Math Mistake: Spotting The Error In Step 2

Hey mathletes! Let's dive into a common homework hiccup that Riley ran into. We've got this expression: $\frac{x}{x^2-5 x+6}+\frac{3}{x+3}$. Riley's trying to add these two fractions, a totally standard task in algebra. She's on the right track by first factoring the denominator of the first fraction. That's a crucial first step, and she nailed it in Step 1, transforming the expression into $\frac{x}{(x-2)(x-3)}+\frac{3}{x+3}$. This factoring is key because it helps us find the least common denominator (LCD), which is essential for adding fractions. Remember, guys, you can only add or subtract fractions when they have the same denominator. If they don't, you gotta make 'em match!

Now, let's zero in on Step 2, because that's where things got a little tangled for Riley. Her goal in this step is to rewrite both fractions so they share the same denominator. The LCD of $(x-2)(x-3)$ and $(x+3)$ is $(x-2)(x-3)(x+3)$. So, the first fraction, $\frac{x}{(x-2)(x-3)}$, needs to be multiplied by $\frac{x+3}{x+3}$. This gives us $\frac{x(x+3)}{(x-2)(x-3)(x+3)}$. The second fraction, $\frac{3}{x+3}$, needs to be multiplied by $\frac{(x-2)(x-3)}{(x-2)(x-3)}$. This would result in $\frac{3(x-2)(x-3)}{(x+3)(x-2)(x-3)}$.

However, Riley's Step 2 looks like this: $\frac{x}{(x-2)(x+3)}+\frac{3(x-2)}{(x-2)(x+3)}$. Let's break down what she did and where the mistake lies. For the first fraction, $\frac{x}{(x-2)(x-3)}$, she changed the denominator from $(x-2)(x-3)$ to $(x-2)(x+3)$. This is where the error pops up. She swapped out the $(x-3)$ term for an $(x+3)$ term. This fundamentally alters the denominator and, consequently, the entire value of the fraction. To get the LCD, she needed to multiply the first fraction's numerator and denominator by $(x+3)$. So, it should have been $\frac{x(x+3)}{(x-2)(x-3)(x+3)}$.

For the second fraction, $\frac{3}{x+3}$, she correctly identified that she needed to introduce the factors $(x-2)$ and $(x-3)$ to match the LCD. But then she only multiplied the numerator by $(x-2)$, giving her $\frac{3(x-2)}{(x-2)(x+3)}$. The denominator she wrote, $(x-2)(x+3)$, is also incorrect. It's missing the $(x-3)$ factor that should have been included to make it the full LCD. Moreover, the second fraction should have been multiplied by $\frac{(x-2)(x-3)}{(x-2)(x-3)}$ to get the correct common denominator, which would yield $\frac{3(x-2)(x-3)}{(x+3)(x-2)(x-3)}$.

So, to sum it up, Riley's mistake in Step 2 is twofold: she incorrectly changed the denominator of the first fraction by swapping a factor, and she failed to multiply the numerator and denominator of the second fraction by the correct factors needed to achieve the least common denominator. This is a super common pitfall, guys, mixing up factors when trying to find that elusive LCD. Always double-check your factored forms and ensure you're multiplying both the numerator and the denominator by the exact same expression to maintain the fraction's value. Keep practicing, and you'll get the hang of it!

The Correct Path Forward: Finding the True LCD

Alright, let's rewind and get Riley's problem back on the right track. We left off with Step 1: $\frac{x}{(x-2)(x-3)}+\frac{3}{x+3}$. The main mission here is to get a common denominator so we can actually add these bad boys. To do this, we need to identify the least common denominator (LCD). Looking at the denominators, we have $(x-2)(x-3)$ and $(x+3)$. The LCD is formed by taking all the unique factors from both denominators and multiplying them together. In this case, the unique factors are $(x-2)$, $(x-3)$, and $(x+3)$. So, our LCD is a beautiful, clean $(x-2)(x-3)(x+3)$.

