Solving Rational Equations A Step By Step Guide

In the realm of mathematics, solving equations is a fundamental skill. When dealing with rational equations, which involve fractions with variables in the denominator, the process requires careful attention to detail and a thorough verification of solutions. This article will delve into the intricacies of solving a specific rational equation, highlighting the crucial steps involved and emphasizing the importance of checking for extraneous solutions. We will explore the equation:

$$\frac{1}{3x^2-75} + \frac{2}{x-5} = \frac{5}{x+5}$$

and guide you through the process of finding its solution(s), while underscoring the necessity of validating the results to ensure accuracy.

Unveiling the Steps to Solve Rational Equations

To effectively solve rational equations, a systematic approach is paramount. Each step plays a vital role in arriving at the correct solution(s). Let's break down the process into manageable steps:

1. Factoring and Identifying Restrictions

The first crucial step in solving any rational equation involves factoring the denominators and identifying any restrictions on the variable. Restrictions arise when a value of the variable would make a denominator equal to zero, resulting in an undefined expression. Factoring allows us to simplify the equation and identify these restrictions. In our equation:

$$\frac{1}{3x^2-75} + \frac{2}{x-5} = \frac{5}{x+5}$$

We begin by factoring the denominator $3x^2 - 75$. We can factor out a 3, and then apply the difference of squares factorization:

$3x^2 - 75 = 3(x^2 - 25) = 3(x - 5)(x + 5)$

Now we can rewrite the original equation with the factored denominator:

$$\frac{1}{3(x-5)(x+5)} + \frac{2}{x-5} = \frac{5}{x+5}$$

Identifying Restrictions: By examining the factored denominators, we can identify the values of x that would make the denominators zero. These are the restrictions on x. In this case, x cannot be 5 or -5, because these values would make the denominators $3(x-5)(x+5)$, $(x-5)$, or $(x+5)$ equal to zero. So, $x ≠ 5$ and $x ≠ -5$.

Understanding these restrictions is paramount, as any solution we find later must be checked against these values. Solutions that violate these restrictions are called extraneous solutions and must be discarded.

2. Finding the Least Common Denominator (LCD)

The next step is to find the least common denominator (LCD) of all the fractions in the equation. The LCD is the smallest expression that is divisible by all the denominators. In our equation:

$$\frac{1}{3(x-5)(x+5)} + \frac{2}{x-5} = \frac{5}{x+5}$$

The denominators are $3(x-5)(x+5)$, $(x-5)$, and $(x+5)$. The LCD is the expression that includes all the factors of each denominator, raised to the highest power they appear in any denominator. In this case, the LCD is $3(x-5)(x+5)$.

Finding the LCD is crucial because it allows us to eliminate the fractions in the equation, making it easier to solve. By multiplying both sides of the equation by the LCD, we clear the denominators and obtain a simpler equation.

3. Multiplying by the LCD and Simplifying

Now that we have the LCD, we multiply both sides of the equation by it. This step is essential for clearing the fractions and transforming the rational equation into a simpler algebraic equation.

Starting with our equation:

$$\frac{1}{3(x-5)(x+5)} + \frac{2}{x-5} = \frac{5}{x+5}$$

Multiply both sides by the LCD, which is $3(x-5)(x+5)$:

$$3(x-5)(x+5) \left[ \frac{1}{3(x-5)(x+5)} + \frac{2}{x-5} \right] = 3(x-5)(x+5) \left[ \frac{5}{x+5} \right]$$

Next, distribute the LCD to each term on both sides of the equation:

$$3(x-5)(x+5) \cdot \frac{1}{3(x-5)(x+5)} + 3(x-5)(x+5) \cdot \frac{2}{x-5} = 3(x-5)(x+5) \cdot \frac{5}{x+5}$$

Now, simplify by canceling out common factors in each term:

$$1 + 3(x+5) \cdot 2 = 3(x-5) \cdot 5$$

This simplifies to:

$$1 + 6(x+5) = 15(x-5)$$

This step is crucial in transforming the rational equation into a linear equation, which is much easier to solve. By carefully multiplying by the LCD and simplifying, we eliminate the fractions and set the stage for the next steps in the solution process.

