Riley's Math Mistake Unveiled Error Analysis In Rational Expression Simplification

Introduction: The Case of the Errant Homework Step

In the realm of mathematics, precision is paramount. A single misstep can lead to an entirely incorrect solution, turning a seemingly straightforward problem into a perplexing puzzle. This is precisely what happened to Riley as she tackled her homework, specifically when simplifying a rational expression. In this article, we'll dissect Riley's work, pinpoint the exact location of her error in step 2, and illuminate the correct path to solving the problem. This exploration is not just about finding a mistake; it's about understanding the underlying principles of algebraic manipulation and the importance of meticulous attention to detail. This exercise is invaluable for anyone learning algebra, as it highlights common pitfalls and reinforces the necessity of careful, step-by-step problem-solving. So, let's dive into the mathematical mystery and uncover where Riley's journey went awry. To begin, we will revisit the original problem Riley was trying to solve, carefully analyzing each step to identify the deviation from the correct methodology. Understanding the nature of the mistake is crucial, not just for correcting this specific problem, but for preventing similar errors in future mathematical endeavors. By the end of this article, you'll not only know what Riley did wrong but also have a strengthened grasp of how to correctly simplify rational expressions. This knowledge will empower you to approach similar problems with confidence and accuracy, making your mathematical journey smoother and more successful.

The Problem Unveiled: A Rational Expression Challenge

The problem Riley was grappling with involves the simplification of a rational expression. Rational expressions, which are essentially fractions with polynomials in the numerator and denominator, require a specific set of rules and techniques to manipulate. The problem at hand is:

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

This expression presents a classic challenge in algebra: how to combine two fractions with different denominators into a single, simplified fraction. The key to solving this lies in finding a common denominator, a fundamental concept in fraction arithmetic that extends seamlessly to algebraic expressions. However, before we can find a common denominator, we often need to factor the polynomials in the denominators, a crucial step that allows us to identify the building blocks of the expressions and determine the least common multiple. Factoring polynomials involves breaking them down into simpler expressions that, when multiplied together, yield the original polynomial. This process is akin to finding the prime factors of a number and is essential for simplifying rational expressions. Without proper factorization, finding the common denominator becomes significantly more complex, potentially leading to errors and a convoluted solution. The next sections will delve into the steps Riley took, highlighting where the process went awry and explaining the correct methodology for simplifying the expression.

Step-by-Step Analysis: Riley's Attempt

To understand Riley's mistake, let's meticulously examine each step she took in attempting to simplify the rational expression. This step-by-step analysis will not only pinpoint the error but also provide a clear contrast between the incorrect approach and the correct methodology. By carefully comparing Riley's steps with the proper procedures, we can gain a deeper understanding of the underlying concepts and avoid similar mistakes in the future. This detailed examination is crucial for solidifying our grasp of algebraic manipulations and ensuring accuracy in problem-solving. So, let's embark on this analytical journey, retracing Riley's steps and illuminating the path to the correct solution.

Step 1: Factoring the Denominator

Riley's first step was to factor the quadratic expression in the denominator of the first fraction. This is a crucial step in simplifying rational expressions, as it helps identify common factors and facilitates finding a common denominator. The original denominator was $x^2 - 5x + 6$. Riley correctly factored this quadratic expression into $(x-2)(x-3)$. This factorization is accurate because $(x-2)(x-3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6$. This demonstrates a solid understanding of factoring quadratic expressions, a fundamental skill in algebra. Riley's ability to correctly factor the denominator in this step sets a good foundation for the rest of the problem. The expression after Step 1 looks like this:

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

This correctly factored form is essential for identifying the least common denominator needed to combine the two fractions. Without this accurate factorization, the subsequent steps would be significantly more challenging, if not impossible. Therefore, Riley's successful completion of Step 1 is a testament to her understanding of basic algebraic principles. However, as we will see, a mistake in a later step will negate this initial correct move. The next step is where the critical error occurs, highlighting the importance of maintaining accuracy throughout the entire problem-solving process.

