Solving Rational Equations A Step By Step Guide

Introduction

In the realm of mathematics, solving equations is a fundamental skill. Among the various types of equations, rational equations often pose a unique challenge due to the presence of variables in the denominators. This article delves into the step-by-step process of solving a specific rational equation: ${\frac{5x}{x^2-9} = \frac{15}{x^2-9} - \frac{4}{x+3}}$. We will explore the underlying principles, identify potential pitfalls, and provide a clear, comprehensive solution. Understanding how to tackle such equations is crucial for anyone studying algebra or related fields. Mastering these techniques not only enhances your problem-solving capabilities but also lays a solid foundation for more advanced mathematical concepts. Our goal is to break down each step, making the process accessible and understandable, so you can confidently solve similar problems in the future. Let's begin by setting the stage and understanding the key components of this equation.

Understanding Rational Equations

Rational equations are equations that contain at least one fraction whose numerator and denominator are polynomials. These equations require a careful approach because we must consider the values of the variable that would make the denominator zero, as division by zero is undefined. In our given equation, ${\frac{5x}{x^2-9} = \frac{15}{x^2-9} - \frac{4}{x+3}}$, we can see that the denominators involve expressions with ${x}$. Before we even begin to solve for ${x}$, it's important to identify any values of ${x}$ that would make the denominators equal to zero. These values are known as restricted values or excluded values, and they must be excluded from our final solution set. Failing to account for these values can lead to incorrect or extraneous solutions. Rational equations appear frequently in various mathematical contexts, including calculus, physics, and engineering, making their mastery essential. Therefore, let's proceed by examining the denominators in our equation to determine the restricted values.

Identifying Restricted Values

The first critical step in solving any rational equation is to identify the values of the variable that make the denominators equal to zero. These values are called restricted values because they make the expression undefined. In our equation, ${\frac{5x}{x^2-9} = \frac{15}{x^2-9} - \frac{4}{x+3}}$, we have two distinct denominators: ${x^2 - 9}$ and ${x + 3}$. To find the restricted values, we set each denominator equal to zero and solve for ${x}$. For the first denominator, we have ${x^2 - 9 = 0}$. This is a difference of squares, which can be factored as ${(x - 3)(x + 3) = 0}$. Setting each factor equal to zero gives us ${x - 3 = 0}$ and ${x + 3 = 0}$, which leads to ${x = 3}$ and ${x = -3}$. For the second denominator, we have ${x + 3 = 0}$, which simply gives us ${x = -3}$. Therefore, the restricted values for this equation are ${x = 3}$ and ${x = -3}$. These values must be excluded from our solution set. By identifying these restricted values early on, we avoid potential errors later in the solving process. Now that we know the values ${x}$ cannot take, we can proceed to clear the fractions and simplify the equation.

Clearing the Fractions

To solve the rational equation ${\frac{5x}{x^2-9} = \frac{15}{x^2-9} - \frac{4}{x+3}}$, the next crucial step is to clear the fractions. This simplifies the equation and makes it easier to work with. To do this, we need to find the least common denominator (LCD) of all the fractions in the equation. The denominators are ${x^2 - 9}$ and ${x + 3}$. As we noted earlier, ${x^2 - 9}$ can be factored as ${(x - 3)(x + 3)}$. Therefore, the LCD is ${(x - 3)(x + 3)}$, which is equivalent to ${x^2 - 9}$. We multiply both sides of the equation by the LCD to eliminate the fractions. Multiplying both sides by ${(x^2 - 9)}$, we get:

${ (x^2 - 9) \cdot \frac{5x}{x^2-9} = (x^2 - 9) \cdot \left(\frac{15}{x^2-9} - \frac{4}{x+3}\right) }$

This simplifies to:

${ 5x = (x^2 - 9) \cdot \frac{15}{x^2-9} - (x^2 - 9) \cdot \frac{4}{x+3} }$

Now, we can cancel out terms:

${ 5x = 15 - 4 \cdot \frac{x^2 - 9}{x+3} }$

Since ${x^2 - 9 = (x - 3)(x + 3)}$, we can further simplify:

${ 5x = 15 - 4 \cdot \frac{(x - 3)(x + 3)}{x+3} }$

${ 5x = 15 - 4(x - 3) }$

Clearing the fractions transforms the original rational equation into a linear equation, which is much easier to solve. This step is critical in simplifying the problem and moving closer to finding the solution. Next, we will simplify and solve the resulting equation.

Simplifying and Solving the Equation

After clearing the fractions in the rational equation ${\frac{5x}{x^2-9} = \frac{15}{x^2-9} - \frac{4}{x+3}}$, we arrived at the simplified equation ${5x = 15 - 4(x - 3)}$. Now, we need to further simplify and solve for ${x}$. First, let's distribute the -4 across the parentheses:

${ 5x = 15 - 4x + 12 }$

Next, combine the constants on the right side of the equation:

${ 5x = 27 - 4x }$

Now, we want to isolate ${x}$ on one side of the equation. Add ${4x}$ to both sides:

${ 5x + 4x = 27 }$

${ 9x = 27 }$

Finally, divide both sides by 9 to solve for ${x}$:

${ x = \frac{27}{9} }$

${ x = 3 }$

So, we have found a potential solution: ${x = 3}$. However, we must remember the restricted values we identified earlier. It's crucial to check whether this solution is valid or if it's an extraneous solution. The next step is to verify our solution against the restricted values.

Checking for Extraneous Solutions

In solving rational equations, it is essential to check for extraneous solutions. These are solutions that we obtain algebraically but do not satisfy the original equation because they make one or more denominators equal to zero. We previously found the potential solution ${x = 3}$ for the equation ${\frac{5x}{x^2-9} = \frac{15}{x^2-9} - \frac{4}{x+3}}$. We also identified the restricted values as ${x = 3}$ and ${x = -3}$. Since our potential solution, ${x = 3}$, is a restricted value, it makes the denominators ${x^2 - 9}$ and ${x + 3}$ equal to zero. Therefore, ${x = 3}$ is an extraneous solution and must be discarded. This means that the original equation has no solution, because the only value we found that could be a solution is actually a restricted value. This underscores the importance of checking for extraneous solutions in rational equations. If we had not checked, we might have incorrectly concluded that ${x = 3}$ is a valid solution. In this case, the correct conclusion is that there is no solution to the equation. This can be a common occurrence with rational equations, making the verification step absolutely necessary. With this understanding, let's summarize our findings.

Conclusion: No Solution

In summary, we embarked on a journey to solve the rational equation ${\frac{5x}{x^2-9} = \frac{15}{x^2-9} - \frac{4}{x+3}}$. We began by understanding the nature of rational equations and the importance of identifying restricted values. We found that the restricted values for this equation are ${x = 3}$ and ${x = -3}$, as these values would make the denominators zero, resulting in an undefined expression. We then proceeded to clear the fractions by multiplying both sides of the equation by the least common denominator, which was ${x^2 - 9}$. This transformed the rational equation into a simpler linear equation. Solving the simplified equation, we arrived at a potential solution of ${x = 3}$. However, we emphasized the critical step of checking for extraneous solutions. Upon verifying our potential solution against the restricted values, we found that ${x = 3}$ is indeed an extraneous solution because it is one of the values that make the denominators zero. Therefore, we conclude that the original equation has no solution. This comprehensive approach highlights the key steps in solving rational equations: identifying restricted values, clearing fractions, solving the resulting equation, and, most importantly, checking for extraneous solutions. By following these steps, you can confidently tackle rational equations and avoid common pitfalls.