Missing Numerator In Rational Expression Equation A Step-by-Step Solution

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In the realm of mathematics, rational expressions often present a unique challenge, requiring a blend of algebraic manipulation and conceptual understanding to solve effectively. This article delves into a specific problem involving rational expressions, where the task is to identify a missing numerator in an equation. Our focus will be on providing a step-by-step solution, ensuring clarity and accessibility for learners of all levels. We will explore the fundamental principles governing rational expressions, such as finding common denominators, combining fractions, and simplifying the resulting expressions. By carefully examining the given equation and applying these principles, we will systematically work towards uncovering the missing numerator. This process not only yields the correct answer but also deepens our understanding of the underlying mathematical concepts. So, let's embark on this journey of mathematical exploration and unveil the missing numerator with confidence and precision.

The given equation presents a fascinating puzzle involving rational expressions. Our primary objective is to determine the missing numerator that completes the equation. To achieve this, we'll embark on a step-by-step journey, carefully dissecting each component of the equation and applying the fundamental principles of algebraic manipulation. We will begin by examining the denominators of the fractions, seeking to identify common factors and establish a common denominator for the entire expression. This crucial step will enable us to combine the fractions effectively, paving the way for simplification and ultimately revealing the missing numerator. Throughout this process, we'll emphasize clarity and precision, ensuring that each step is logically sound and mathematically accurate. Our goal is not only to arrive at the correct answer but also to cultivate a deeper understanding of the concepts governing rational expressions and their manipulation. So, let us proceed with diligence and unravel the mystery of the missing numerator, gaining valuable insights along the way.

The task at hand involves deciphering the missing piece of a mathematical equation centered around rational expressions. Specifically, we are presented with an equation containing two fractions, each possessing its own numerator and denominator, and our objective is to determine the missing numerator that satisfies the equation's balance. To accomplish this, we will need to employ a strategic approach, drawing upon our knowledge of algebraic principles and techniques. We will begin by scrutinizing the denominators of the given fractions, seeking to identify common factors or patterns that will aid in the process of finding a common denominator. Once we have successfully established a common denominator, we can proceed to combine the fractions, carefully adjusting the numerators to reflect the changes made to the denominators. This process will yield a single fraction, which we can then simplify and compare to the given form of the equation, ultimately revealing the missing numerator. Throughout this endeavor, we will prioritize clarity and accuracy, ensuring that each step is meticulously executed and logically justified. Our aim is not only to solve the problem but also to enhance our understanding of rational expressions and their manipulation, empowering us to tackle similar challenges with confidence and competence.

Problem Statement

We are given the following equation:

2cc2−2c−1+4cc2+3c−4=?(c−1)(c−1)(c+4)\frac{2 c}{c^2-2 c-1}+\frac{4 c}{c^2+3 c-4}=\frac{?}{(c-1)(c-1)(c+4)}

The question is: Which of the following is missing from the numerator?

  1. 6c2−8c−46 c^2-8 c-4
  2. 2c2+4c+8c2 c^2+4 c+8 c
  3. 2c+4c2 c+4 c

Step-by-Step Solution

1. Factor the Denominators

First, we need to factor the denominators of the fractions on the left side of the equation.

  • c2−2c−1c^2 - 2c - 1 cannot be factored easily using integers. We'll address this term later.
  • c2+3c−4c^2 + 3c - 4 can be factored as (c−1)(c+4)(c - 1)(c + 4).

So the equation becomes:

2cc2−2c−1+4c(c−1)(c+4)=?(c−1)(c−1)(c+4)\frac{2 c}{c^2-2 c-1}+\frac{4 c}{(c-1)(c+4)}=\frac{?}{(c-1)(c-1)(c+4)}

2. Find the Common Denominator

The common denominator for the left side of the equation is (c2−2c−1)(c−1)(c+4)(c^2 - 2c - 1)(c - 1)(c + 4). We need to rewrite each fraction with this common denominator.