Step 2: The Mistake Emerges

This is where Riley's solution veers off course. In Step 2, she attempts to find a common denominator to combine the two fractions. However, a critical error is made in the process. Let's examine what Riley did:

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

The mistake is evident: Riley incorrectly changed the factor $(x-3)$ in the denominator of the first fraction to $(x+3)$. This is a significant error because it fundamentally alters the expression and leads to an incorrect common denominator. The correct denominator from Step 1 was $(x-2)(x-3)$, and it should have remained as such. By changing it to $(x+3)$, Riley introduced an error that invalidates the subsequent steps. This type of error, a simple sign change, is a common pitfall in algebra, emphasizing the importance of carefully reviewing each step to ensure accuracy. To correctly find a common denominator, Riley needed to multiply the second fraction by $\frac{(x-2)(x-3)}{(x-2)(x-3)}$ and the first fraction by $\frac{x+3}{x+3}$. Instead, she only multiplied the numerator of the second fraction and incorrectly altered the denominator of the first. This misstep highlights the importance of understanding the fundamental principles of fraction addition and the necessity of maintaining consistency throughout the problem-solving process. The incorrect denominator in Step 2 will propagate through the rest of the solution, leading to a wrong answer. This underscores the critical nature of each step in algebraic manipulation and the need for meticulous attention to detail.

Step 3: (Consequence of the Error)

Due to the error in Step 2, the subsequent steps will inevitably lead to an incorrect result. We can see that Step 3 builds upon the flawed foundation laid in the previous step, further compounding the initial mistake. Because the denominator in the first fraction was incorrectly changed, the combination of the numerators will not yield the correct simplified expression. It's important to recognize that in mathematics, errors tend to propagate. One small mistake can have a cascading effect, rendering the final answer completely wrong. This is why it's so crucial to double-check each step and ensure that all operations are performed accurately. In Riley's case, the error in Step 2 means that the terms being added in Step 3 are not equivalent to the original expression. Therefore, even if the addition and simplification in Step 3 are performed correctly based on the incorrect premise, the final result will still be wrong. This highlights the interconnectedness of mathematical steps and the importance of maintaining accuracy throughout the entire problem-solving process. To illustrate, let's look at the expression Riley has at the beginning of Step 3:

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

This expression is a direct result of the error in Step 2. The numerator is a combination of the numerators from the fractions in Step 2, and the denominator is the incorrect common denominator. If Riley were to continue simplifying this expression, she would obtain an answer that is not equivalent to the original problem. This underscores the crucial role of Step 2 in setting the stage for the rest of the solution. Because Step 2 contains a fundamental error, the subsequent steps are essentially working with a different problem than the one originally presented.

The Correct Approach: A Step-by-Step Solution

To rectify Riley's mistake and arrive at the correct solution, let's walk through the proper steps for simplifying the given rational expression. This will not only correct the error but also provide a clear and concise methodology for tackling similar problems in the future. By understanding the correct approach, we can avoid common pitfalls and develop a stronger foundation in algebraic manipulation. This step-by-step solution will serve as a guide, illustrating the importance of each step and the logical flow of the problem-solving process. So, let's embark on this journey to mathematical accuracy, where we'll see how to correctly combine and simplify rational expressions.

Step 1: Factoring the Denominator (Revisited)

As Riley correctly did, the first step is to factor the denominator of the first fraction:

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

This step is crucial because it allows us to identify the factors needed to create a common denominator. Factoring the quadratic expression into its linear factors is a fundamental skill in algebra, and it's essential for simplifying rational expressions. By correctly factoring the denominator, we set the stage for finding the least common multiple, which is the key to combining the fractions. This step demonstrates the importance of recognizing patterns in algebraic expressions and applying the appropriate factoring techniques. Without this initial factorization, the subsequent steps would be significantly more complex. So, ensuring the accuracy of this step is paramount to the overall success of the solution.

