Solving Rational Equations The Equation 5x/(x^2-25) = 25/(x^2-25) - 3/(x+5)

In mathematics, solving equations is a fundamental skill. Rational equations, which involve fractions with polynomials in the numerator and denominator, can seem daunting, but they become manageable with a systematic approach. This comprehensive guide delves into the process of solving the rational equation $\frac{5x}{x^2-25} = \frac{25}{x^2-25} - \frac{3}{x+5}$, providing a step-by-step solution and insightful explanations to enhance your understanding.

1. Understanding Rational Equations

To begin, it's crucial to understand what rational equations are. A rational equation is an equation that contains at least one fraction whose numerator and denominator are polynomials. Solving these equations involves finding the values of the variable that make the equation true. However, it's essential to be mindful of values that would make the denominator zero, as these values are excluded from the solution set.

The equation we're tackling, $\frac{5x}{x^2-25} = \frac{25}{x^2-25} - \frac{3}{x+5}$, is a classic example of a rational equation. It features fractions with polynomial expressions, and our goal is to isolate x and find its possible values.

2. Identifying Restrictions and the Domain

The first critical step in solving any rational equation is identifying the restrictions on the variable. Restrictions arise from the denominators of the fractions. We must determine which values of x would make any denominator equal to zero, as division by zero is undefined.

In our equation, we have two distinct denominators: $x^2 - 25$ and $x + 5$. Let's analyze each:

• $x^2 - 25$: This is a difference of squares, which can be factored as $(x - 5)(x + 5)$. Setting this equal to zero, we get $(x - 5)(x + 5) = 0$. This yields two restrictions: $x = 5$ and $x = -5$.
• $x + 5$: Setting this equal to zero, we get $x + 5 = 0$, which gives us the restriction $x = -5$. Notice that this restriction is already covered by the previous case.

Therefore, the restrictions on our variable are $x \neq 5$ and $x \neq -5$. These values must be excluded from our final solution set. Understanding these restrictions from the outset is crucial to avoid incorrect solutions later on. The domain of the equation, then, is all real numbers except 5 and -5. This means we are looking for solutions within the set $(-\infty, -5) \cup (-5, 5) \cup (5, \infty)$. Remember, any solution we find must be checked against these restrictions.

3. Finding the Least Common Denominator (LCD)

To effectively solve the rational equation, our next step is to eliminate the fractions. This is achieved by multiplying both sides of the equation by the least common denominator (LCD). The LCD is the smallest expression that is divisible by all the denominators in the equation.

In our case, the denominators are $x^2 - 25$ and $x + 5$. As we identified earlier, $x^2 - 25$ can be factored as $(x - 5)(x + 5)$. Thus, our denominators are essentially $(x - 5)(x + 5)$ and $(x + 5)$.

The LCD is the product of the unique factors, each raised to the highest power that appears in any denominator. Here, the factors are $(x - 5)$ and $(x + 5)$. Therefore, the LCD is $(x - 5)(x + 5)$, which is equivalent to $x^2 - 25$.

Finding the correct LCD is a critical step. A wrong LCD can lead to incorrect simplifications and, ultimately, the wrong solution. So, take your time and ensure you've correctly identified the LCD before proceeding.

4. Multiplying by the LCD and Simplifying

With the LCD identified as $x^2 - 25$, we now multiply both sides of the rational equation by this expression. This step is crucial for clearing the fractions and transforming the equation into a more manageable form.

Our original equation is: $\frac{5x}{x^2-25} = \frac{25}{x^2-25} - \frac{3}{x+5}$.

Multiplying both sides by the LCD, $x^2 - 25$, we get:

$(x^2 - 25) \cdot \frac{5x}{x^2-25} = (x^2 - 25) \cdot \left( \frac{25}{x^2-25} - \frac{3}{x+5} \right)$.

Now, we distribute the LCD on the right side:

$5x = (x^2 - 25) \cdot \frac{25}{x^2-25} - (x^2 - 25) \cdot \frac{3}{x+5}$.

Next, we simplify each term. On the left side, we simply have $5x$. On the right side:

• The first term simplifies to 25 because the $(x^2 - 25)$ terms cancel out.
• For the second term, we can rewrite $x^2 - 25$ as $(x - 5)(x + 5)$. So, we have $-(x - 5)(x + 5) \cdot \frac{3}{x+5}$. The $(x + 5)$ terms cancel out, leaving us with $-3(x - 5)$.

Our equation now looks like this: $5x = 25 - 3(x - 5)$.

This step of multiplying by the LCD and simplifying is often the most mechanically challenging part of solving rational equations. It requires careful distribution and cancellation of terms. A small error here can propagate through the rest of the solution, so double-check your work.

5. Solving the Resulting Equation

After multiplying by the LCD and simplifying, we're left with a much simpler equation: $5x = 25 - 3(x - 5)$. This is now a linear equation, which we can solve using standard algebraic techniques. This transformation is why finding the LCD and multiplying through is so effective in dealing with rational equations.

First, distribute the -3 on the right side: $5x = 25 - 3x + 15$.

Combine the constants on the right side: $5x = 40 - 3x$.

Next, add $3x$ to both sides to get all the x terms on one side: $5x + 3x = 40$, which simplifies to $8x = 40$.

Finally, divide both sides by 8 to solve for x: $x = \frac{40}{8}$, which gives us $x = 5$.

Solving the resulting equation is often the easiest part of the process, but it's crucial to perform each step carefully. A mistake in solving the linear equation can lead to an incorrect solution for the original rational equation.

6. Checking for Extraneous Solutions

The final, and arguably most important, step in solving rational equations is checking for extraneous solutions. Extraneous solutions are values that satisfy the transformed equation (in our case, the linear equation) but not the original rational equation. They arise because multiplying both sides of an equation by an expression involving x can introduce solutions that don't actually work in the original equation.

Recall that in Step 2, we identified restrictions on our variable: $x \neq 5$ and $x \neq -5$. Our solution, $x = 5$, coincides with one of these restrictions. This means that $x = 5$ is an extraneous solution because it would make the denominators $x^2 - 25$ and $x + 5$ equal to zero, rendering the original equation undefined.

Since our only potential solution is an extraneous one, we conclude that the original rational equation has no solution. This highlights the critical importance of checking for extraneous solutions. Had we skipped this step, we would have incorrectly identified $x = 5$ as a solution.

7. Conclusion: No Solution

After meticulously solving the rational equation $\frac{5x}{x^2-25} = \frac{25}{x^2-25} - \frac{3}{x+5}$, we've arrived at a definitive answer: there is no solution. This outcome underscores the significance of each step in the process, from identifying restrictions to checking for extraneous solutions. Each step plays a vital role in arriving at the correct conclusion.

Here's a recap of the key steps we followed:

1. Understanding Rational Equations: Recognizing the equation type and the importance of handling denominators.
2. Identifying Restrictions and the Domain: Determining values of x that make denominators zero and excluding them.
3. Finding the Least Common Denominator (LCD): Determining the smallest expression divisible by all denominators.
4. Multiplying by the LCD and Simplifying: Eliminating fractions and transforming the equation.
5. Solving the Resulting Equation: Solving the simplified equation for x.
6. Checking for Extraneous Solutions: Verifying that solutions satisfy the original equation and restrictions.
7. Conclusion: Determining the solution set or lack thereof.

By mastering these steps, you can confidently tackle a wide range of rational equations. Remember to always be mindful of restrictions and extraneous solutions, as they are crucial for arriving at the correct answer. The solution set is B. There is no solution.