Riley's Homework Error A Step-by-Step Analysis

#H1 Riley's Homework Error A Step-by-Step Analysis

Let's delve into Riley's mathematical journey and pinpoint the exact location where she took a wrong turn. In this article, we'll meticulously analyze each step of her work, highlighting the error and providing a comprehensive explanation to ensure a clear understanding of the correct approach. Our aim is to not only identify the mistake but also to reinforce the fundamental principles of fraction manipulation in algebra.

The Problem Unveiled

Riley's task involves simplifying the expression:

$\frac{x}{x^2-5 x+6}+\frac{3}{x+3}$

Her initial steps are crucial in setting the stage for a successful solution. Let's break down each step to identify where the error lies. Understanding the problem is the first key step in finding a solution. The expression involves adding two fractions, which requires a common denominator. This means we need to factor the quadratic expression in the denominator of the first fraction and then find a common denominator for both fractions. The steps involved in simplifying this expression are critical for mastering algebraic manipulations.

Step 1: Factoring the Denominator

Riley's first move is to factor the quadratic expression in the denominator of the first fraction. This is a standard technique in simplifying rational expressions. Factoring allows us to identify common factors and simplify the expression more easily. In this case, the quadratic expression is $x^2 - 5x + 6$. We need to find two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. Therefore, the factored form of the quadratic expression is $(x - 2)(x - 3)$.

Step 1: $\frac{x}{(x-2)(x-3)}+\frac{3}{x+3}$

This step is correct. Riley has accurately factored the denominator of the first fraction. The factored form $(x - 2)(x - 3)$ is indeed the correct representation of $x^2 - 5x + 6$. This factorization is essential for finding a common denominator in the subsequent steps. Factoring the quadratic expression correctly is a critical step, and Riley's accurate work here lays the groundwork for further simplification. This demonstrates a solid understanding of factoring techniques, which are fundamental in algebra. The ability to factor quadratic expressions is a cornerstone of algebraic manipulation, and Riley's success in this step indicates a good grasp of this concept.

Step 2: The Mistake Emerges

Now, let's examine Step 2, where Riley's error occurs. This is the crucial point in our analysis, as identifying the mistake here is the primary objective. We need to carefully compare this step with the previous one to pinpoint the exact deviation from the correct procedure. The process of finding a common denominator involves manipulating the fractions so that they have the same denominator. This often involves multiplying the numerator and denominator of one or both fractions by a suitable expression.

Step 2: $\frac{x}{(x-2)(x+3)}+\frac{3(x-2)}{(x-2)(x+3)}$

Here's where the mistake lies. Instead of keeping the original factored form $(x-2)(x-3)$ in the denominator of the first fraction, Riley incorrectly wrote $(x-2)(x+3)$. This is a crucial error because it changes the entire expression and leads to an incorrect simplification. The denominator of the first fraction should remain $(x-2)(x-3)$ as established in Step 1. By changing the denominator, Riley has fundamentally altered the problem and will not arrive at the correct solution. This error highlights the importance of careful and accurate transcription in mathematical problem-solving. A seemingly small mistake like this can have significant consequences on the final result.

To correctly find a common denominator, Riley needed to multiply the numerator and denominator of the second fraction by $(x-2)(x-3)$ to match the denominator of the first fraction. The correct common denominator should involve all the unique factors present in the denominators of both fractions, which in this case are $(x-2)$, $(x-3)$, and $(x+3)$. Therefore, the common denominator should be $(x-2)(x-3)(x+3)$. The second fraction should have been manipulated to have this denominator, but Riley's error in the first fraction's denominator complicates the process.

Correcting the Course

To rectify Riley's mistake, we need to rewind to Step 2 and apply the correct procedure. The goal is to find a common denominator for the two fractions. As we've established, the correct common denominator is $(x-2)(x-3)(x+3)$. This means we need to multiply the numerator and denominator of each fraction by the factors that are missing from its current denominator.

The first fraction, $\frac{x}{(x-2)(x-3)}$, needs to be multiplied by $\frac{(x+3)}{(x+3)}$. This will give us a common denominator of $(x-2)(x-3)(x+3)$. The second fraction, $\frac{3}{x+3}$, needs to be multiplied by $\frac{(x-2)(x-3)}{(x-2)(x-3)}$. This will also result in the common denominator of $(x-2)(x-3)(x+3)$.

