Solutions For The Equation W/(2w-3) = 4/w

In the realm of algebra, solving equations is a fundamental skill. Among the various types of equations, rational equations present a unique challenge due to the presence of variables in the denominators. This article delves into the process of solving a specific rational equation: w/(2_w_-3) = 4/w. We aim to provide a step-by-step guide to finding the solutions, while also discussing the underlying concepts and potential pitfalls. Understanding how to solve rational equations like this one is crucial for success in higher-level mathematics and various applications in science and engineering. The equation w/(2_w_-3) = 4/w is a classic example of a rational equation, where we have algebraic fractions set equal to each other. Solving such equations involves a series of algebraic manipulations to isolate the variable and find its possible values. However, it's not just about crunching numbers; it's also about understanding the domain of the equation and checking for extraneous solutions. These are solutions that arise during the solving process but do not actually satisfy the original equation, often due to division by zero. This article will methodically guide you through each stage, from the initial setup to the final verification of solutions, ensuring a comprehensive understanding of how to tackle rational equations. We'll explore why certain steps are necessary, and how to avoid common mistakes. Whether you're a student learning algebra, or someone brushing up on their math skills, this guide will provide you with the tools and knowledge to confidently solve rational equations.

1. Setting the Stage: Understanding Rational Equations

Before we dive into the specific equation, it's essential to grasp the nature of rational equations. A rational equation is essentially an equation containing one or more fractions where the numerator and/or the denominator are polynomials. Our given equation, w/(2_w_-3) = 4/w, perfectly fits this description. The key challenge in solving these equations lies in dealing with the denominators. Unlike linear or quadratic equations, rational equations can sometimes lead to solutions that aren't valid because they make the denominator zero, which is undefined in mathematics. These invalid solutions are called extraneous solutions, and we must be vigilant in identifying and discarding them. To effectively solve rational equations, a solid foundation in algebraic manipulations is essential. This includes skills like cross-multiplication, simplifying fractions, and solving polynomial equations. We also need to understand the concept of the domain of a function. In the context of rational equations, the domain is the set of all possible values of the variable that don't make any denominator equal to zero. Identifying the domain is a crucial first step in solving rational equations because it helps us anticipate potential extraneous solutions. For the equation w/(2_w_-3) = 4/w, we can immediately see that w cannot be 0, because that would make the denominator of the right-hand side zero. Also, 2_w_-3 cannot be 0, which means w cannot be 3/2. These restrictions are critical to keep in mind as we proceed with solving the equation. By understanding these fundamental concepts, we can approach rational equations with a clear strategy and avoid common pitfalls. The following sections will build upon this foundation, guiding you through the steps needed to find the valid solutions to the equation w/(2_w_-3) = 4/w.

2. The Cross-Multiplication Technique: A Key Step

To solve the equation w/(2_w_-3) = 4/w, our first step is to eliminate the fractions. This is often achieved using a technique called cross-multiplication. Cross-multiplication is a shortcut for multiplying both sides of the equation by the denominators, effectively clearing the fractions. In our case, we multiply both sides by (w) and by (2_w_-3). This gives us w * w = 4 * (2_w_-3). Cross-multiplication is a powerful tool, but it's essential to remember why it works. It's essentially a combination of two separate multiplications: multiplying both sides by the denominator of the left-hand side and then multiplying both sides by the denominator of the right-hand side. By performing cross-multiplication, we transform the rational equation into a simpler polynomial equation that is easier to solve. However, we must always keep in mind the restrictions on the variable that we identified earlier. These restrictions, which are w ≠ 0 and w ≠ 3/2, will be crucial in the final step when we check for extraneous solutions. After cross-multiplication, the equation w/(2_w_-3) = 4/w becomes w^2 = 4(2_w_-3). This new equation is a quadratic equation, which we can solve using various methods, such as factoring, completing the square, or using the quadratic formula. The transformation from a rational equation to a quadratic equation is a significant step forward in the solving process. It allows us to apply familiar techniques to find the possible values of w. However, the solutions we obtain at this stage are only potential solutions. We must still verify that they satisfy the original equation and do not violate the restrictions on the variable. The next section will delve into the process of simplifying and solving the resulting quadratic equation.