3. Rewrite the Fractions

  • For the first fraction, we multiply the numerator and denominator by (c−1)(c+4)(c - 1)(c + 4):

    2cc2−2c−1⋅(c−1)(c+4)(c−1)(c+4)=2c(c−1)(c+4)(c2−2c−1)(c−1)(c+4)\frac{2 c}{c^2-2 c-1} \cdot \frac{(c-1)(c+4)}{(c-1)(c+4)} = \frac{2c(c-1)(c+4)}{(c^2-2c-1)(c-1)(c+4)}

  • For the second fraction, we multiply the numerator and denominator by (c2−2c−1)(c^2 - 2c - 1):

    4c(c−1)(c+4)⋅c2−2c−1c2−2c−1=4c(c2−2c−1)(c−1)(c+4)(c2−2c−1)\frac{4 c}{(c-1)(c+4)} \cdot \frac{c^2-2 c-1}{c^2-2 c-1} = \frac{4c(c^2-2c-1)}{(c-1)(c+4)(c^2-2c-1)}

4. Combine the Fractions

Now, we add the two fractions:

2c(c−1)(c+4)+4c(c2−2c−1)(c2−2c−1)(c−1)(c+4)\frac{2c(c-1)(c+4) + 4c(c^2-2c-1)}{(c^2-2c-1)(c-1)(c+4)}

5. Expand the Numerator

Expand the terms in the numerator:

  • 2c(c−1)(c+4)=2c(c2+4c−c−4)=2c(c2+3c−4)=2c3+6c2−8c2c(c-1)(c+4) = 2c(c^2 + 4c - c - 4) = 2c(c^2 + 3c - 4) = 2c^3 + 6c^2 - 8c
  • 4c(c2−2c−1)=4c3−8c2−4c4c(c^2 - 2c - 1) = 4c^3 - 8c^2 - 4c

Adding these together:

2c3+6c2−8c+4c3−8c2−4c=6c3−2c2−12c2c^3 + 6c^2 - 8c + 4c^3 - 8c^2 - 4c = 6c^3 - 2c^2 - 12c

So the combined fraction is:

6c3−2c2−12c(c2−2c−1)(c−1)(c+4)\frac{6c^3 - 2c^2 - 12c}{(c^2-2c-1)(c-1)(c+4)}

6. Analyze the Target Denominator

We want the denominator to be (c−1)(c−1)(c+4)(c-1)(c-1)(c+4). Notice that this is different from the denominator we have, (c2−2c−1)(c−1)(c+4)(c^2-2c-1)(c-1)(c+4). This suggests there may be an error in the problem statement or the provided options. However, let's proceed assuming there's a simplification we've missed.

7. Re-examine the Original Problem

Let's go back to the original equation:

2cc2−2c−1+4cc2+3c−4=?(c−1)(c−1)(c+4)\frac{2 c}{c^2-2 c-1}+\frac{4 c}{c^2+3 c-4}=\frac{?}{(c-1)(c-1)(c+4)}

We factored c2+3c−4c^2 + 3c - 4 as (c−1)(c+4)(c - 1)(c + 4). The target denominator is (c−1)(c−1)(c+4)(c-1)(c-1)(c+4). This implies that we might have to manipulate the first fraction to have a denominator of (c−1)2(c+4)(c-1)^2(c+4). However, c2−2c−1c^2 - 2c - 1 does not easily factor into (c−1)2=c2−2c+1(c-1)^2 = c^2 - 2c + 1.

8. Correcting a Possible Misunderstanding

It appears there might be a slight error in the problem statement. If the first denominator were c2−2c+1c^2 - 2c + 1, it would factor to (c−1)2(c-1)^2. Let's assume for a moment that the original equation was:

2cc2−2c+1+4cc2+3c−4=?(c−1)(c−1)(c+4)\frac{2 c}{c^2-2 c + 1}+\frac{4 c}{c^2+3 c-4}=\frac{?}{(c-1)(c-1)(c+4)}

In this case, the first denominator factors to (c−1)2(c-1)^2, and the second denominator factors to (c−1)(c+4)(c-1)(c+4).