Now, we need to adjust each fraction so that it has this LCD. Let's tackle the first fraction: $\frac{x}{(x-2)(x-3)}$. To get the LCD, we are missing the $(x+3)$ factor in the denominator. So, we multiply both the numerator and the denominator by $(x+3)$. This keeps the fraction's value exactly the same, which is super important. Think of it like multiplying by 1 in a fancy disguise! This gives us: $\frac{x \times (x+3)}{(x-2)(x-3) \times (x+3)}$, which simplifies to $\frac{x(x+3)}{(x-2)(x-3)(x+3)}$.

Next up is the second fraction: $\frac{3}{x+3}$. To get our LCD, $(x-2)(x-3)(x+3)$, we need to add the $(x-2)$ and $(x-3)$ factors to the denominator. Therefore, we must multiply both the numerator and the denominator by the product of these missing factors, which is $(x-2)(x-3)$. This looks like: $\frac{3 \times (x-2)(x-3)}{(x+3) \times (x-2)(x-3)}$. After multiplying, the numerator becomes $3(x-2)(x-3)$, and the denominator becomes $(x+3)(x-2)(x-3)$, which is exactly our LCD, $(x-2)(x-3)(x+3)$.

So, after correctly adjusting both fractions, our expression now looks like this: $\frac{x(x+3)}{(x-2)(x-3)(x+3)} + \frac{3(x-2)(x-3)}{(x-2)(x-3)(x+3)}$. See how both fractions now have the same denominator? That's the magic of finding the LCD! This setup is what Riley should have achieved in her Step 2. It might seem like extra work, but this step is absolutely fundamental for combining the numerators correctly in the next stage.

Moving On: Combining the Numerators

With the fractions properly aligned with their common denominator, the next logical step, Step 3, is to combine the numerators. Riley's initial problem gave us $\frac{x}{x^2-5 x+6}+\frac{3}{x+3}$. After factoring and finding the LCD, we correctly rewrote this as $\frac{x(x+3)}{(x-2)(x-3)(x+3)} + \frac{3(x-2)(x-3)}{(x-2)(x-3)(x+3)}$. Now that the denominators are identical, we can simply add the numerators together while keeping the common denominator.

The numerator of the first fraction is $x(x+3)$. Expanding this gives us $x^2 + 3x$. The numerator of the second fraction is $3(x-2)(x-3)$. We need to expand this part carefully. First, multiply $(x-2)(x-3)$ to get $x^2 - 3x - 2x + 6$, which simplifies to $x^2 - 5x + 6$. Then, multiply this entire expression by 3: $3(x^2 - 5x + 6) = 3x^2 - 15x + 18$.

Now, we add the two expanded numerators: $(x^2 + 3x) + (3x^2 - 15x + 18)$. Combine like terms: $(x^2 + 3x^2) + (3x - 15x) + 18$. This results in $4x^2 - 12x + 18$. This is our combined numerator.

So, the expression after combining the numerators over the common denominator is $\frac{4x^2 - 12x + 18}{(x-2)(x-3)(x+3)}$. This is the correct result for Step 3, assuming Riley had correctly set up her fractions in Step 2. It's crucial to be meticulous with algebraic expansions and combining terms. A small slip-up here, like a sign error or missing a term, can lead to a completely different final answer.

It's also worth noting that sometimes, after combining the numerators, the resulting polynomial might be factorable, and that new factor might cancel out with one of the factors in the denominator. This would lead to a simplified form of the expression. For our current numerator, $4x^2 - 12x + 18$, we can factor out a 2 to get $2(2x^2 - 6x + 9)$. The quadratic $2x^2 - 6x + 9$ does not have real roots (its discriminant $b^2-4ac = (-6)^2 - 4(2)(9) = 36 - 72 = -36$ is negative), so it cannot be factored further over the real numbers. Therefore, this expression is in its simplest form.

Remember, guys, the journey through algebraic fractions involves several key stages: factoring denominators, finding the LCD, adjusting numerators, combining numerators, and finally, simplifying. Each step builds upon the previous one, so accuracy is paramount. Riley's mistake highlights how a single error in finding the common denominator can throw off the entire problem. Keep your eyes peeled, check your work, and you'll conquer these problems like a pro!