3. Simplifying and Solving the Quadratic Equation: Finding Potential Solutions

Having applied cross-multiplication, we arrived at the equation w^2 = 4(2_w_-3). The next step is to simplify this equation and solve for w. First, we distribute the 4 on the right-hand side, which gives us w^2 = 8_w_ - 12. To solve this quadratic equation, we need to rearrange it into the standard form ax^2 + bx + c = 0. Subtracting 8_w_ and adding 12 to both sides, we get w^2 - 8_w_ + 12 = 0. Now, we have a quadratic equation in standard form. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, the equation is easily factorable. We look for two numbers that multiply to 12 and add up to -8. These numbers are -2 and -6. Therefore, we can factor the quadratic as (w - 2)(w - 6) = 0. According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. So, either w - 2 = 0 or w - 6 = 0. Solving these linear equations, we find two potential solutions: w = 2 and w = 6. These are the values of w that make the quadratic equation equal to zero. However, it's crucial to remember that these are just potential solutions to the original rational equation. We still need to check if they satisfy the original equation and if they violate any restrictions on w. We identified earlier that w cannot be 0 or 3/2. Since neither 2 nor 6 is 0 or 3/2, they are potential candidates for solutions. The next section will focus on the critical step of verifying these potential solutions to ensure they are valid.

4. Verifying the Solutions: Eliminating Extraneous Roots

After solving the quadratic equation, we obtained two potential solutions: w = 2 and w = 6. However, before we declare these as the final solutions, we must verify them in the original rational equation: w/(2_w_-3) = 4/w. This step is crucial because solving rational equations can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. Extraneous solutions typically arise when we perform operations that can change the domain of the equation, such as multiplying both sides by an expression containing the variable. Let's first check w = 2. Substituting w = 2 into the original equation, we get 2/(2(2)-3) = 4/2, which simplifies to 2/1 = 2. This is true, so w = 2 is a valid solution. Next, we check w = 6. Substituting w = 6 into the original equation, we get 6/(2(6)-3) = 4/6, which simplifies to 6/9 = 2/3. This is also true, so w = 6 is also a valid solution. We also need to recall the restrictions we identified earlier: w cannot be 0 and w cannot be 3/2. Neither of our solutions, 2 and 6, violates these restrictions. Therefore, both w = 2 and w = 6 are valid solutions to the original rational equation. By meticulously verifying our solutions, we ensure that we have not included any extraneous roots and that our final answer is accurate. This step highlights the importance of careful and thorough work when solving rational equations. The next section will summarize our findings and present the final solution.

5. Conclusion: The Final Answer

In this article, we embarked on a journey to solve the rational equation w/(2_w_-3) = 4/w. We began by understanding the nature of rational equations and the importance of identifying potential restrictions on the variable. We then employed the cross-multiplication technique to eliminate the fractions and transform the equation into a quadratic form. We solved the resulting quadratic equation by factoring, which yielded two potential solutions: w = 2 and w = 6. Crucially, we verified these potential solutions in the original equation to ensure they were valid and not extraneous. Both w = 2 and w = 6 satisfied the original equation and did not violate any restrictions on the variable. Therefore, we can confidently conclude that the solutions to the equation w/(2_w_-3) = 4/w are w = 2 and w = 6. This exercise demonstrates the step-by-step process of solving rational equations, emphasizing the importance of understanding the underlying concepts, performing algebraic manipulations accurately, and verifying solutions to avoid extraneous roots. Solving rational equations is a fundamental skill in algebra, with applications in various fields, including physics, engineering, and economics. By mastering this skill, you'll be well-equipped to tackle more complex mathematical problems and real-world applications. The key takeaways from this guide are the importance of identifying restrictions on the variable, using cross-multiplication to eliminate fractions, solving the resulting polynomial equation, and, most importantly, verifying the solutions in the original equation. With practice and attention to detail, you can confidently solve a wide range of rational equations.

Therefore, the correct answer is D. w = 2 and w = 6