Riley's Homework Error: Spot The Mistake In Step 2

Let's dive into a common algebra mistake and see if we can spot where Riley went wrong! This article will walk you through a step-by-step breakdown of Riley's homework problem, highlighting the error in Step 2 and explaining the correct approach. Understanding these kinds of mistakes is crucial for mastering algebraic manipulations. Whether you are a student tackling similar problems or just brushing up on your math skills, this guide will help you sharpen your problem-solving abilities. So, let’s put on our detective hats and figure out what went wrong in Riley's work. We'll explore the initial problem, dissect each step, and pinpoint the exact moment where the error occurred. By the end of this article, you'll not only understand Riley's mistake but also be better equipped to avoid similar pitfalls in your own mathematical journey.

The Problem

Here's the problem Riley was working on:

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

Riley's goal was to add these two fractions together. To do that, she needed to find a common denominator. Let's follow her steps and see where things went sideways.

Riley's Steps

Here's how Riley approached the problem:

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

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

Step 3: $\frac{x+3}{...}$

Identifying the Error

At first glance, the problem might seem a bit intimidating, but let's break it down. The key to solving this problem lies in correctly finding the common denominator. In Step 1, Riley correctly factored the quadratic expression in the denominator of the first fraction. Factoring the quadratic $x^2 - 5x + 6$ into $(x-2)(x-3)$ is a crucial first step because it helps reveal the common factors needed to combine the fractions. This is a fundamental technique in algebra, and Riley nailed it! By breaking down the quadratic expression into its factors, she set the stage for finding a common denominator and simplifying the expression. The ability to factor quadratics accurately is essential for solving a wide range of algebraic problems, making this a critical skill to master. So far, so good!

Now, let’s zoom in on Step 2, which is where the mistake occurs. Can you spot it? Riley attempts to find a common denominator, but there’s a subtle yet significant error. When looking at the original expression in Step 1, $\frac{x}{(x-2)(x-3)} + \frac{3}{x+3}$, the denominators are $(x-2)(x-3)$ and $(x+3)$. To combine these fractions, we need a common denominator that includes all unique factors. The least common denominator (LCD) should be $(x-2)(x-3)(x+3)$. Riley incorrectly changes the denominator of the first fraction. Instead of keeping $(x-2)(x-3)$ she writes $(x-2)(x+3)$ which is incorrect. She should have focused on making both denominators the same by multiplying the second fraction by the missing factors from the first fraction’s denominator. This is a classic mistake, highlighting the importance of carefully tracking each step in algebraic manipulations. Recognizing and avoiding this kind of error is key to successfully navigating more complex problems in algebra.

The Correct Approach

Let's walk through the correct way to solve this problem. Starting from Step 1:

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

To get a common denominator, we need to multiply the first fraction by $\frac{x+3}{x+3}$ and the second fraction by $\frac{(x-2)(x-3)}{(x-2)(x-3)}$. This will give us the least common denominator (LCD) of $(x-2)(x-3)(x+3)$.

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

Now we can combine the numerators:

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

Expand the numerator:

Step 4: $\frac{x^2 + 3x + 3(x^2 - 5x + 6)}{(x-2)(x-3)(x+3)}$

Step 5: $\frac{x^2 + 3x + 3x^2 - 15x + 18}{(x-2)(x-3)(x+3)}$

Combine like terms:

Step 6: $\frac{4x^2 - 12x + 18}{(x-2)(x-3)(x+3)}$

We can factor out a 2 from the numerator:

Step 7: $\frac{2(2x^2 - 6x + 9)}{(x-2)(x-3)(x+3)}$

This is the simplified form of the expression. The most important aspect of this process is ensuring that each fraction is correctly multiplied by the necessary factors to achieve the common denominator, without altering the original value of the expression. This involves careful distribution and simplification, which are key skills in algebraic manipulation.

Why This Mistake Matters

Understanding why Riley's mistake matters is crucial for preventing similar errors in the future. The core issue here is a misunderstanding of how to create equivalent fractions when finding a common denominator. The golden rule is that whatever you multiply the denominator by, you must also multiply the numerator by the same factor. This ensures that the value of the fraction remains unchanged. In Riley's Step 2, she altered the denominator of the first fraction incorrectly, which completely changed the expression and led to an incorrect solution. This type of mistake can have significant consequences in more complex algebraic problems, so mastering the technique of finding common denominators is essential.

Furthermore, this example highlights the importance of double-checking each step in your work. Math is a sequential process, and a small error early on can snowball into a major problem later. By carefully reviewing each step and ensuring that you are applying the correct operations, you can catch mistakes before they lead to incorrect answers. Practicing these skills not only improves your accuracy but also deepens your understanding of the underlying mathematical principles. So, remember to take your time, be meticulous, and always double-check your work!

Key Takeaways

• Factoring is your friend: Factoring the denominators makes it easier to identify common factors and the least common denominator.
• Multiply both numerator and denominator: When finding a common denominator, ensure you multiply both the numerator and the denominator by the same factor.
• Double-check each step: A small mistake early on can lead to a big problem later. Always review your work.

By understanding these key concepts and carefully applying them, you can avoid making similar mistakes and excel in algebra. Math might seem daunting at times, but with practice and a clear understanding of the fundamentals, you can tackle even the most challenging problems. Keep practicing, stay curious, and don't be afraid to ask for help when you need it. You've got this!

Conclusion

Riley's mistake in Step 2 is a common one, but by understanding the correct process for finding a common denominator, we can avoid similar errors. Remember to factor, multiply both the numerator and denominator, and double-check your work. Mastering these skills will help you confidently tackle algebraic problems. For further learning on fractions and algebraic expressions, you can explore resources like Khan Academy's Algebra I section which offers comprehensive lessons and practice exercises.