Solving Rational Equations Identifying Valid Solutions

In this article, we will delve into the process of solving the rational equation $\frac{x}{x+2} + \frac{1}{x} = 1$. Our primary goal is to identify the correct statement that accurately describes the solutions of this equation. This involves not only finding the potential solutions but also verifying their validity by checking for extraneous solutions. Extraneous solutions are those that arise during the solving process but do not satisfy the original equation. These often occur in rational equations due to the presence of variables in the denominators, which can lead to undefined expressions if the solutions make the denominators equal to zero.

The question asks us to determine whether the equation has two valid solutions, no valid solutions, or if extraneous solutions are present. To address this, we will systematically solve the equation, identify any potential solutions, and then meticulously check each one to ensure it satisfies the original equation. This step-by-step approach will allow us to confidently determine the correct statement about the nature of the solutions.

By the end of this article, you will have a comprehensive understanding of how to solve rational equations, identify extraneous solutions, and accurately describe the solution set. This knowledge is crucial for success in algebra and calculus, where rational equations frequently appear. We will break down each step, providing clear explanations and justifications to ensure a thorough understanding of the process.

To solve the rational equation $\frac{x}{x+2} + \frac{1}{x} = 1$, our first step is to eliminate the fractions. This is achieved by multiplying both sides of the equation by the least common denominator (LCD) of the fractions involved. In this case, the denominators are $x+2$ and $x$, so the LCD is $x(x+2)$. Multiplying both sides by this LCD will clear the fractions and transform the equation into a more manageable form.

Multiplying both sides of the equation by $x(x+2)$, we get:

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

Distributing $x(x+2)$ to both terms on the left side, we have:

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

Now, we can cancel out the common factors:

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

This simplifies to:

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

Now, we have a quadratic equation. To solve it, we want to set the equation to zero. We can do this by subtracting $x^2$ and $2x$ from both sides:

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

This simplifies to:

$-x + 2 = 0$

Adding $x$ to both sides, we get:

$2 = x$

So, we have found a potential solution: $x = 2$. However, we must check if this solution is valid by substituting it back into the original equation to ensure it does not result in any undefined expressions.

Now that we have found a potential solution, $x=2$, it is crucial to check whether this solution is valid or extraneous. To do this, we substitute $x=2$ 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:

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

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

$1 = 1$

Since the equation holds true when $x=2$, this solution is valid and not extraneous. This confirms that $x=2$ is indeed a solution to the original equation.

However, it's essential to remember that rational equations can sometimes have more than one solution or even no solutions. To ensure we have considered all possibilities, we need to revisit our steps and check for any other potential solutions. In this case, we arrived at a linear equation after clearing the fractions, which typically yields only one solution. However, if we had obtained a quadratic equation, we would need to solve it using methods such as factoring, completing the square, or the quadratic formula to find all possible solutions.

In this specific problem, since we found one valid solution and our algebraic manipulations led us to a linear equation, we can confidently conclude that there is only one solution. The next step is to consider the restrictions on the variable $x$ imposed by the denominators in the original equation. The denominators are $x+2$ and $x$. We must ensure that these denominators are not equal to zero, as division by zero is undefined.

This means we must have $x+2 \neq 0$ and $x \neq 0$. Solving these inequalities, we find that $x \neq -2$ and $x \neq 0$. Our solution $x=2$ satisfies these conditions, further confirming its validity. Now, let's analyze the implications of these restrictions and their impact on the solution set.

As we identified in the previous section, the original equation $\frac{x}{x+2} + \frac{1}{x} = 1$ has restrictions on the variable $x$ due to the denominators $x+2$ and $x$. These restrictions are $x \neq -2$ and $x \neq 0$, because if $x$ were equal to either of these values, the denominators would become zero, resulting in undefined expressions.

Understanding these restrictions is crucial for identifying extraneous solutions. An extraneous solution is a value that we obtain through the algebraic solving process but does not satisfy the original equation because it violates these restrictions. In other words, an extraneous solution makes one or more of the denominators in the original equation equal to zero.

In our case, we found a potential solution $x=2$. Since $2$ is not equal to $-2$ or $0$, it does not violate any of the restrictions. Therefore, $x=2$ is a valid solution. However, if we had found a solution like $x=-2$ or $x=0$, we would have had to discard it as an extraneous solution because it would make the original equation undefined.

Now, let's consider what would happen if we had obtained a quadratic equation during the solving process, which could potentially lead to two solutions. In such cases, we would need to check both solutions against the restrictions to determine their validity. It is possible to have two valid solutions, one valid and one extraneous solution, or even two extraneous solutions, depending on the specific equation and the solutions obtained.

In this particular problem, our algebraic manipulations resulted in a linear equation, which gave us only one potential solution. After checking this solution against the restrictions, we confirmed its validity. This simplifies our analysis, but it's important to understand the general principle of checking for extraneous solutions whenever dealing with rational equations.

With the solution $x=2$ confirmed and the restrictions analyzed, we can now confidently determine the correct statement that describes the solutions of the equation. This will involve evaluating the options provided in the original problem and selecting the one that accurately reflects our findings.

Having solved the equation $\frac{x}{x+2} + \frac{1}{x} = 1$, we found one potential solution, $x=2$. We then verified that this solution is valid by substituting it back into the original equation and confirming that it satisfies the equation. Additionally, we checked for extraneous solutions by considering the restrictions imposed by the denominators, which are $x \neq -2$ and $x \neq 0$. Since $x=2$ does not violate these restrictions, it is indeed a valid solution.

Now, we can definitively state that the equation has one valid solution and no extraneous solutions. This conclusion allows us to evaluate the given options and select the one that accurately describes the solution set.

Let's revisit the options:

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

Based on our analysis, neither of these statements is correct. We found one valid solution, $x=2$, and we did not encounter any extraneous solutions.

Therefore, the correct statement should reflect that the equation has one valid solution and no extraneous solutions.

In this article, we undertook a comprehensive exploration of the rational equation $\frac{x}{x+2} + \frac{1}{x} = 1$. We meticulously solved the equation, identified the potential solution $x=2$, and rigorously checked its validity. Our analysis included verifying that the solution satisfies the original equation and ensuring that it does not violate any restrictions imposed by the denominators.

We emphasized the importance of checking for extraneous solutions, which can arise in rational equations due to the presence of variables in the denominators. Extraneous solutions are values that emerge during the solving process but do not satisfy the original equation, often because they make the denominators equal to zero.

Through our step-by-step approach, we demonstrated how to eliminate fractions by multiplying both sides of the equation by the least common denominator, simplify the resulting equation, and solve for the variable. We also highlighted the significance of analyzing restrictions on the variable to identify and discard any extraneous solutions.

Our findings revealed that the equation has one valid solution, $x=2$, and no extraneous solutions. This conclusion allows us to accurately describe the solution set and understand the nature of the equation's solutions.

The skills and concepts covered in this article are fundamental to solving rational equations and are applicable in various areas of mathematics, including algebra, calculus, and beyond. By mastering these techniques, you can confidently tackle rational equations and ensure accurate solutions.

In summary, the key takeaways from this article are:

• Solving rational equations involves eliminating fractions and simplifying the resulting equation.
• Checking for extraneous solutions is crucial to ensure the validity of the solutions.
• Restrictions on the variable, imposed by the denominators, must be considered.
• The solution set should accurately reflect the valid solutions and exclude any extraneous solutions.

With these principles in mind, you are well-equipped to solve rational equations and analyze their solutions effectively.