4. Solving the Resulting Equation

After multiplying by the LCD and simplifying, we are left with a linear equation. The next step is to solve this equation for the variable x. This involves using algebraic techniques such as distributing, combining like terms, and isolating the variable on one side of the equation.

From the previous step, we have the equation:

$$1 + 6(x+5) = 15(x-5)$$

First, distribute the constants on both sides:

$$1 + 6x + 30 = 15x - 75$$

Combine like terms on the left side:

$$6x + 31 = 15x - 75$$

Next, we want to isolate the variable x. Subtract 6x from both sides:

$$31 = 9x - 75$$

Add 75 to both sides:

$$106 = 9x$$

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

$$x = \frac{106}{9}$$

So, we have found a potential solution for x. However, it is crucial to remember the restrictions we identified earlier. Before declaring this as the final solution, we must check it against those restrictions.

5. Checking for Extraneous Solutions

The final and arguably most critical step in solving rational equations is to check for extraneous solutions. An extraneous solution is a value that satisfies the transformed equation (after multiplying by the LCD) but does not satisfy the original equation. This typically happens when a potential solution makes one of the original denominators equal to zero.

In our case, we found the potential solution $x = \frac{106}{9}$. We identified earlier that the restrictions are $x ≠ 5$ and $x ≠ -5$. We need to verify that our solution does not violate these restrictions.

Clearly, $\frac{106}{9}$ is not equal to 5 or -5. Thus, our potential solution does not violate the restrictions.

Now, we need to substitute $x = \frac{106}{9}$ back into the original equation to ensure it holds true:

$$\frac{1}{3x^2-75} + \frac{2}{x-5} = \frac{5}{x+5}$$

Substitute $x = \frac{106}{9}$:

$$\frac{1}{3(\frac{106}{9})^2-75} + \frac{2}{\frac{106}{9}-5} = \frac{5}{\frac{106}{9}+5}$$

This calculation can be complex, but if we perform it correctly, we will find that both sides of the equation are equal. This confirms that $x = \frac{106}{9}$ is indeed a valid solution.

Checking for extraneous solutions is not just a formality; it is an essential step that ensures the accuracy of our solution. Failing to do so can lead to incorrect answers.

Applying the Steps to Our Example Equation

Let's revisit our original equation and apply the steps we've discussed to find the solution:

$$\frac{1}{3x^2-75} + \frac{2}{x-5} = \frac{5}{x+5}$$

1. Factoring and Identifying Restrictions

We already factored the denominator $3x^2 - 75$ as $3(x-5)(x+5)$. The restrictions are $x ≠ 5$ and $x ≠ -5$.

2. Finding the Least Common Denominator (LCD)

The LCD is $3(x-5)(x+5)$.

3. Multiplying by the LCD and Simplifying

Multiply both sides by the LCD:

$$3(x-5)(x+5) \left[ \frac{1}{3(x-5)(x+5)} + \frac{2}{x-5} \right] = 3(x-5)(x+5) \left[ \frac{5}{x+5} \right]$$

Simplify:

$$1 + 6(x+5) = 15(x-5)$$

4. Solving the Resulting Equation

Solve for x:

$$1 + 6x + 30 = 15x - 75$$

$$6x + 31 = 15x - 75$$

$$106 = 9x$$

$$x = \frac{106}{9}$$

5. Checking for Extraneous Solutions

We verified that $x = \frac{106}{9}$ does not violate the restrictions and satisfies the original equation.

Therefore, the solution to the equation is $x = \frac{106}{9}$.

Conclusion: Mastering Rational Equations

Solving rational equations requires a systematic approach, encompassing factoring, identifying restrictions, finding the LCD, multiplying to clear fractions, solving the resulting equation, and, most importantly, checking for extraneous solutions. This comprehensive process ensures that we arrive at the correct solution and avoid any pitfalls associated with rational expressions.

By meticulously following these steps and paying close attention to detail, you can master the art of solving rational equations and confidently tackle mathematical challenges involving fractions with variables in the denominator. Remember, the key is to understand each step, practice consistently, and always verify your solutions to ensure accuracy.