Simplifying Algebraic Fractions: A Step-by-Step Guide

Hey guys! Let's dive into the world of simplifying algebraic fractions. It might sound a little intimidating at first, but trust me, with a few simple steps, you'll be cruising through these problems like a pro. We're going to break down each problem, making sure you understand every single move. So, grab your pencils and let's get started! We'll be simplifying the following expressions, turning them into their most basic fraction form. This is super important for all sorts of algebra problems, so pay close attention!

a) Simplifying $\frac{14p^4}{q^6} \cdot \frac{q^5}{56p^4}$

Alright, let's tackle the first one: $\frac{14p^4}{q^6} \cdot \frac{q^5}{56p^4}$. The key here is to remember that when multiplying fractions, you multiply the numerators (the top numbers) together and the denominators (the bottom numbers) together. So, let's do that first. We will start by multiplying our numerators together, which is $14p^4$ and $q^5$. Then we will multiply the denominators together, which are $q^6$ and $56p^4$. This gives us $\frac{14p^4 \cdot q^5}{q^6 \cdot 56p^4}$. Now we have it all in one fraction. The next step is to simplify! This is where the fun begins!

First, let's look at the numbers. We have 14 in the numerator and 56 in the denominator. Both of these numbers are divisible by 14. So, we can divide both by 14. 14 divided by 14 is 1, and 56 divided by 14 is 4. This simplifies the fraction to $\frac{1p^4 \cdot q^5}{q^6 \cdot 4p^4}$. Now, let's look at the variables. We have $p^4$ in the numerator and $p^4$ in the denominator. When we have the same variable raised to the same power in both the numerator and the denominator, they cancel each other out. So, the $p^4$ terms disappear. This leaves us with $\frac{1 \cdot q^5}{q^6 \cdot 4}$.

Next, consider the $q$ terms. We have $q^5$ in the numerator and $q^6$ in the denominator. When we divide exponents with the same base, we subtract the powers. So, we subtract the exponent in the numerator (5) from the exponent in the denominator (6). This gives us $q^{6-5} = q^1$ or simply $q$ in the denominator. Remember, since the larger exponent was in the denominator, the remaining $q$ stays in the denominator. Now we have $\frac{1}{4q}$. We're almost there! Finally, multiply the remaining terms. This gives us our final answer of $\frac{1}{4q}$. We simplified it! Great job, everyone! That's our final answer for part a!

b) Simplifying $45a^3b \cdot \frac{c^2}{30a^4b}$

Now, let's move on to the next problem: $45a^3b \cdot \frac{c^2}{30a^4b}$. This one looks a bit different, but the same principles apply. When a term is not a fraction, we can think of it as being over 1. Therefore, we can rewrite the expression as $\frac{45a^3b}{1} \cdot \frac{c^2}{30a^4b}$. Now, just like before, we can multiply the numerators and denominators together. This gives us $\frac{45a^3b \cdot c^2}{1 \cdot 30a^4b}$, which simplifies to $\frac{45a^3bc^2}{30a^4b}$. So, let's get simplifying. Remember, it's all about breaking it down step-by-step!

First, let's look at the numbers 45 and 30. Both are divisible by 15. 45 divided by 15 is 3, and 30 divided by 15 is 2. This simplifies the fraction to $\frac{3a^3bc^2}{2a^4b}$. Next, we deal with the variables. Let's start with the $a$ terms. We have $a^3$ in the numerator and $a^4$ in the denominator. When dividing exponents with the same base, we subtract the powers. Since the larger exponent is in the denominator, the 'a' will stay in the denominator. This becomes $a^{4-3} = a^1$ or simply $a$ in the denominator. This leaves us with $\frac{3bc^2}{2ab}$.

Now, let's look at the $b$ terms. We have $b$ in the numerator and $b$ in the denominator. Since we have the same variable raised to the same power in both the numerator and the denominator, these cancel each other out. Thus, the $b$s disappear. We are so close now! This simplifies to $\frac{3c^2}{2a}$. There is nothing left to simplify. We have now arrived at our final answer. Congratulations, you have solved part b!

c) Simplifying $\frac{3a-9}{a+2} : \frac{a^2-9}{a^2-4}$

Alright, let's tackle the final problem: $\frac{3a-9}{a+2} : \frac{a^2-9}{a^2-4}$. This one involves division, but don't worry! Dividing fractions is just like multiplying by the reciprocal. The reciprocal of a fraction is just flipping it over (switching the numerator and the denominator). So, our first step is to rewrite the division as multiplication by the reciprocal. This gives us $\frac{3a-9}{a+2} \cdot \frac{a^2-4}{a^2-9}$. Keep going, you are doing great!

Now we have a multiplication problem! Let's factor where we can. In the numerator of the first fraction ($3a-9$), we can factor out a 3. This leaves us with $3(a-3)$. In the denominator of the second fraction ($a^2-9$), this is a difference of squares, which can be factored into $(a-3)(a+3)$. And in the numerator of the second fraction ($a^2-4$), this is also a difference of squares, which factors into $(a-2)(a+2)$. Now we can rewrite the expression as $\frac{3(a-3)}{a+2} \cdot \frac{(a-2)(a+2)}{(a-3)(a+3)}$.

Let's put all these steps into one easy-to-read line. Now we have $\frac{3(a-3)(a-2)(a+2)}{(a+2)(a-3)(a+3)}$. Now we can start cancelling out terms. In the numerator and denominator, we can cancel out the $(a-3)$ term and also the $(a+2)$ term. This leaves us with $\frac{3(a-2)}{a+3}$. Since nothing else can be simplified, we have our final answer. You have officially completed simplifying the algebraic fractions, and have solved part c!

Summary and Key Takeaways

Wow, guys! You did it! We have successfully simplified three different algebraic fractions. Remember these key takeaways:

• Multiplying Fractions: Multiply numerators and denominators. Then, simplify. Great job!
• Dividing Fractions: Multiply by the reciprocal. Remember to flip the second fraction.
• Simplifying: Factor expressions where possible, cancel common factors in the numerator and denominator, simplify the numbers. Always check your work, and remember that practice makes perfect! Congratulations on completing the problem!

Keep practicing, and these types of problems will become second nature to you. You're well on your way to mastering algebra. Keep up the great work, and I'll see you next time!