Subtracting Rational Expressions Step-by-Step Guide

Hey guys! Let's dive into a super interesting math problem today where we're going to break down the difference between rational expressions. It looks a little intimidating at first, but trust me, we'll tackle it together step by step. We're going to take a look at a specific problem involving fractions with polynomials and see how to simplify it. So, grab your thinking caps, and let's get started!

Deconstructing the Initial Problem

Our main goal here is to understand the difference between two rational expressions. Rational expressions, as you might remember, are just fractions where the numerator and the denominator are polynomials. In this case, we're starting with the expression:

$\frac{x}{x^2-2 x-15}-\frac{4}{x^2+2 x-35}$

This looks a bit complex, right? But don't worry! The first thing we need to do when dealing with subtraction (or addition) of fractions is to find a common denominator. It's just like when you're adding \frac{1}{2} and \frac{1}{3}; you need a common denominator before you can combine them. The same principle applies here, but instead of simple numbers, we have polynomials. To find this common denominator, we'll need to factor the denominators of both fractions.

Let's start by factoring $x^2 - 2x - 15$. We need to find two numbers that multiply to -15 and add up to -2. Those numbers are -5 and +3. So, we can factor the first denominator as $(x - 5)(x + 3)$.

Now, let's factor the second denominator, $x^2 + 2x - 35$. We need two numbers that multiply to -35 and add up to +2. Those numbers are +7 and -5. So, the second denominator factors into $(x + 7)(x - 5)$.

So, now our expression looks like this:

$\frac{x}{(x-5)(x+3)} - \frac{4}{(x+7)(x-5)}$

Notice that both denominators share a common factor of $(x - 5)$. This is great news! To get our common denominator, we need to include all the unique factors from both denominators. That means our common denominator will be $(x - 5)(x + 3)(x + 7)$.

Getting to a Common Denominator

Now that we know our common denominator, we need to rewrite each fraction with this new denominator. For the first fraction, $\frac{x}{(x-5)(x+3)}$, we're missing the $(x + 7)$ factor in the denominator. So, we multiply both the numerator and the denominator by $(x + 7)$:

$\frac{x}{(x-5)(x+3)} * \frac{(x+7)}{(x+7)} = \frac{x(x+7)}{(x-5)(x+3)(x+7)}$

This gives us $\frac{x(x+7)}{(x-5)(x+3)(x+7)}$.

For the second fraction, $\frac{4}{(x+7)(x-5)}$, we're missing the $(x + 3)$ factor in the denominator. So, we multiply both the numerator and the denominator by $(x + 3)$:

$\frac{4}{(x+7)(x-5)} * \frac{(x+3)}{(x+3)} = \frac{4(x+3)}{(x-5)(x+7)(x+3)}$

This results in $\frac{4(x+3)}{(x-5)(x+7)(x+3)}$.

Now we can rewrite our original expression with the common denominator:

$\frac{x(x+7)}{(x-5)(x+3)(x+7)} - \frac{4(x+3)}{(x-5)(x+3)(x+7)}$

Combining the Fractions

With a common denominator in place, we can now subtract the fractions. This means we subtract the numerators and keep the denominator the same:

$\frac{x(x+7) - 4(x+3)}{(x-5)(x+3)(x+7)}$

Now, let's expand the numerator. First, $x(x + 7)$ becomes $x^2 + 7x$. Then, $4(x + 3)$ becomes $4x + 12$. So, our expression now looks like:

$\frac{x^2 + 7x - (4x + 12)}{(x-5)(x+3)(x+7)}$

Be super careful with that minus sign! We're subtracting the entire expression $(4x + 12)$, so we need to distribute the negative sign:

$\frac{x^2 + 7x - 4x - 12}{(x-5)(x+3)(x+7)}$

Now, let's combine like terms in the numerator. We have $7x$ and $-4x$, which combine to $3x$. So, our numerator becomes $x^2 + 3x - 12$. Our entire expression is now:

$\frac{x^2 + 3x - 12}{(x-5)(x+3)(x+7)}$

Analyzing the Provided Steps

Okay, so we've worked through the problem from scratch. Now, let's compare our result to the steps that were provided. This is where we'll really see how things come together (or where there might be a little hiccup!).

The steps you provided include the expression:

$\frac{x^2+3 x+12}{(x-3)(x-5)(x+7)}$

This seems to be a slightly different numerator than what we arrived at, which was $x^2 + 3x - 12$. Let's hold onto that thought for a moment and look at the other steps.

There's also:

$\frac{x(x+3-12)}{(x+3)(x-5)(x+7)}$

This step looks a little unusual. It seems like there might have been a small error in how the numerator was manipulated. Let's keep moving and see if we can pinpoint exactly where things might have diverged.

Then we see:

$\frac{x^2+3 x+12}{(x+3)(x-5)(x+7)}$

Again, we've got that $x^2 + 3x + 12$ in the numerator, which differs from our calculated $x^2 + 3x - 12$. This suggests that there might be a consistent error in the steps provided, specifically in handling the constant term in the numerator.

Finally, we have:

$\frac{x^2+3 x-12}{(x+3)(x-5)(x+7)}$

Aha! This expression matches the result we obtained by working through the problem ourselves! So, it seems like there was a small error in one of the intermediate steps, but the final answer presented here is consistent with our calculation.

Spotting the Discrepancy

It's super important in math to not just get to an answer, but also to double-check your work (and the work of others!). In this case, we noticed that the numerator in one of the intermediate steps, $x^2 + 3x + 12$, was different from what we calculated, $x^2 + 3x - 12$. This kind of discrepancy can arise from a simple sign error or a mistake in combining like terms. Always, always, always double-check your work!

The Final Verdict and Key Takeaways

So, what's the real difference here? The key takeaway is that when subtracting rational expressions, precision is crucial. One tiny mistake in distributing a negative sign or combining like terms can lead to a different result. By working through the problem step by step and comparing our result to the provided steps, we were able to identify a potential error and confirm the correct simplified expression.

The correct simplified expression for $\frac{x}{x^2-2 x-15}-\frac{4}{x^2+2 x-35}$ is:

$\frac{x^2 + 3x - 12}{(x-5)(x+3)(x+7)}$

Remember, guys, math isn't just about getting the right answer; it's about understanding the process and being able to spot errors along the way. Keep practicing, keep questioning, and you'll become math superstars in no time!

Summing It Up

In this article, we've walked through a problem involving the subtraction of rational expressions. We started by factoring the denominators, finding a common denominator, and rewriting each fraction. Then, we combined the fractions, simplified the numerator, and arrived at our final expression. Along the way, we compared our steps to a set of provided steps, identified a potential discrepancy, and confirmed the correct solution.

The most important things to remember are:

• Factoring is your friend! It's essential for finding common denominators.
• Pay close attention to signs, especially when distributing negative signs.
• Always double-check your work. It can save you from making small errors that lead to big differences.

By following these tips and practicing regularly, you'll be able to tackle even the most challenging rational expression problems with confidence. Keep up the great work, and I'll see you in the next math adventure!