Simplify This Algebraic Fraction Product!

Dec 16, 2025 by Andrew McMorgan 42 views

Hey math whizzes and number crunchers! Today, we've got a super cool problem that's all about simplifying products of algebraic fractions. You know, those fractions with letters in them? We're diving deep into the world of algebra to crack this one. So grab your calculators, your notebooks, and let's get this done!

Understanding the Product of Algebraic Fractions

Alright guys, let's talk about what we're actually doing here. When we talk about the product of algebraic fractions, we're essentially multiplying two or more fractions that contain variables. Think of it like multiplying regular fractions, but with an extra layer of complexity because of those pesky variables. The core principle remains the same: you multiply the numerators together and the denominators together. However, the real magic, and where a lot of the simplification happens, is in factoring the expressions within the fractions. Factoring allows us to spot common factors in the numerator and denominator, which we can then cancel out. This is crucial because it can drastically simplify the expression, making it much easier to work with. So, for our specific problem, we have:

rac{2 y}{y-3} \cdot \frac{4 y-12}{2 y+6}

Our goal is to find the simplified form of this product. This means we need to look for opportunities to factor each part of the fractions. We'll be applying our factoring skills to the numerators and denominators. Remember, the goal of simplification is to get the expression into its most basic form, where no further cancellation is possible. This often involves techniques like finding the greatest common factor (GCF) or recognizing special factoring patterns like the difference of squares or sum/difference of cubes, though for this problem, we'll likely be focusing on the GCF.

The strategy is always to factor first! Don't be tempted to just multiply straight across without factoring. You'll end up with a much more complicated expression that's harder to simplify. By factoring each polynomial (the expressions on top and bottom), we can rewrite the fraction. Once factored, we can look for any terms that appear in both the numerator and the denominator. These are our common factors, and they can be cancelled out. This process is fundamental to working with rational expressions, which is the fancy math term for fractions involving variables. So, let's break down each part of our expression and see what we can factor.

We have the first fraction $\frac{2y}{y-3}$ . The numerator, $2y$ , is already in its simplest factored form. The denominator, $y-3$ , is also as simple as it gets; you can't factor out any common terms from $y$ and $-3$ . So, the first fraction is ready to go.

Now, let's look at the second fraction: $\frac{4y-12}{2y+6}$ .

In the numerator, $4y-12$ , we can see that both $4y$ and $-12$ are divisible by 4. So, we can factor out a 4. This gives us $4(y-3)$ .

In the denominator, $2y+6$ , both $2y$ and $6$ are divisible by 2. So, we can factor out a 2. This gives us $2(y+3)$ .

So, our original product now looks like this after factoring:

\frac{2 y}{y-3} \cdot \frac{4(y-3)}{2(y+3)}

See how factoring made things clearer? We can now see a $(y-3)$ term in the numerator of the second fraction and also in the denominator of the first fraction. This is exactly what we were looking for!

Step-by-Step Simplification Process

Alright guys, now that we've done the crucial factoring step, let's put it all together and simplify this expression. We've successfully rewritten the product as:

\frac{2 y}{y-3} \cdot \frac{4(y-3)}{2(y+3)}

Remember the rule for multiplying fractions: you multiply the numerators together and the denominators together. So, this becomes:

\frac{(2y) \cdot 4(y-3)}{(y-3) \cdot 2(y+3)}

Now comes the exciting part – cancellation! Look closely at the numerator and the denominator. Do you see any common factors? Yes, you do! We have a $(y-3)$ term in the numerator and another $(y-3)$ term in the denominator. These are identical, so we can cancel them out. Think of it like having $\frac{5}{5}$ – it equals 1, right? So, $\frac{y-3}{y-3}$ also equals 1, as long as $y \neq 3$ (because we can't have a zero in the denominator).

After cancelling out the $(y-3)$ terms, our expression simplifies to:

\frac{2y \cdot 4}{2(y+3)}

Now, let's simplify the remaining parts. In the numerator, $2y \cdot 4$ equals $8y$ . So we have:

\frac{8y}{2(y+3)}

We can also simplify the numbers. We have an 8 in the numerator and a 2 in the denominator. $8 \div 2$ is 4. So, we can simplify the fraction further:

\frac{4y}{(y+3)}

And there you have it! We've successfully simplified the product of the algebraic fractions. The final simplified expression is $\frac{4y}{y+3}$ .

