in algebra, applicable to a wide range of problems involving fractions. It not only simplifies the addition and subtraction of fractions but also lays the groundwork for more advanced algebraic manipulations. Understanding how to identify and apply the least common denominator is essential for anyone looking to master algebra. Furthermore, it highlights the importance of careful and systematic work in mathematics, where each step builds upon the previous one. In the next step, we will combine the fractions on the left-hand side, further simplifying the equation and bringing us closer to the solution. By mastering these techniques, you'll be well-equipped to tackle even more complex equations in the future.
2. Combining Fractions and Simplifying
With a common denominator established, we can now combine the fractions on the left-hand side of the equation. We have (x-3)/((x-1)(x-2)(x-3)) + (x-1)/((x-1)(x-2)(x-3)) = 2/3. Since the denominators are the same, we can add the numerators directly. This gives us ((x-3) + (x-1))/((x-1)(x-2)(x-3)) = 2/3. Now, we simplify the numerator by combining like terms. The expression (x-3) + (x-1) simplifies to 2x - 4. Therefore, the equation becomes (2x - 4)/((x-1)(x-2)(x-3)) = 2/3.
Next, we can further simplify the numerator by factoring out a 2, which gives us 2(x - 2)/((x-1)(x-2)(x-3)) = 2/3. At this point, we notice that the factor (x-2) appears in both the numerator and the denominator. We can cancel this common factor, but it's crucial to remember the initial restriction that x ≠2. Cancelling (x-2) simplifies the equation to 2/((x-1)(x-3)) = 2/3. This simplification is a significant step, as it reduces the complexity of the equation and makes it easier to solve. However, it's essential to keep track of the restrictions on x, as they will be important when we verify our solutions later.
The process of combining fractions and simplifying is a cornerstone of algebraic manipulation. It demonstrates the power of reducing complex expressions to their simplest forms, making them more manageable and easier to work with. This step not only involves basic arithmetic operations but also requires a keen eye for spotting common factors and applying simplification rules. By carefully combining fractions and simplifying, we pave the way for the next phase of solving the equation, which involves cross-multiplication and further algebraic manipulations. Understanding these techniques is vital for success in algebra and beyond, enabling you to tackle a wide range of mathematical problems with confidence and efficiency.
3. Cross-Multiplication and Rearranging
Having simplified the equation to 2/((x-1)(x-3)) = 2/3, the next step involves cross-multiplication. This technique allows us to eliminate the fractions and transform the equation into a more manageable form. Cross-multiplication is based on the principle that if a/b = c/d, then ad = bc. Applying this to our equation, we multiply the numerator of the left side by the denominator of the right side, and vice versa. This gives us 2 * 3 = 2 * (x-1)(x-3).
Simplifying this, we get 6 = 2(x-1)(x-3). We can divide both sides of the equation by 2 to further simplify it, resulting in 3 = (x-1)(x-3). Now, we need to expand the right side of the equation. Multiplying (x-1) by (x-3) gives us x² - 3x - x + 3, which simplifies to x² - 4x + 3. So, our equation now looks like 3 = x² - 4x + 3. To solve this, we need to rearrange the equation into a standard quadratic form, which is ax² + bx + c = 0. Subtracting 3 from both sides, we get 0 = x² - 4x.
Rearranging the equation into this standard form is a crucial step in solving quadratic equations. It allows us to apply various methods such as factoring, completing the square, or using the quadratic formula. In this particular case, the equation is relatively simple and can be solved by factoring. However, understanding how to rearrange equations and put them into standard forms is a fundamental skill in algebra, applicable to a wide range of mathematical problems. Cross-multiplication, as demonstrated here, is a powerful technique for dealing with equations involving fractions, and it often simplifies the problem significantly. By mastering these techniques, you'll be well-prepared to tackle more complex equations and mathematical challenges.
4. Solving the Quadratic Equation
We've now arrived at the quadratic equation 0 = x² - 4x. To solve this quadratic equation, we can use the factoring method, which is particularly effective when the equation can be easily factored. Factoring involves expressing the quadratic expression as a product of two binomials. In this case, we can factor out an x from both terms on the right side of the equation. This gives us 0 = x(x - 4).
