Solving Rational Equations Valid And Extraneous Solutions

Rational equations, a fascinating area within mathematics, often present unique challenges when solving for unknown variables. These equations, characterized by fractions with polynomials in the numerator and denominator, demand a careful approach to avoid potential pitfalls. One such pitfall is the emergence of extraneous solutions – values that satisfy the transformed equation but not the original one. In this comprehensive guide, we will explore the intricacies of solving rational equations, focusing on the crucial distinction between valid and extraneous solutions. We will dissect the given equation:

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

to determine its solutions and understand why some might be extraneous. Let's embark on this mathematical journey to master the art of solving rational equations with confidence.

Understanding Rational Equations and Extraneous Solutions

Before diving into the solution process, it's crucial to grasp the fundamental concept of rational equations and the phenomenon of extraneous solutions. Rational equations are algebraic equations containing one or more fractions where the numerator and/or the denominator are polynomials. When solving these equations, our primary goal is to isolate the variable, typically by eliminating the fractions. This often involves multiplying both sides of the equation by the least common denominator (LCD). However, this multiplication can sometimes introduce solutions that don't actually satisfy the original equation. These are what we call extraneous solutions.

The appearance of extraneous solutions stems from the fact that multiplying both sides of an equation by an expression containing a variable can change the domain of the equation. Specifically, if the expression we multiply by becomes zero for some value of the variable, that value might appear as a solution in the transformed equation, even if it makes the original equation undefined (e.g., by resulting in division by zero). For example, consider a simple equation like x = 2. If we multiply both sides by (x - 3), we get x(x - 3) = 2(x - 3), which simplifies to x^2 - 3x = 2x - 6, and further to x^2 - 5x + 6 = 0. Factoring gives us (x - 2)(x - 3) = 0, leading to solutions x = 2 and x = 3. However, x = 3 is an extraneous solution because it doesn't satisfy the original equation x = 2. This illustrates the importance of checking solutions obtained after multiplying by an expression containing a variable.

Therefore, the key takeaway is that while the process of eliminating fractions is essential for solving rational equations, we must always verify our solutions by substituting them back into the original equation. Any solution that results in an undefined expression (like division by zero) is an extraneous solution and must be discarded. Only solutions that make the original equation true are considered valid.

Solving the Equation: Step-by-Step

Now, let's tackle the given equation:

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

Our first step in solving this rational equation is to eliminate the fractions. To do this, we need to find the least common denominator (LCD) of the fractions involved. In this case, the denominators are (x + 2) and x. Therefore, the LCD is x(x + 2). We'll multiply both sides of the equation by this LCD:

$$x(x+2)\left(\frac{x}{x+2}+\frac{1}{x}\right)=1 \cdot x(x+2)$$

Distributing the LCD on the left side gives us:

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

Now, we can cancel out common factors in each term:

$$x \cdot x + (x+2) \cdot 1 = x(x+2)$$

Simplifying further, we get:

$$x^2 + x + 2 = x^2 + 2x$$

Next, we want to get all terms on one side of the equation to set it equal to zero. Subtracting x² from both sides, we have:

$$x + 2 = 2x$$

Subtracting x from both sides gives:

$$2 = x$$

So, we have found a potential solution: x = 2. However, as we discussed earlier, it's crucial to check for extraneous solutions. We need to substitute x = 2 back into the original equation to see if it holds true.

Checking for Extraneous Solutions

To determine if x = 2 is a valid solution or an extraneous one, we substitute it back into the original equation:

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

Substituting x = 2, we get:

$$\frac{2}{2+2}+\frac{1}{2}=1$$

Simplifying the fractions:

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

$$\frac{1}{2}+\frac{1}{2}=1$$

$$1 = 1$$

The equation holds true when x = 2. This means that x = 2 is a valid solution. However, we need to be thorough and ensure that there aren't other potential solutions. To do this, let's go back to the simplified equation before we isolated x:

$$x^2 + x + 2 = x^2 + 2x$$

We can rearrange this equation to get a quadratic equation equal to zero. Subtracting (x² + 2x) from both sides, we get:

$$-x + 2 = 0$$

This is a linear equation, and we've already solved it, finding x = 2. Since the equation simplified to a linear form, there will be only one solution. Therefore, we can confidently say that x = 2 is the only potential solution.

