Solve And Check $\frac{c-4}{c-2}=\frac{c-2}{c+2}-\frac{1}{2-c}$.1. The Solution Is $c =$ $\square$.2. How Many Extraneous Solutions Are There? $\square$.

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Introduction

In mathematics, solving and checking equations is a crucial step in understanding the behavior of functions and relationships between variables. In this article, we will focus on solving and checking the equation c−4c−2=c−2c+2−12−c\frac{c-4}{c-2}=\frac{c-2}{c+2}-\frac{1}{2-c}. We will break down the solution into manageable steps and provide a clear explanation of each step.

Step 1: Simplify the Equation

To simplify the equation, we need to get rid of the fractions. We can do this by multiplying both sides of the equation by the least common multiple (LCM) of the denominators. In this case, the LCM is (c−2)(c+2)(2−c)(c-2)(c+2)(2-c).

\frac{(c-4)(c+2)(2-c)}{(c-2)(c+2)(2-c)} = \frac{(c-2)(c+2)(2-c)}{(c+2)(2-c)} - \frac{(c-2)(c+2)(2-c)}{(c-2)(2-c)}

Step 2: Expand and Simplify

Now that we have eliminated the fractions, we can expand and simplify the equation.

(c-4)(c+2)(2-c) = (c-2)(c+2)(2-c) - (c-2)(c+2)(2-c)

Expanding the left-hand side of the equation, we get:

(c^2 - 2c - 8)(2-c) = (c-2)(c+2)(2-c) - (c-2)(c+2)(2-c)

Simplifying the right-hand side of the equation, we get:

(c^2 - 2c - 8)(2-c) = (c-2)(c+2)(2-c) - (c-2)(c+2)(2-c)

Step 3: Solve for c

Now that we have simplified the equation, we can solve for c.

(c^2 - 2c - 8)(2-c) = 0

This equation can be factored as:

(c^2 - 2c - 8)(2-c) = (c-4)(c+2)(2-c) = 0

Therefore, we have three possible solutions:

c = 4, c = -2, c = 2

Step 4: Check the Solutions

Now that we have found the possible solutions, we need to check them to see if they are extraneous.

c = 4: \frac{4-4}{4-2} = \frac{4-2}{4+2}-\frac{1}{2-4} = 0 = 0 - \frac{1}{-2} = \frac{1}{2}

Since the left-hand side of the equation is not equal to the right-hand side, c = 4 is an extraneous solution.

c = -2: \frac{-2-4}{-2-2} = \frac{-2-2}{-2+2}-\frac{1}{2-(-2)} = \frac{-6}{-4} = \frac{3}{2} = \frac{3}{2} - \frac{1}{2} = \frac{2}{2} = 1

Since the left-hand side of the equation is not equal to the right-hand side, c = -2 is an extraneous solution.

c = 2: \frac{2-4}{2-2} = \frac{2-2}{2+2}-\frac{1}{2-2} = \frac{-2}{0} = \frac{0}{4} - \frac{1}{0}

Since the left-hand side of the equation is not defined, c = 2 is an extraneous solution.

Conclusion

In this article, we have solved and checked the equation c−4c−2=c−2c+2−12−c\frac{c-4}{c-2}=\frac{c-2}{c+2}-\frac{1}{2-c}. We have found that the only valid solution is c = 2, but it is an extraneous solution. Therefore, there are no valid solutions to the equation.

Discussion

The equation c−4c−2=c−2c+2−12−c\frac{c-4}{c-2}=\frac{c-2}{c+2}-\frac{1}{2-c} is a rational equation, which means that it contains fractions with variables in the numerator and denominator. To solve this equation, we need to eliminate the fractions by multiplying both sides of the equation by the least common multiple (LCM) of the denominators.

The LCM of the denominators is (c−2)(c+2)(2−c)(c-2)(c+2)(2-c). Multiplying both sides of the equation by this expression, we get:

\frac{(c-4)(c+2)(2-c)}{(c-2)(c+2)(2-c)} = \frac{(c-2)(c+2)(2-c)}{(c+2)(2-c)} - \frac{(c-2)(c+2)(2-c)}{(c-2)(2-c)}

Expanding and simplifying the equation, we get:

(c^2 - 2c - 8)(2-c) = (c-2)(c+2)(2-c) - (c-2)(c+2)(2-c)

Solving for c, we get:

(c^2 - 2c - 8)(2-c) = 0

Factoring the equation, we get:

(c-4)(c+2)(2-c) = 0

Therefore, we have three possible solutions:

c = 4, c = -2, c = 2

Checking these solutions, we find that c = 4, c = -2, and c = 2 are all extraneous solutions.

Final Answer

Introduction

In our previous article, we solved and checked the equation c−4c−2=c−2c+2−12−c\frac{c-4}{c-2}=\frac{c-2}{c+2}-\frac{1}{2-c}. We found that the only valid solution is c = 2, but it is an extraneous solution. In this article, we will answer some frequently asked questions about solving and checking equations.

Q: What is the first step in solving a rational equation?

A: The first step in solving a rational equation is to eliminate the fractions by multiplying both sides of the equation by the least common multiple (LCM) of the denominators.

Q: How do I find the LCM of the denominators?

A: To find the LCM of the denominators, you need to list the factors of each denominator and find the product of the highest power of each factor.

Q: What is the next step after eliminating the fractions?

A: After eliminating the fractions, you need to expand and simplify the equation.

Q: How do I expand and simplify the equation?

A: To expand and simplify the equation, you need to multiply out the terms and combine like terms.

Q: What is the final step in solving a rational equation?

A: The final step in solving a rational equation is to check the solutions to see if they are extraneous.

Q: How do I check the solutions?

A: To check the solutions, you need to plug each solution back into the original equation and see if it is true.

Q: What is an extraneous solution?

A: An extraneous solution is a solution that is not valid because it makes one or more of the denominators equal to zero.

Q: Why is it important to check the solutions?

A: It is important to check the solutions because an extraneous solution can be a solution that is not valid.

Q: What are some common mistakes to avoid when solving rational equations?

A: Some common mistakes to avoid when solving rational equations include:

  • Not eliminating the fractions
  • Not expanding and simplifying the equation
  • Not checking the solutions
  • Not being careful when multiplying out the terms

Q: How can I practice solving rational equations?

A: You can practice solving rational equations by working through examples and exercises in a textbook or online resource.

Q: What are some real-world applications of solving rational equations?

A: Solving rational equations has many real-world applications, including:

  • Physics: Solving rational equations is used to describe the motion of objects and the behavior of physical systems.
  • Engineering: Solving rational equations is used to design and optimize systems, such as electrical circuits and mechanical systems.
  • Economics: Solving rational equations is used to model economic systems and make predictions about economic behavior.

Conclusion

In this article, we have answered some frequently asked questions about solving and checking equations. We have also provided some tips and examples to help you practice solving rational equations. Remember to always eliminate the fractions, expand and simplify the equation, and check the solutions to ensure that you are getting valid solutions.