Now, we apply the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. In other words, if ab = 0, then either a = 0 or b = 0 (or both). Applying this property to our equation, x(x - 4) = 0, we have two possibilities: either x = 0 or x - 4 = 0. Solving the second equation for x, we add 4 to both sides, giving us x = 4. So, we have found two potential solutions: x = 0 and x = 4.
Factoring is a fundamental technique in algebra and is often the quickest way to solve quadratic equations when it's applicable. It requires recognizing common factors and applying algebraic identities to rewrite the expression. The zero-product property is a powerful tool that allows us to break down a single equation into multiple simpler equations, each of which can be solved independently. However, it's essential to remember that not all quadratic equations can be easily factored, and in those cases, other methods such as completing the square or using the quadratic formula may be necessary. In this instance, factoring provided a straightforward path to the solutions, but the next crucial step is to verify these solutions against the original equation and any restrictions we identified earlier.
5. Verifying the Solutions
Now that we have obtained the potential solutions, x = 0 and x = 4, it is crucial to verify these solutions to ensure they are valid and do not violate any restrictions imposed by the original equation. The original equation was 1/((x-1)(x-2)) + 1/((x-2)(x-3)) = 2/3, with the restrictions x ≠1, 2, 3. These restrictions are in place because if x were equal to 1, 2, or 3, the denominators of the fractions would become zero, leading to undefined expressions.
Let's first check x = 0. Substituting x = 0 into the original equation, we get: 1/((-1)(-2)) + 1/((-2)(-3)) = 2/3. This simplifies to 1/2 + 1/6 = 2/3. To add these fractions, we find a common denominator, which is 6. So, we have 3/6 + 1/6 = 4/6, which simplifies to 2/3. Thus, x = 0 is a valid solution because it satisfies the original equation and does not violate any restrictions.
Next, we check x = 4. Substituting x = 4 into the original equation, we get: 1/((4-1)(4-2)) + 1/((4-2)(4-3)) = 2/3. This simplifies to 1/(32) + 1/(21) = 2/3, which is 1/6 + 1/2 = 2/3. Again, we find a common denominator, which is 6. So, we have 1/6 + 3/6 = 4/6, which simplifies to 2/3. Therefore, x = 4 is also a valid solution because it satisfies the original equation and does not violate any restrictions.
The process of verifying solutions is an essential step in solving any equation, especially rational equations where restrictions on the variable are common. This step ensures that the solutions we obtain are mathematically sound and applicable within the context of the problem. By substituting the potential solutions back into the original equation, we can confirm whether they satisfy the equation and respect the domain restrictions. In this case, both x = 0 and x = 4 have been verified as valid solutions, completing our solution process.
Conclusion
In conclusion, we have successfully solved the equation 1/((x-1)(x-2)) + 1/((x-2)(x-3)) = 2/3, with the restrictions x ≠1, 2, 3. Through a systematic approach, we identified the need for a common denominator, combined the fractions, simplified the equation, and ultimately solved the resulting quadratic equation. We found two potential solutions, x = 0 and x = 4, and rigorously verified each of them against the original equation and the given restrictions. This verification step is crucial to ensure the validity of the solutions, and in this case, both solutions were confirmed to be correct.
The process of solving this equation highlights several key concepts in algebra, including finding common denominators, combining fractions, simplifying expressions, solving quadratic equations, and verifying solutions. Each of these steps is fundamental to algebraic problem-solving, and mastering them is essential for success in mathematics. The equation we tackled is a representative example of rational equations, which often require careful manipulation and attention to detail to solve correctly. By working through this example, we have demonstrated a comprehensive approach that can be applied to similar problems. Furthermore, we have emphasized the importance of understanding the underlying principles and techniques, rather than just memorizing steps. This deeper understanding will enable you to tackle a wider range of mathematical challenges with confidence and proficiency.
Ultimately, the journey through solving this equation has not only provided us with the solutions but also reinforced a systematic and methodical approach to problem-solving. By breaking down the problem into smaller, manageable steps, we were able to navigate the complexities of the equation and arrive at the correct answers. This approach is applicable not only in mathematics but also in various other fields, highlighting the importance of critical thinking and problem-solving skills. As you continue your mathematical journey, remember to apply these principles and techniques, and you will be well-equipped to tackle even the most challenging problems.