9. Solving the Corrected Problem

With the corrected equation, we have:

2c(c−1)2+4c(c−1)(c+4)=?(c−1)2(c+4)\frac{2 c}{(c-1)^2}+\frac{4 c}{(c-1)(c+4)}=\frac{?}{(c-1)^2(c+4)}

  • The common denominator is (c−1)2(c+4)(c-1)^2(c+4).
  • Rewrite the first fraction:

    2c(c−1)2⋅(c+4)(c+4)=2c(c+4)(c−1)2(c+4)\frac{2 c}{(c-1)^2} \cdot \frac{(c+4)}{(c+4)} = \frac{2c(c+4)}{(c-1)^2(c+4)}

  • Rewrite the second fraction:

    4c(c−1)(c+4)⋅(c−1)(c−1)=4c(c−1)(c−1)2(c+4)\frac{4 c}{(c-1)(c+4)} \cdot \frac{(c-1)}{(c-1)} = \frac{4c(c-1)}{(c-1)^2(c+4)}

  • Combine the fractions:

    2c(c+4)+4c(c−1)(c−1)2(c+4)\frac{2c(c+4) + 4c(c-1)}{(c-1)^2(c+4)}

  • Expand the numerator:

    2c2+8c+4c2−4c=6c2+4c2c^2 + 8c + 4c^2 - 4c = 6c^2 + 4c

So, the combined fraction is:

6c2+4c(c−1)2(c+4)\frac{6c^2 + 4c}{(c-1)^2(c+4)}

10. Final Answer (Corrected Problem)

If the problem was intended to be:

2cc2−2c+1+4cc2+3c−4=?(c−1)(c−1)(c+4)\frac{2 c}{c^2-2 c + 1}+\frac{4 c}{c^2+3 c-4}=\frac{?}{(c-1)(c-1)(c+4)}

Then the missing numerator would be 6c2+4c6c^2 + 4c.

11. Addressing the Original Problem (with the uncorrected denominator)

Going back to our work with the original problem, we had:

6c3−2c2−12c(c2−2c−1)(c−1)(c+4)\frac{6c^3 - 2c^2 - 12c}{(c^2-2c-1)(c-1)(c+4)}

If we are forced to choose from the given options, we need to see if any of the options are a factor of our numerator or if there was a missed step in simplification. The options provided are:

  1. 6c2−8c−46 c^2-8 c-4
  2. 2c2+4c+8c2 c^2+4 c+8 c
  3. 2c+4c2 c+4 c

None of these match our derived numerator. However, option 3 simplifies to 6c6c. If we factor a 2c2c from our corrected numerator 6c2+4c6c^2 + 4c, we get 2c(3c+2)2c(3c + 2). If we factor a 6c6c from our derived numerator of the original problem, 6c3−2c2−12c6c^3 - 2c^2 - 12c, we get 2c(3c2−c−6)2c(3c^2 - c - 6). None of these gives us a direct match.

12. Conclusion and Final Answer

Based on our comprehensive analysis, it seems there's a high probability of an error in the original problem statement, specifically in the first denominator. If the first denominator was intended to be c2−2c+1c^2 - 2c + 1 instead of c2−2c−1c^2 - 2c - 1, the missing numerator would be 6c2+4c6c^2 + 4c.

However, analyzing the original problem as stated leads to a complex numerator that does not directly correspond to any of the provided options. Therefore, with the original problem, it is not possible to determine which of the given options is correct.

Final Answer (to the corrected problem): If the problem was intended to have c2−2c+1c^2 - 2c + 1 as the first denominator, the missing numerator is 6c2+4c6c^2 + 4c.

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In conclusion, unraveling the mystery of the missing numerator in the given rational expression equation required a meticulous step-by-step approach. We began by factoring the denominators, a crucial step in identifying the common denominator necessary for combining the fractions. We then expanded the numerators and combined like terms, leading us to a complex expression. However, the discrepancy between our derived denominator and the target denominator highlighted a potential error in the problem statement. By assuming a corrected version of the problem with a slightly altered denominator, we were able to arrive at a plausible solution for the missing numerator. Throughout this process, we emphasized the importance of understanding the fundamental principles of rational expression manipulation, including factoring, simplifying, and finding common denominators. We also addressed common mistakes to avoid, such as incorrect factoring and overlooking extraneous solutions. Ultimately, this exercise demonstrates the power of systematic problem-solving and the value of careful attention to detail in mathematics.

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