Now, let's consider the restrictions on the variable x. In the original equation,

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

we have two denominators: (x + 2) and x. The denominators cannot be equal to zero, as division by zero is undefined. Therefore, we have two restrictions:

1. x + 2 ≠ 0, which means x ≠ -2
2. x ≠ 0

Our potential solution, x = 2, does not violate these restrictions. Therefore, it remains a valid solution.

Analyzing Potential Extraneous Solutions

To further solidify our understanding, let's consider what values of x would make the original equation undefined and thus be potential extraneous solutions. As we identified earlier, the denominators in the equation are (x + 2) and x. Setting each of these equal to zero gives us:

1. x + 2 = 0 => x = -2
2. x = 0

These values, x = -2 and x = 0, are not in the domain of the original equation. If we were to obtain either of these as potential solutions during the solving process, we would immediately discard them as extraneous.

Let's imagine, for the sake of illustration, that during our solution process, we had arrived at a quadratic equation that factored to give us solutions x = 2 and x = -2. We would then need to check both solutions in the original equation. We already know that x = 2 is a valid solution. However, if we were to substitute x = -2 into the original equation, we would get:

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

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

The term -2/0 is undefined, indicating that x = -2 is an extraneous solution. Therefore, we would discard x = -2 and only accept x = 2 as the valid solution.

This exercise highlights the importance of not only solving the equation but also carefully considering the domain of the original equation and checking for extraneous solutions.

Determining the Correct Answer

Having thoroughly solved the equation and checked for extraneous solutions, we can now confidently determine the correct answer to the original question: Which statement describes the solutions of this equation?

The equation we solved is:

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

We found that the equation has one valid solution, x = 2, and no extraneous solutions. We arrived at this conclusion by:

1. Finding the least common denominator (LCD) and multiplying both sides of the equation by it to eliminate fractions.
2. Simplifying the resulting equation and solving for x.
3. Substituting the potential solution(s) back into the original equation to check for validity.
4. Identifying any restrictions on the variable x based on the denominators in the original equation.

Based on our analysis, we can evaluate the provided options:

A. The equation has two valid solutions and no extraneous solutions. B. The equation has no valid solutions and two

Our solution clearly shows that option A is incorrect, as we found only one valid solution. Option B is also incorrect since we found one valid solution.

Therefore, to accurately answer the question, we need to formulate a statement that reflects our findings. The correct statement is:

The equation has one valid solution and no extraneous solutions.

Key Takeaways and Best Practices

Solving rational equations requires a systematic approach and a keen awareness of potential pitfalls. To ensure accuracy and avoid extraneous solutions, consider these key takeaways and best practices:

1. Identify the Least Common Denominator (LCD): The first step in solving a rational equation is to find the LCD of all the fractions involved. This is crucial for eliminating the fractions and simplifying the equation.

2. Multiply Both Sides by the LCD: Multiply both sides of the equation by the LCD. This will eliminate the fractions, making the equation easier to solve.

3. Simplify and Solve: After eliminating the fractions, simplify the equation and solve for the variable. This may involve combining like terms, factoring, or using the quadratic formula.

4. Check for Extraneous Solutions: This is the most critical step. Always substitute your potential solutions back into the original equation. If a solution makes any denominator equal to zero, it is an extraneous solution and must be discarded.

5. Identify Restrictions on the Variable: Before solving, identify any values of the variable that would make the denominators in the original equation equal to zero. These values are not in the domain of the equation and cannot be valid solutions.

6. Be Thorough: Ensure you have considered all potential solutions. If the equation simplifies to a quadratic, expect up to two solutions. If it simplifies to a linear equation, expect one solution.

7. Practice Regularly: The more you practice solving rational equations, the more comfortable and confident you will become in identifying and avoiding extraneous solutions.

By adhering to these best practices, you can master the art of solving rational equations and confidently navigate the challenges they present.

Conclusion

Solving rational equations is a fundamental skill in algebra, with applications extending to various fields of mathematics and beyond. The key to success lies in understanding the underlying principles, following a systematic approach, and diligently checking for extraneous solutions. By multiplying by the LCD, simplifying, and verifying solutions, we can accurately solve these equations and avoid the trap of extraneous results. In the specific equation we explored,

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

we found that x = 2 is the sole valid solution, with no extraneous solutions present. This comprehensive guide has equipped you with the knowledge and tools necessary to tackle rational equations with confidence and precision. Remember to always prioritize checking your solutions and understanding the domain of the equation to ensure accurate and meaningful results in your mathematical endeavors.