Solving For U A Step-by-Step Guide

In the realm of algebra, solving for variables is a fundamental skill. This article delves into the process of solving the equation (2)/(u-5) = (6)/(3u-15) - 4 for the variable u. We will explore the steps involved, the underlying principles, and potential pitfalls to avoid. Whether you're a student grappling with algebraic equations or simply seeking to refresh your knowledge, this guide will provide a comprehensive understanding of the solution.

1. Understanding the Equation

Before we embark on the solution, let's dissect the equation. We have a rational equation, which means it involves fractions with variables in the denominator. The equation is: (2)/(u-5) = (6)/(3u-15) - 4.

Our goal is to isolate the variable u on one side of the equation. To achieve this, we'll employ a series of algebraic manipulations, ensuring that we maintain the equality throughout the process. We will delve into each of these steps in detail, providing clear explanations and justifications.

Identifying the Key Components

It's crucial to identify the key components of the equation. We have two fractions: (2)/(u-5) and (6)/(3u-15). The denominators, (u-5) and (3u-15), play a significant role in determining the solution. We also have a constant term, -4, which needs to be addressed appropriately. Understanding these components is the foundation for solving the equation effectively.

The Importance of Domain

In rational equations, the domain of the variable is critical. We need to ensure that the denominators are not equal to zero, as division by zero is undefined. This means u cannot be 5, as this would make the denominator (u-5) zero. Similarly, u cannot be 5 for the denominator (3u-15). This constraint will be crucial when we check our solution later.

2. Simplifying the Equation

The first step towards solving for u is to simplify the equation. This involves eliminating fractions and combining like terms. Simplifying the equation makes it easier to manipulate and solve.

Finding the Least Common Denominator (LCD)

To eliminate the fractions, we need to find the least common denominator (LCD) of the denominators (u-5) and (3u-15). Notice that (3u-15) can be factored as 3*(u-5). Therefore, the LCD is 3(u-5)*. Identifying the LCD is a critical step in simplifying rational equations.

Multiplying by the LCD

Next, we multiply both sides of the equation by the LCD, 3*(u-5). This will eliminate the fractions: 3(u-5)* * [(2)/(u-5)] = 3*(u-5)* * [(6)/(3u-15) - 4]. When we multiply each term by the LCD, we are essentially clearing the fractions, which simplifies the equation significantly. This step is based on the principle that multiplying both sides of an equation by the same non-zero value maintains the equality.

Distributing and Cancelling

After multiplying by the LCD, we distribute and cancel common factors. On the left side, the (u-5) terms cancel out, leaving us with 3 * 2 = 6. On the right side, the 3*(u-5)* terms cancel out in the first fraction, leaving us with 6. We also need to distribute 3*(u-5)* to the -4 term, resulting in -12*(u-5). This simplification process transforms the equation into a more manageable form. The equation now looks like this: 6 = 6 - 12(u-5)*.

3. Solving for u

Now that we've simplified the equation, we can proceed to solve for u. This involves isolating u on one side of the equation using algebraic manipulations.

Further Simplification

Let's simplify the equation further. We have 6 = 6 - 12*(u-5). Subtracting 6 from both sides gives us 0 = -12(u-5)*. This step isolates the term containing u, bringing us closer to the solution.

Isolating the Variable

To isolate (u-5), we divide both sides by -12. This gives us 0 = u - 5. Dividing both sides of the equation by -12 isolates the variable term, making it easier to solve for u. This step is valid as long as we are not dividing by zero, which is not the case here.

Final Solution

Finally, we add 5 to both sides to isolate u. This gives us u = 5. Adding 5 to both sides isolates u, providing the solution to the equation. This is the final step in solving for u.

4. Checking the Solution

It's crucial to check our solution to ensure it is valid. In rational equations, we need to verify that our solution does not make any of the denominators zero. This is particularly important because division by zero is undefined.

Substituting the Value of u

We substitute u = 5 back into the original equation: (2)/(u-5) = (6)/(3u-15) - 4. Substituting u = 5 gives us (2)/(5-5) = (6)/(3*5-15) - 4, which simplifies to (2)/0 = (6)/0 - 4. This substitution highlights the importance of checking our solution, as it reveals a potential issue.

Identifying Extraneous Solutions

We encounter division by zero, which is undefined. This means that u = 5 is an extraneous solution. An extraneous solution is a value that satisfies the transformed equation but not the original equation. In this case, the solution u = 5 makes the denominators zero, rendering the equation undefined. This means that u = 5 cannot be a valid solution to the original equation.

The Absence of a Solution

Since u = 5 is an extraneous solution and there are no other potential solutions, the equation has no solution. This can happen in rational equations when the solution obtained makes the denominator zero. Recognizing the absence of a solution is an important outcome in solving equations. It's essential to acknowledge when an equation has no solution, as this indicates a fundamental incompatibility within the equation itself.

5. Conclusion

In this comprehensive guide, we've walked through the process of solving the equation (2)/(u-5) = (6)/(3u-15) - 4. We've covered simplifying the equation, solving for u, and, most importantly, checking the solution. We discovered that u = 5 is an extraneous solution, leading us to the conclusion that the equation has no solution.

Solving rational equations requires a careful approach, paying close attention to the domain of the variable and the potential for extraneous solutions. This guide has provided a detailed framework for tackling such equations, equipping you with the knowledge and skills to solve them effectively. Remember to always check your solutions and be mindful of the potential for extraneous solutions.

By understanding these steps and principles, you can confidently approach a wide range of algebraic equations and solve for the variables involved. This skill is essential not only in mathematics but also in various fields that rely on quantitative analysis and problem-solving.

Key Takeaways

• Simplifying the equation by eliminating fractions is a crucial first step.
• Finding the LCD is essential for clearing fractions in rational equations.
• Checking the solution is vital to identify extraneous solutions.
• Understanding the domain of the variable helps avoid division by zero.
• Recognizing the absence of a solution is an important outcome in equation-solving.

This comprehensive guide should empower you to tackle similar problems with confidence and accuracy. Remember, practice is key to mastering these concepts and developing your problem-solving skills.