Simplify $\frac{35 V W^3}{20 V^2 W^3}$: A Detailed Guide

Simplifying algebraic fractions might seem daunting at first, but trust me, it's totally manageable! We're going to break down the process of simplifying the fraction $\frac{35 v w^3}{20 v^2 w^3}$ into easy-to-follow steps. By the end of this guide, you'll be a pro at handling these types of problems. Let’s dive in and make those fractions look simpler, shall we?

Understanding the Basics of Simplifying Fractions

Before we jump into the specifics of the given fraction, let’s quickly recap the fundamental principles of simplifying fractions. Simplifying a fraction means reducing it to its lowest terms. In simpler terms, we want to make the numbers as small as possible while keeping the fraction equivalent to its original form. This involves finding common factors in both the numerator (the top part of the fraction) and the denominator (the bottom part) and then canceling them out. Think of it like this: if you have the fraction $\frac{4}{6}$, both 4 and 6 are divisible by 2. So, you can divide both by 2 to get $\frac{2}{3}$, which is the simplified form. This concept is crucial when we deal with algebraic fractions, where variables are involved. Remember, the goal is always to find the greatest common factor (GCF) and divide both the numerator and the denominator by it. This ensures that the fraction is in its simplest form. Simplifying fractions not only makes them easier to work with but also helps in understanding the relationships between numbers and variables. It's a foundational skill in algebra, and mastering it will significantly help in more advanced topics. So, let’s keep this basic principle in mind as we move forward and tackle our specific problem.

Breaking Down the Fraction $\frac{35 v w^3}{20 v^2 w^3}$

Okay, guys, let's get to the heart of the matter! We need to simplify the fraction $\frac{35 v w^3}{20 v^2 w^3}$. The first thing we're going to do is look at the coefficients, which are the numerical parts of the terms. In this case, we have 35 in the numerator and 20 in the denominator. What's the greatest common factor (GCF) of 35 and 20? If you think about it, both numbers are divisible by 5. So, let’s divide both 35 and 20 by 5. This gives us 7 in the numerator (since 35 ÷ 5 = 7) and 4 in the denominator (since 20 ÷ 5 = 4). Now our fraction looks like $\frac{7 v w^3}{4 v^2 w^3}$. See, we’ve already made progress! Next, we need to tackle the variables. We have 'v' and 'w' in both the numerator and the denominator, each raised to certain powers. When simplifying variables in fractions, remember the rule: when dividing like bases, you subtract the exponents. This is where things get interesting and where we can really simplify this fraction. Stay with me, and we’ll break down how to handle those variables step by step.

Simplifying the Variable 'v'

Alright, let's zoom in on the variable 'v' in our fraction $\frac{7 v w^3}{4 v^2 w^3}$. In the numerator, we have 'v' which is essentially $v^1$ (remember, if there's no exponent written, it's understood to be 1). In the denominator, we have $v^2$. Now, to simplify, we apply the rule of exponents which states that when you divide like bases, you subtract the exponents. So, we have $v^1$ divided by $v^2$, which can be written as $v^{1-2}$. What's 1 minus 2? It's -1! So, we get $v^{-1}$. But what does a negative exponent mean? A negative exponent means we need to move the variable to the denominator to make the exponent positive. So, $v^{-1}$ is the same as $\frac{1}{v}$. This means that after simplifying the 'v' terms, we'll have 'v' in the denominator. Another way to think about it is that since we have one 'v' in the numerator and two 'v's in the denominator, one 'v' will cancel out, leaving one 'v' in the denominator. Either way, we arrive at the same conclusion. Understanding this step is crucial because it’s a common operation in simplifying algebraic expressions. Now that we've handled 'v', let's move on to the next variable, 'w'.

Simplifying the Variable 'w'

Now, let’s shift our focus to the variable 'w' in the fraction $\frac{7 v w^3}{4 v^2 w^3}$. We've already simplified the numerical coefficients and the 'v' terms, so 'w' is the last piece of the puzzle. Looking at the fraction, we see that we have $w^3$ in both the numerator and the denominator. This makes our job a little easier! When we have the same power of a variable in both the numerator and the denominator, they simply cancel each other out. This is because when dividing like bases, you subtract the exponents. In this case, we have $w^3$ divided by $w^3$, which can be written as $w^{3-3}$. What's 3 minus 3? It's 0! So, we get $w^0$. Anything raised to the power of 0 is 1. Therefore, $w^0$ is equal to 1. This means that the $w^3$ terms completely cancel each other out, leaving no 'w' terms in our simplified fraction. This is a fantastic simplification because it eliminates a variable altogether, making the fraction cleaner and easier to understand. Remember, this only happens when the exponents are the same. If the exponents were different, we'd have to subtract them just like we did with the 'v' terms. Now that we’ve simplified both 'v' and 'w', we’re ready to put everything together and see the final simplified fraction.

Putting It All Together: The Simplified Fraction

Okay, everyone, let's bring it all together! We've simplified the numerical coefficients, the 'v' terms, and the 'w' terms in the fraction $\frac{35 v w^3}{20 v^2 w^3}$. Let's recap what we found: First, we simplified the coefficients 35 and 20 by dividing both by their greatest common factor, 5. This gave us $\frac{7}{4}$. Then, we tackled the 'v' terms. We had $v^1$ in the numerator and $v^2$ in the denominator. After subtracting the exponents, we ended up with 'v' in the denominator, so we have $\frac{1}{v}$. Next, we looked at the 'w' terms. We had $w^3$ in both the numerator and the denominator, which canceled each other out completely. So, there's no 'w' in our simplified fraction. Now, let's combine all these simplified parts. We have the fraction $\frac{7}{4}$, the 'v' term $\frac{1}{v}$, and no 'w' term. Multiplying these together, we get our final simplified fraction: $\frac{7}{4v}$. And there you have it! We've taken a potentially intimidating fraction and broken it down into its simplest form. This process demonstrates the power of understanding and applying the basic rules of algebra. Remember, practice makes perfect, so keep simplifying those fractions!

Final Simplified Fraction: $\frac{7}{4v}$

So, there you have it, guys! The simplified form of the fraction $\frac{35 v w^3}{20 v^2 w^3}$ is $\frac{7}{4v}$. We walked through each step, from finding the greatest common factor of the coefficients to simplifying the variables using the rules of exponents. Remember, the key to simplifying algebraic fractions is to break down the problem into smaller, manageable parts. First, tackle the numbers by finding their GCF. Then, handle the variables one by one, using the rule of subtracting exponents when dividing like bases. And don't forget, if you end up with a negative exponent, move the variable to the other side of the fraction (numerator to denominator or vice versa) to make the exponent positive. This entire process not only simplifies the fraction but also gives you a deeper understanding of the algebraic principles at play. Keep practicing these steps, and you'll find that simplifying fractions becomes second nature. Great job, and keep up the awesome work!