Let's perform these operations:

• First Fraction: $\frac{x}{(x-2)(x-3)} \cdot \frac{(x+3)}{(x+3)} = \frac{x(x+3)}{(x-2)(x-3)(x+3)} = \frac{x^2 + 3x}{(x-2)(x-3)(x+3)}$
• Second Fraction: $\frac{3}{x+3} \cdot \frac{(x-2)(x-3)}{(x-2)(x-3)} = \frac{3(x-2)(x-3)}{(x-2)(x-3)(x+3)} = \frac{3(x^2 - 5x + 6)}{(x-2)(x-3)(x+3)} = \frac{3x^2 - 15x + 18}{(x-2)(x-3)(x+3)}$

Now that both fractions have the same denominator, we can add them together. This involves adding the numerators and keeping the common denominator.

Step 3: Combining the Fractions

With the fractions now sharing a common denominator, we can proceed to combine them. This involves adding the numerators while keeping the denominator the same. This is a fundamental step in adding fractions, and it's crucial to ensure that the numerators are added correctly. The process of combining fractions with a common denominator is a core concept in algebra, and mastering this skill is essential for solving more complex problems.

$\frac{x^2 + 3x}{(x-2)(x-3)(x+3)} + \frac{3x^2 - 15x + 18}{(x-2)(x-3)(x+3)} = \frac{x^2 + 3x + 3x^2 - 15x + 18}{(x-2)(x-3)(x+3)}$

Now, we simplify the numerator by combining like terms:

$\frac{4x^2 - 12x + 18}{(x-2)(x-3)(x+3)}$

We can factor out a 2 from the numerator:

$\frac{2(2x^2 - 6x + 9)}{(x-2)(x-3)(x+3)}$

The quadratic expression $2x^2 - 6x + 9$ does not factor easily, so we leave it as is. This is an important observation, as it prevents us from attempting to factor it further unnecessarily. Recognizing when a quadratic expression cannot be factored is a key skill in algebraic simplification. It saves time and effort by avoiding fruitless attempts at factorization.

The Final Simplified Expression

Therefore, the simplified expression is:

$\frac{2(2x^2 - 6x + 9)}{(x-2)(x-3)(x+3)}$

This is the correct simplified form of the original expression. We have successfully navigated through the steps, identified Riley's mistake, and corrected it to arrive at the accurate answer. This process highlights the importance of attention to detail and the careful application of algebraic principles. The final simplified expression represents the most concise form of the original expression, and it is the result of a series of accurate algebraic manipulations.

Key Takeaways

• Accuracy in Factoring: Factoring is a crucial step in simplifying rational expressions. Ensure that the factors are correct, as errors in factoring can propagate through the entire solution.
• Common Denominator: Finding the correct common denominator is essential for adding or subtracting fractions. The common denominator must include all unique factors from the original denominators.
• Careful Transcription: Mistakes often occur during transcription. Double-check each step to ensure that the expressions are copied correctly.
• Simplify Completely: After combining fractions, simplify the resulting expression as much as possible. This may involve factoring and canceling common factors.

By understanding these key takeaways, students can avoid common mistakes and improve their problem-solving skills in algebra. The process of simplifying rational expressions is a fundamental skill, and mastering it is essential for success in higher-level mathematics. This article has provided a detailed analysis of Riley's mistake and the correct approach to solving the problem, offering valuable insights for students learning algebraic manipulation.

Conclusion

In conclusion, Riley's mistake occurred in Step 2 when she incorrectly factored the denominator. By carefully identifying and correcting this error, we were able to simplify the expression accurately. This exercise underscores the importance of precision in each step of algebraic manipulation. By focusing on accuracy and understanding the fundamental principles, students can confidently tackle similar problems and achieve success in mathematics. The ability to identify and correct mistakes is a crucial skill in problem-solving, and this analysis provides a clear example of how to approach such situations effectively. The journey through Riley's homework error serves as a valuable lesson in the importance of attention to detail and the application of correct algebraic techniques.