Understanding Valid and Extraneous Solutions

Oct 13, 2025 by recoveryshake.com 75 views

Introduction

When it comes to solving equations, it's essential to understand the different types of solutions that can arise. In this article, we'll delve into the world of equations and explore the concept of valid and extraneous solutions. We'll examine a specific equation and determine which statement accurately describes its solutions.

The Equation

The given equation is:

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

Understanding Valid and Extraneous Solutions

Before we dive into solving the equation, let's define what valid and extraneous solutions are.

Valid solutions: These are the solutions that satisfy the original equation and are not introduced by the process of solving the equation.
Extraneous solutions: These are the solutions that are introduced by the process of solving the equation but do not satisfy the original equation.

Solving the Equation

To solve the equation, we can start by combining the fractions on the left-hand side:

\frac{x}{x+2} + \frac{1}{x} = \frac{x^2 + (x+2)}{x(x+2)}

Simplifying the expression, we get:

\frac{x^2 + x + 2}{x(x+2)} = 1

Multiplying both sides by $x(x+2)$ to eliminate the fraction, we get:

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

Subtracting $x^2 + 2x$ from both sides, we get:

x + 2 = 2x

Subtracting $x$ from both sides, we get:

2 = x

Analyzing the Solutions

Now that we have the solution, let's analyze it to determine if it's a valid or extraneous solution.

Substituting $x = 2$ into the original equation, we get:

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

Simplifying the expression, we get:

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

Further simplifying, we get:

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

Which is true. Therefore, $x = 2$ is a valid solution.

Conclusion

In conclusion, the equation $\frac{x}{x+2} + \frac{1}{x} = 1$ has one valid solution, $x = 2$ , and no extraneous solutions.

Final Answer

The correct statement that describes the solutions of the equation is:

A. The equation has one valid solution and no extraneous solutions.

This statement accurately describes the solutions of the equation, as we have found one valid solution and no extraneous solutions.

Introduction

The Equation

The given equation is:

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

Understanding Valid and Extraneous Solutions

Before we dive into solving the equation, let's define what valid and extraneous solutions are.

Valid solutions: These are the solutions that satisfy the original equation and are not introduced by the process of solving the equation.
Extraneous solutions: These are the solutions that are introduced by the process of solving the equation but do not satisfy the original equation.

Solving the Equation

To solve the equation, we can start by combining the fractions on the left-hand side:

\frac{x}{x+2} + \frac{1}{x} = \frac{x^2 + (x+2)}{x(x+2)}

Simplifying the expression, we get:

\frac{x^2 + x + 2}{x(x+2)} = 1

Multiplying both sides by $x(x+2)$ to eliminate the fraction, we get:

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

Subtracting $x^2 + 2x$ from both sides, we get:

x + 2 = 2x

Subtracting $x$ from both sides, we get:

2 = x

Analyzing the Solutions

Now that we have the solution, let's analyze it to determine if it's a valid or extraneous solution.

Substituting $x = 2$ into the original equation, we get:

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

Simplifying the expression, we get:

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

Further simplifying, we get:

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

Which is true. Therefore, $x = 2$ is a valid solution.

Conclusion

In conclusion, the equation $\frac{x}{x+2} + \frac{1}{x} = 1$ has one valid solution, $x = 2$ , and no extraneous solutions.

Final Answer

The correct statement that describes the solutions of the equation is:

A. The equation has one valid solution and no extraneous solutions.

This statement accurately describes the solutions of the equation, as we have found one valid solution and no extraneous solutions.

Q&A

Q: What is the difference between a valid and extraneous solution?

A: A valid solution is a solution that satisfies the original equation and is not introduced by the process of solving the equation. An extraneous solution is a solution that is introduced by the process of solving the equation but does not satisfy the original equation.

Q: How do you determine if a solution is valid or extraneous?

A: To determine if a solution is valid or extraneous, you need to substitute the solution into the original equation and check if it satisfies the equation. If the solution satisfies the equation, it is a valid solution. If the solution does not satisfy the equation, it is an extraneous solution.

Q: Can an equation have multiple valid solutions?

A: Yes, an equation can have multiple valid solutions. However, in the case of the equation $\frac{x}{x+2} + \frac{1}{x} = 1$ , we found only one valid solution, $x = 2$ .

Q: Can an equation have no valid solutions?

A: Yes, an equation can have no valid solutions. This occurs when the equation has no real solutions or when the solutions are complex numbers.

Q: What is the importance of understanding valid and extraneous solutions?

A: Understanding valid and extraneous solutions is crucial in solving equations, as it helps you determine the accuracy of the solutions and avoid incorrect conclusions.

Q: Can you provide an example of an equation with extraneous solutions?

A: Consider the equation $x^2 = 4$ . The solutions to this equation are $x = 2$ and $x = -2$ . However, if we substitute $x = -2$ into the original equation, we get $(-2)^2 = 4$ , which is true. Therefore, $x = -2$ is a valid solution. However, if we substitute $x = 3$ into the original equation, we get $3^2 = 4$ , which is not true. Therefore, $x = 3$ is an extraneous solution.

Q: Can you provide an example of an equation with no valid solutions?

A: Consider the equation $x^2 + 1 = 0$ . This equation has no real solutions, as the square of any real number is non-negative. Therefore, this equation has no valid solutions.

Q: Can you provide an example of an equation with multiple valid solutions?

A: Consider the equation $x^2 - 4x + 4 = 0$ . This equation can be factored as $(x - 2)^2 = 0$ . Therefore, the solution to this equation is $x = 2$ , which is a valid solution. However, if we substitute $x = 2$ into the original equation, we get $(2 - 2)^2 = 0$ , which is true. Therefore, $x = 2$ is a valid solution.