Solving For 'w': A Step-by-Step Guide

Hey Plastik Magazine readers! Let's dive into some math, shall we? Today, we're going to solve for 'w' in an equation that might look a bit intimidating at first glance. But don't worry, we'll break it down into easy-to-understand steps. This guide will walk you through the process, making sure you grasp every concept. We're going to take this problem and make it our own. Ready? Let's get started!

Understanding the Problem: The Equation Unveiled

Alright, guys, here’s the equation we’re tackling: rac{1}{w-5}+rac{5}{w+3}=rac{2}{w^2-2 w-15}. First things first, what are we actually looking at? We've got an equation with fractions, and our goal is to find the value(s) of 'w' that make this equation true. The equation includes fractions, and the denominators contain 'w'. The main goal is to isolate 'w' and find its value. Before we jump into solving, let's take a look at the given equation again to understand each part. The left side has two fractions added together, each with a different denominator containing 'w'. On the right side, there's another fraction, and its denominator is a quadratic expression. The presence of a quadratic expression in the denominator hints that we might end up with a quadratic equation, which could have two solutions. This problem requires us to understand how to manipulate fractions, factor quadratic expressions, and solve for variables. Also, it’s important to note any restrictions on 'w'. We can't have a zero in the denominator because that would make the fractions undefined. So, before we get to the solution, we must consider the values of 'w' that make the denominators zero. We'll need to keep these restrictions in mind as we work through the steps.

Identifying Key Components

Let’s break down the components of the equation. We have three fractions, each playing a crucial role. The first fraction has a denominator of w-5. The second has a denominator of w+3. The third fraction's denominator is w^2 - 2w - 15. This quadratic expression is important because it is a combination of the other two denominators. Notice anything? That quadratic in the denominator on the right side looks familiar? That's because $w^2 - 2w - 15$ can actually be factored into $(w-5)(w+3)$. Recognizing this relationship is key to simplifying the equation. This will become clearer as we move forward. Now that we understand the equation and its components, let’s move on to the next step, which is to simplify this problem.

Simplifying the Equation: Finding Common Ground

Now, here’s where the magic happens. We're going to simplify this equation by getting rid of those pesky fractions. The most effective way to do this is to find a common denominator. Lucky for us, as we saw earlier, the quadratic expression in the denominator is simply a product of the other two denominators. This means our common denominator is $(w-5)(w+3)$. We will multiply both sides of the equation by this common denominator. Why do we do this? Because it eliminates the fractions. Multiplying both sides by the same expression ensures that the equation remains balanced. It’s like using a seesaw; to keep it level, you must do the same thing on both sides. Once we multiply each term by the common denominator, we can cancel out the denominators of each fraction. This will leave us with a much simpler equation to work with. Remember, our aim here is to isolate 'w', and simplifying the equation is the first big step in that direction. This is a game changer, guys.

Multiplying by the Common Denominator

So, let’s take the original equation: rac{1}{w-5}+rac{5}{w+3}=rac{2}{w^2-2 w-15}. We multiply both sides by $(w-5)(w+3)$. On the left side, the first fraction rac{1}{w-5} gets multiplied by $(w-5)(w+3)$. The $(w-5)$ in the denominator cancels out with the $(w-5)$ in the multiplier, leaving us with $1*(w+3)$, which simplifies to $w+3$. The second fraction rac{5}{w+3} gets multiplied by $(w-5)(w+3)$. The $(w+3)$ in the denominator cancels out with the $(w+3)$ in the multiplier, leaving us with $5*(w-5)$, which is equal to $5w - 25$. On the right side, the fraction rac{2}{w^2-2 w-15} is multiplied by $(w-5)(w+3)$. Because we know that $(w-5)(w+3)$ equals $w^2-2 w-15$, they cancel each other out, leaving us with just 2. We are left with this new equation: $w+3+5 w-25=2$. This is much easier to work with than the original equation, right?

Solving for 'w': The Grand Finale

We're now at the final stage – solving for 'w'. With the fractions gone, we can now combine like terms and isolate 'w'. First, let's combine the 'w' terms and the constant terms on the left side of the equation $w+3+5 w-25=2$. Combining 'w' terms, $w + 5w$ gives us $6w$. Combining the constant terms, $3 - 25$ gives us $-22$. Now, our equation is simplified to $6w - 22 = 2$. Now, let's move the constant term to the other side of the equation. We add 22 to both sides of the equation. This gives us $6w = 24$. Finally, we divide both sides by 6 to isolate 'w'. This gives us $w = 4$. So, we have solved for 'w' and found that $w = 4$. But hold on, are we completely done? Not quite!

Checking for Extraneous Solutions

Before we declare victory, we need to check if our solution is valid. Remember those values of 'w' that would make the denominator zero? They are $w = 5$ and $w = -3$. Our solution, $w = 4$, does not match either of these. Therefore, $w = 4$ is a valid solution. We always have to check the possible solutions against the values that make the denominators zero to ensure that our answer is mathematically sound. In this case, the solution is valid, meaning we did the job correctly.

Conclusion: You Did It!

Awesome work, guys! We successfully solved for 'w' in what seemed like a complex equation. We simplified the equation, found a common denominator, eliminated the fractions, combined like terms, isolated 'w', and verified our answer. This process highlights the importance of understanding the basics of algebra, including factoring, finding common denominators, and solving linear equations. Remember, practice makes perfect. The more problems you solve, the more comfortable you'll become with these concepts. Keep practicing, and you will become math masters in no time.

I hope this step-by-step guide was helpful. If you have any questions or want to try another problem, drop a comment below. Until next time, keep exploring the world of math!