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Which of the Following Is an Extraneous Solution of mc020-1.jpg?

When solving equations, it is crucial to identify the solutions that satisfy the original equation and discard any extraneous solutions. An extraneous solution is a seemingly valid solution that does not actually satisfy the original equation. In this article, we will explore the concept of extraneous solutions and discuss how to identify them. Additionally, we will examine an example equation and determine which of the given options is an extraneous solution.

Understanding Extraneous Solutions:

An extraneous solution often arises when we perform certain operations during the process of solving an equation. These operations may introduce additional constraints or conditions that restrict the validity of certain solutions. As a result, some solutions that initially seem valid may not satisfy the original equation.

To identify an extraneous solution, we must substitute the obtained solution back into the original equation and verify its validity. If the substituted value does not satisfy the original equation, it is an extraneous solution and should be discarded.

Example Equation:

Let’s consider the equation mc020-1.jpg, where x represents an unknown variable. To determine which of the given options is an extraneous solution, we will analyze each option by substituting it back into the original equation.

Option A: x = -5

Substituting -5 into the equation, we have (-5 + 3) / (-5 – 1) = (-2) / (-6) = 1/3. However, when we substitute x = -5 into the original equation, we obtain (-5 + 3) / (-5 – 1) = -2 / -6 = 1/3. Since the substituted value satisfies the original equation, option A is not an extraneous solution.

Option B: x = -1

Substituting -1 into the equation, we have (-1 + 3) / (-1 – 1) = 2 / -2 = -1. However, when we substitute x = -1 into the original equation, we obtain (-1 + 3) / (-1 – 1) = 2 / -2 = -1. Thus, option B is also not an extraneous solution.

Option C: x = 0

Substituting 0 into the equation, we have (0 + 3) / (0 – 1) = 3 / -1 = -3. However, when we substitute x = 0 into the original equation, we obtain (0 + 3) / (0 – 1) = 3 / -1 = -3. Consequently, option C is not an extraneous solution.

Option D: x = 2

Substituting 2 into the equation, we have (2 + 3) / (2 – 1) = 5 / 1 = 5. However, when we substitute x = 2 into the original equation, we obtain (2 + 3) / (2 – 1) = 5 / 1 = 5. Therefore, option D is not an extraneous solution.

Based on our analysis, we conclude that none of the given options (A, B, C, or D) is an extraneous solution of the equation mc020-1.jpg.

FAQs:

Q: What is an extraneous solution?

A: An extraneous solution is a solution that appears to be valid but does not satisfy the original equation when substituted back into it.

Q: How can we identify an extraneous solution?

A: To identify an extraneous solution, we need to substitute the obtained solution back into the original equation and check if it satisfies the equation.

Q: Why do extraneous solutions occur?

A: Extraneous solutions often occur when certain operations, such as squaring both sides of an equation, introduce additional constraints that restrict the validity of certain solutions.

Q: Are all equations prone to having extraneous solutions?

A: No, not all equations have extraneous solutions. The occurrence of extraneous solutions depends on the specific operations performed during the equation-solving process.

Q: Why is it important to identify and discard extraneous solutions?

A: Identifying and discarding extraneous solutions is important to ensure that we only consider solutions that satisfy the original equation. Including extraneous solutions can lead to incorrect conclusions or solutions that do not make sense in the given context.

In conclusion, extraneous solutions can be misleading and may not satisfy the original equation. It is crucial to perform the necessary checks and discard any solutions that do not satisfy the equation. In the example provided, none of the given options were found to be extraneous solutions. By understanding extraneous solutions and following the steps to identify them, we can ensure accurate and valid solutions when solving equations.

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