Step 2: Finding the Common Denominator (Corrected)

The crucial step is to correctly identify and create a common denominator. The denominators we have are $(x-2)(x-3)$ and $(x+3)$. Therefore, the least common denominator (LCD) is $(x-2)(x-3)(x+3)$. To achieve this common denominator, we need to multiply each fraction by a suitable form of 1:

• Multiply the first fraction by $\frac{x+3}{x+3}$:

  $$\frac{x}{(x-2)(x-3)} \cdot \frac{x+3}{x+3} = \frac{x(x+3)}{(x-2)(x-3)(x+3)}$$

• Multiply the second fraction by $\frac{(x-2)(x-3)}{(x-2)(x-3)}$:

  $$\frac{3}{x+3} \cdot \frac{(x-2)(x-3)}{(x-2)(x-3)} = \frac{3(x-2)(x-3)}{(x-2)(x-3)(x+3)}$$

Now, both fractions have the same denominator, allowing us to combine them. This step highlights the importance of multiplying both the numerator and the denominator by the same expression to maintain the value of the fraction. It also demonstrates the systematic approach to finding a common denominator, which involves identifying the missing factors in each denominator and multiplying accordingly. This process ensures that the fractions are equivalent to their original forms while having a common denominator, which is essential for performing addition or subtraction. The result of this step is two fractions with a common denominator, ready for the numerators to be combined.

Step 3: Combining the Fractions

Now that we have a common denominator, we can combine the fractions:

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

This step involves adding the numerators while keeping the common denominator. It's a straightforward application of fraction addition, but it's crucial to ensure that the numerators are combined correctly. This step sets the stage for simplifying the expression by expanding and combining like terms in the numerator. The common denominator remains unchanged throughout this process, emphasizing its role as the foundation for the combined fraction. The next step will focus on simplifying the numerator, which will ultimately lead to the simplified form of the rational expression.

Step 4: Simplifying the Numerator

Next, we simplify the numerator by expanding and combining like terms:

$$\begin{aligned} x(x+3) + 3(x-2)(x-3) &= x^2 + 3x + 3(x^2 - 5x + 6) \\ &= x^2 + 3x + 3x^2 - 15x + 18 \\ &= 4x^2 - 12x + 18 \end{aligned}$$

This step involves careful application of the distributive property and combining like terms. Accuracy is paramount here, as any error in the expansion or combination process will lead to an incorrect simplified expression. This step demonstrates the importance of algebraic manipulation skills, including the ability to expand products and identify like terms. The simplified numerator is now a quadratic expression, which may or may not be further factorable. The next step will involve placing this simplified numerator back over the common denominator and looking for opportunities to simplify the fraction further.

Step 5: The Final Simplified Expression

Now, we place the simplified numerator back over the common denominator:

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

We can factor out a 2 from the numerator:

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

At this point, we check to see if the quadratic in the numerator can be factored further. In this case, $2x^2 - 6x + 9$ does not factor nicely using integers. Also, there are no common factors between the numerator and the denominator that can be canceled. Therefore, this is the simplified form of the expression. This final step demonstrates the importance of checking for any further simplifications, such as factoring and canceling common factors. In this case, the expression is simplified as much as possible, leaving us with the final answer. This concludes the correct solution to the problem, highlighting the importance of each step in the process and the need for accuracy and attention to detail.

Conclusion: Learning from Mistakes

Riley's mistake in Step 2 serves as a valuable lesson in the importance of careful attention to detail when working with algebraic expressions. A seemingly small error, such as incorrectly changing a sign, can have significant consequences and lead to an incorrect solution. By analyzing Riley's work, we've not only identified the specific error but also reinforced the correct steps for simplifying rational expressions. This process of error analysis is crucial for learning mathematics effectively. It allows us to understand not just the “what” but also the “why” behind mathematical procedures. By recognizing common pitfalls, we can develop strategies to avoid them and build a stronger foundation in algebraic concepts. This exercise highlights the iterative nature of learning mathematics, where mistakes are opportunities for growth and deeper understanding. The correct solution presented here provides a clear roadmap for tackling similar problems in the future, emphasizing the importance of factoring, finding a common denominator, combining fractions, and simplifying the result. Ultimately, Riley's mistake, when analyzed and understood, becomes a powerful tool for learning and improvement in mathematics. This underscores the value of embracing errors as part of the learning process and using them as stepping stones to mastery.

By understanding where Riley went wrong and reviewing the correct methodology, we can all become more adept at simplifying rational expressions and tackling algebraic challenges with confidence. Remember, mathematics is a journey of learning and discovery, and every mistake is an opportunity to grow.