Solving Rational Equations A Step-by-Step Guide To X/(x-5) - 5/(x+5) = 10x/(x^2-25)

Introduction

In the realm of mathematics, rational equations often present a unique set of challenges. These equations, which involve fractions with polynomials in the numerator and denominator, require careful manipulation and a solid understanding of algebraic principles to solve effectively. This comprehensive guide will delve into the step-by-step process of solving the rational equation ${\frac{x}{x-5}-\frac{5}{x+5}=\frac{10x}{x^2-25}}$. We will explore the underlying concepts, common pitfalls, and strategies for verifying solutions to ensure accuracy. By the end of this discussion, you will have a clear understanding of how to tackle similar problems and enhance your problem-solving skills in algebra.

Understanding Rational Equations

Rational equations are essentially equations that contain one or more rational expressions. A rational expression is a fraction where the numerator and denominator are polynomials. Solving these equations involves eliminating the fractions to obtain a more manageable algebraic form, typically a polynomial equation. However, this process introduces a critical consideration: the potential for extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but not the original rational equation. This usually occurs because certain values of the variable can make the denominator of a fraction equal to zero, which is undefined.

Therefore, the process of solving rational equations requires a methodical approach that includes: 1) identifying any restrictions on the variable, 2) clearing the fractions by multiplying through by the least common denominator (LCD), 3) solving the resulting equation, and 4) verifying the solutions against the original equation to exclude extraneous roots. This structured approach ensures that only valid solutions are retained, maintaining the integrity of the mathematical process.

Step 1: Identify Restrictions

Before diving into the algebraic manipulations, the first crucial step in solving any rational equation is to identify any restrictions on the variable. Restrictions occur when the denominator of any fraction in the equation equals zero, as division by zero is undefined. In our equation, ${\frac{x}{x-5}-\frac{5}{x+5}=\frac{10x}{x^2-25}}$, we have three denominators: ${x-5}$, ${x+5}$, and ${x^2-25}$. To find the restrictions, we set each denominator equal to zero and solve for ${x}$.

For ${x-5=0}$, we add 5 to both sides, resulting in ${x=5}$. This means that ${x}$ cannot be 5, as this would make the denominator of the first fraction zero.

Similarly, for ${x+5=0}$, we subtract 5 from both sides, resulting in ${x=-5}$. Thus, ${x}$ cannot be -5, as this would make the denominator of the second fraction zero.

Finally, for ${x^2-25=0}$, we recognize this as a difference of squares, which can be factored as ${(x-5)(x+5)=0}$. Setting each factor equal to zero gives us ${x=5}$ and ${x=-5}$, which we have already identified. Therefore, the restrictions on ${x}$ are ${x \neq 5}$ and ${x \neq -5}$. These restrictions are critical because any solutions we find later must be checked against these values. If a solution matches a restriction, it is an extraneous solution and must be discarded. Identifying these restrictions at the outset is a fundamental step in accurately solving rational equations.

Step 2: Find the Least Common Denominator (LCD)

The next critical step in solving the rational equation ${\frac{x}{x-5}-\frac{5}{x+5}=\frac{10x}{x^2-25}}$ is to determine the least common denominator (LCD) of the fractions involved. The LCD is the smallest expression that is divisible by all denominators in the equation. This allows us to clear the fractions by multiplying both sides of the equation by the LCD, thereby simplifying the equation into a more manageable form.

In our case, the denominators are ${x-5}$, ${x+5}$, and ${x^2-25}$. Notice that ${x^2-25}$ can be factored as ${(x-5)(x+5)}$, which is a difference of squares. Therefore, the denominators are ${x-5}$, ${x+5}$, and ${(x-5)(x+5)}$. The LCD is the expression that includes each factor the greatest number of times it appears in any denominator. In this scenario, the LCD is ${(x-5)(x+5)}$ because it contains both factors, ${x-5}$ and ${x+5}$, and thus is divisible by each individual denominator.

Finding the LCD correctly is essential for solving rational equations efficiently. A clear understanding of factoring and the ability to recognize patterns such as the difference of squares can greatly aid in this process. Once the LCD is identified, we can proceed to the next step, which involves multiplying both sides of the equation by the LCD to eliminate the fractions.

Step 3: Multiply Both Sides by the LCD

With the LCD identified as ${(x-5)(x+5)}$, the next step in solving the rational equation ${\frac{x}{x-5}-\frac{5}{x+5}=\frac{10x}{x^2-25}}$ is to multiply both sides of the equation by this LCD. This process is crucial because it eliminates the fractions, transforming the equation into a simpler algebraic form that is easier to solve. Multiplying both sides by the same expression maintains the equality of the equation.