Important Note on Restrictions: When simplifying algebraic fractions, it's super important to keep track of any values of the variable that would make the original denominator zero. In our original problem, the denominators were $(y-3)$ and $(2y+6)$ .

For $(y-3)$ , if $y=3$ , the denominator is zero, which is undefined.
For $(2y+6)$ , if $2y = -6$ , so $y = -3$ , the denominator is zero, which is undefined.

So, our simplified expression $\frac{4y}{y+3}$ is valid for all values of $y$ except $y=3$ and $y=-3$ . While the simplified form itself only has a restriction at $y=-3$ , the original expression was undefined at $y=3$ as well. These are called the excluded values.

So, to recap the steps:

Factor all numerators and denominators completely.
Identify common factors in the numerator and denominator across the entire product.
Cancel out these common factors.
Multiply the remaining terms.
State any restrictions on the variable.

This systematic approach will help you tackle any similar problems you encounter. It's all about breaking it down and using those factoring skills!

Analyzing the Options

Okay team, we've done the heavy lifting and simplified our expression to $\frac{4y}{y+3}$ . Now, let's look at the multiple-choice options provided to see which one matches our answer. This is where we check our work and confirm we're on the right track.

Our options are:

A. $\frac{2}{3}$ B. $\frac{10}{9}$ C. $\frac{4 y}{y-3}$ D. $\frac{4 y}{y+3}$

Let's compare our simplified result with each option:

Option A: $\frac{2}{3}$ - This is a simple numerical fraction. Our result contains the variable $y$ , so this is definitely not it. It's likely a distractor that comes from incorrect cancellation or multiplication.
Option B: $\frac{10}{9}$ - Similar to option A, this is a numerical fraction. Our result has variables, so this option is also incorrect.
Option C: $\frac{4 y}{y-3}$ - This looks somewhat similar to our answer because it has $4y$ in the numerator. However, the denominator is $(y-3)$ . Let's think back to our simplification steps. We had the term $(y-3)$ in the denominator of the first fraction and also in the numerator of the second fraction. When we cancelled these out, the $(y-3)$ in the denominator was removed. If this option were correct, it would imply that the $(y-3)$ term in the numerator didn't cancel with anything, or that the $(y-3)$ in the denominator remained. This is not what happened in our calculation.
Option D: $\frac{4 y}{y+3}$ - This option perfectly matches our simplified expression! We found that after factoring and cancelling, the remaining terms resulted in $\frac{4y}{y+3}$ . The $4y$ came from $2y \cdot 4$ divided by 2 (effectively $(2 \cdot 4)/2 = 4$ , and the $y$ remained), and the $(y+3)$ came from the denominator of the second fraction after cancelling the $(y-3)$ terms. This is exactly what we calculated.

Therefore, the correct answer is Option D. It's always satisfying when your calculated answer is one of the choices, right? It confirms that our method was sound and our calculations were accurate. Remember, each of these options represents a potential outcome based on different ways one might incorrectly approach the problem, so understanding why the other options are wrong is just as important as knowing why the correct one is right.

Common Mistakes to Avoid:

Not Factoring: Jumping straight to multiplication without factoring leads to much harder simplification.
Incorrect Factoring: Making mistakes when factoring polynomials, like missing a common factor or factoring incorrectly.
Cancellation Errors: Cancelling terms that are not identical (e.g., cancelling $y$ from $y-3$ ) or cancelling terms that are in the same part of the fraction (e.g., cancelling $4y$ with $2y$ without considering the rest of the expression).
Ignoring Restrictions: Forgetting about the values of the variable that make the original denominators zero. While not always explicitly asked for in multiple-choice, it's a critical part of understanding rational expressions.

By carefully following the steps of factoring, multiplying, and cancelling, and by checking our work against the given options, we've confidently arrived at the correct answer. Keep practicing these types of problems, and you'll become a simplification pro in no time!