Solving (1/2) + (3/(x-3)) = - (3/(x-3)) A Step-by-Step Guide

Introduction

In this article, we will delve into the step-by-step process of solving the equation $\frac{1}{2} + \frac{3}{x-3} = -\frac{3}{x-3}$ for the variable $x$. This type of equation, involving fractions and variables in the denominator, requires a methodical approach to ensure accuracy. We will cover the crucial steps, including identifying restrictions on the variable, combining like terms, eliminating fractions, and isolating $x$. By the end of this detailed explanation, you will gain a comprehensive understanding of how to tackle similar algebraic problems. Mastering these techniques is fundamental for success in algebra and related fields.

1. Identifying Restrictions on x

Before we begin manipulating the equation, it is critical to identify any values of $x$ that would make the denominators of the fractions equal to zero. This is because division by zero is undefined in mathematics. In the given equation, $\frac{1}{2} + \frac{3}{x-3} = -\frac{3}{x-3}$, we have the denominator $x-3$. To find the restricted values, we set the denominator equal to zero:

$x - 3 = 0$

Solving for $x$, we get:

$x = 3$

Therefore, $x$ cannot be equal to 3. This restriction is crucial because if we obtain $x = 3$ as a solution during our calculations, we must discard it as an extraneous solution. The domain of the equation is all real numbers except 3. Understanding and noting these restrictions at the beginning of the problem is an essential step in solving rational equations correctly. Failing to do so can lead to incorrect final answers.

2. Combining Like Terms

The next step in solving the equation $\frac{1}{2} + \frac{3}{x-3} = -\frac{3}{x-3}$ is to combine the terms that have the same denominator. In this case, the terms $\frac{3}{x-3}$ and $-\frac{3}{x-3}$ can be combined. To do this, we add them together:

$\frac{3}{x-3} + \frac{3}{x-3} = -\frac{3}{x-3} + \frac{3}{x-3}$

This simplifies to:

$\frac{3}{x-3} + \frac{3}{x-3} = 0$

Now, the equation looks like this:

$\frac{1}{2} + \frac{3}{x-3} + \frac{3}{x-3} = 0$

Combining the fractions on the left side gives us:

$\frac{1}{2} + \frac{6}{x-3} = 0$

This step is important because it simplifies the equation, making it easier to work with. By combining like terms, we reduce the number of fractions, which will help in the subsequent steps of solving for $x$. Ensuring that like terms are correctly combined is vital for arriving at the correct solution.

3. Eliminating the Fractions

To eliminate the fractions in the equation $\frac{1}{2} + \frac{6}{x-3} = 0$, we need to find the least common denominator (LCD) of the fractions. The denominators in our equation are 2 and $x-3$. The LCD is the smallest expression that both denominators divide into evenly. In this case, the LCD is $2(x-3)$.

Now, we multiply both sides of the equation by the LCD, $2(x-3)$:

$2(x-3) \left( \frac{1}{2} + \frac{6}{x-3} \right) = 2(x-3) \cdot 0$

Distribute $2(x-3)$ to each term on the left side:

$2(x-3) \cdot \frac{1}{2} + 2(x-3) \cdot \frac{6}{x-3} = 0$

Simplify each term:

$(x-3) + 12 = 0$

By multiplying by the LCD, we have successfully eliminated the fractions from the equation. This step is crucial because it transforms the equation into a more manageable form, allowing us to solve for $x$ without the complexities of dealing with fractions. This technique is a cornerstone in solving rational equations and is frequently used in various algebraic problems.

4. Solving for x

Now that we have eliminated the fractions, our equation is:

$x - 3 + 12 = 0$

Combine the constant terms:

$x + 9 = 0$

To isolate $x$, subtract 9 from both sides of the equation:

$x + 9 - 9 = 0 - 9$

This simplifies to:

$x = -9$

So, we have found a potential solution for $x$. It is essential to check whether this solution satisfies the restrictions we identified earlier. Recall that $x$ cannot be equal to 3. Since $-9$ is not equal to 3, it is a valid solution. The process of solving for $x$ involved simplifying the equation and isolating the variable, which are fundamental techniques in algebra. Verifying the solution against any restrictions ensures the correctness of our final answer.

5. Checking the Solution

To ensure that $x = -9$ is indeed the correct solution, we substitute it back into the original equation:

$\frac{1}{2} + \frac{3}{x-3} = -\frac{3}{x-3}$

Substitute $x = -9$:

$\frac{1}{2} + \frac{3}{-9-3} = -\frac{3}{-9-3}$

Simplify the denominators:

$\frac{1}{2} + \frac{3}{-12} = -\frac{3}{-12}$

Reduce the fractions:

$\frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

Find a common denominator for the left side (which is 4):

$\frac{2}{4} - \frac{1}{4} = \frac{1}{4}$

Simplify the left side:

$\frac{1}{4} = \frac{1}{4}$

Since the left side equals the right side, our solution $x = -9$ is correct. This verification step is crucial in solving equations, especially those involving fractions, as it confirms that our solution satisfies the original equation. By checking our solution, we can have confidence in our answer.

6. Final Answer

After going through all the steps, we have confidently found the solution to the equation $\frac{1}{2} + \frac{3}{x-3} = -\frac{3}{x-3}$. We identified the restriction on $x$, combined like terms, eliminated fractions, solved for $x$, and checked our solution. The solution we found is:

$x = -9$

This value satisfies the original equation and does not violate any restrictions. Therefore, we can confidently state that the solution to the given equation is $x = -9$. The final answer, expressed as a reduced improper fraction, is -9.

Conclusion

In this article, we have demonstrated a detailed, step-by-step method for solving the equation $\frac{1}{2} + \frac{3}{x-3} = -\frac{3}{x-3}$. We began by identifying restrictions on the variable $x$, which is a critical step to avoid undefined expressions. We then combined like terms to simplify the equation, followed by eliminating fractions by multiplying through by the least common denominator. Next, we isolated $x$ and found a potential solution, which we meticulously checked against our initial restrictions and by substituting back into the original equation. This comprehensive approach ensures accuracy and a thorough understanding of the problem. Mastering these techniques is invaluable for tackling various algebraic equations and further mathematical challenges. Remember, precision and verification are the cornerstones of successful problem-solving in mathematics.