Starting with the left side of the equation, we multiply ${\frac{x}{x-5}}$ by ${(x-5)(x+5)}$. The factor ${(x-5)}$ in the denominator cancels with the ${(x-5)}$ in the LCD, leaving us with ${x(x+5)}$.

Next, we multiply ${-\frac{5}{x+5}}$ by ${(x-5)(x+5)}$. The factor ${(x+5)}$ in the denominator cancels with the ${(x+5)}$ in the LCD, leaving us with ${-5(x-5)}$.

On the right side of the equation, we multiply ${\frac{10x}{x^2-25}}$ by ${(x-5)(x+5)}$. Since ${x^2-25}$ is equivalent to ${(x-5)(x+5)}$, the entire denominator cancels with the LCD, leaving us with ${10x}$.

Thus, after multiplying both sides by the LCD and simplifying, our equation transforms from ${\frac{x}{x-5}-\frac{5}{x+5}=\frac{10x}{x^2-25}}$ to ${x(x+5) - 5(x-5) = 10x}$. This new equation is free of fractions and can be further simplified by expanding the terms and combining like terms. This step is pivotal in the process of solving rational equations as it sets the stage for solving a standard polynomial equation.

Step 4: Simplify and Solve the Resulting Equation

After multiplying both sides of the rational equation by the LCD, we obtained the simplified equation ${x(x+5) - 5(x-5) = 10x}$. The next step is to expand the terms, combine like terms, and solve for ${x}$. This process involves applying the distributive property, simplifying the resulting quadratic equation, and then using factoring or the quadratic formula to find the solutions.

First, we expand the terms:

${x(x+5)}$ becomes ${x^2 + 5x}$, and

${-5(x-5)}$ becomes ${-5x + 25}$.

So, the equation becomes ${x^2 + 5x - 5x + 25 = 10x}$.

Next, we combine like terms on the left side:

The ${5x}$ and ${-5x}$ cancel each other out, leaving us with ${x^2 + 25 = 10x}$.

To solve this quadratic equation, we need to set it equal to zero. Subtracting ${10x}$ from both sides gives us:

${x^2 - 10x + 25 = 0}$.

Now, we can solve this quadratic equation by factoring. We look for two numbers that multiply to 25 and add to -10. These numbers are -5 and -5. Thus, we can factor the quadratic as:

${(x - 5)(x - 5) = 0}$ or ${(x-5)^2 = 0}$.

Setting the factor equal to zero, we get:

${x - 5 = 0}$, which gives us ${x = 5}$.

So, we have found a potential solution ${x = 5}$. However, this is not the final answer. We must check this solution against the restrictions we identified in Step 1 to ensure it is not an extraneous solution. Solving the resulting equation is a crucial part of handling rational equations, and attention to detail in the algebraic manipulations is essential to arrive at the correct solution(s).

Step 5: Check for Extraneous Solutions

After solving the simplified equation, we found a potential solution of ${x=5}$ for the rational equation ${\frac{x}{x-5}-\frac{5}{x+5}=\frac{10x}{x^2-25}}$. However, this is not the final answer until we verify that this solution is not extraneous. Extraneous solutions are values that satisfy the simplified equation but do not satisfy the original equation because they make one or more denominators equal to zero.

In Step 1, we identified the restrictions on ${x}$ as ${x \neq 5}$ and ${x \neq -5}$. This means that ${x}$ cannot be 5 or -5, as these values would make the denominators ${x-5}$ and ${x^2-25}$ equal to zero, resulting in undefined expressions. Our potential solution, ${x=5}$, exactly matches one of these restrictions.

Therefore, ${x=5}$ is an extraneous solution. This means that it is not a valid solution to the original rational equation. Since we have only one potential solution and it turned out to be extraneous, the original equation has no solution. The importance of this step cannot be overstated; failing to check for extraneous solutions can lead to incorrect answers.

Conclusion

In this comprehensive guide, we have walked through the detailed process of solving the rational equation ${\frac{x}{x-5}-\frac{5}{x+5}=\frac{10x}{x^2-25}}$. We began by identifying the restrictions on the variable, which are crucial for determining valid solutions. We then found the least common denominator (LCD) to clear the fractions, transforming the equation into a more manageable algebraic form. After simplifying and solving the resulting equation, we arrived at a potential solution. However, the final and critical step was to check this solution against the restrictions to identify any extraneous solutions.

In this particular case, the potential solution ${x=5}$ was found to be extraneous because it made the denominators of the original equation equal to zero. As a result, the original equation has no solution. This example highlights the importance of each step in solving rational equations, especially the verification step. It is essential to always check potential solutions against the original equation's restrictions to ensure accuracy.

Solving rational equations requires a systematic approach and a thorough understanding of algebraic principles. By following these steps carefully, you can effectively solve rational equations and avoid common pitfalls, ultimately enhancing your mathematical problem-solving skills.