Solving Equations: What To Multiply?

Hey Plastik Magazine readers! Let's dive into a mathematical puzzle today. Janet's got an equation she needs to solve, and we're here to help her figure out the best way to tackle it. The equation is $y+\frac{y^2-5}{y^2-1}=\frac{y^2+y+2}{y+1}$, and the big question is: what should she multiply both sides of the equation by to get rid of those pesky fractions? This is a classic algebra problem, and by understanding the underlying principles, you'll be able to solve similar equations with confidence. So, grab your thinking caps, and let’s break it down together!

Understanding the Problem

To really nail this problem, we need to understand what's going on with the equation. Our main goal here is to eliminate the fractions. Fractions can make equations look intimidating, but they're much easier to handle once you clear them out. So, how do we do that? Well, we need to find a common expression that, when multiplied across the entire equation, will cancel out the denominators (the bottom parts of the fractions). Think of it like finding a common language that all the fractions can understand. When we multiply both sides of the equation by this common expression, we ensure that the equation remains balanced while simultaneously getting rid of those fractions. The key is to identify the denominators and figure out their least common multiple, which will be our magic expression.

Now, let's look at the equation again: $y+\frac{y^2-5}{y^2-1}=\frac{y^2+y+2}{y+1}$. See those denominators? We've got $y^2-1$ and $y+1$. The first step is to factor those denominators. Factoring is like taking a number or expression and breaking it down into smaller pieces that multiply together to give you the original thing. It's a crucial skill in algebra, and it helps us see the structure of expressions more clearly. When we factor, we reveal the building blocks that make up the expression, making it easier to manipulate and simplify. In this case, factoring the denominators will help us identify the common factors and determine the least common multiple.

Factoring the Denominators

Okay, let's get our factoring hats on! We need to factor the denominators in Janet's equation. The first denominator we encounter is $y^2-1$. Does this expression look familiar? It should! It's a classic example of what we call a "difference of squares." A difference of squares is when you have something squared (like $y^2$) minus another thing squared (like $1$, which is $1^2$). There's a handy formula for factoring these: $a^2 - b^2 = (a + b)(a - b)$. This formula is your best friend when you spot a difference of squares. It allows you to quickly and easily break down the expression into its factors. So, applying this formula to $y^2-1$, we get $(y + 1)(y - 1)$.

Now, let’s look at the other denominator, $y+1$. Can we factor this any further? Nope, it's already in its simplest form! It's a linear expression, meaning the highest power of $y$ is 1, and it can't be broken down into smaller polynomial factors. So, we can leave it as is. Now that we've factored the denominators, we have a much clearer picture of what we're working with. We've transformed the original denominators into their component parts, making it easier to identify common factors and determine the expression we need to multiply both sides of the equation by. This is a crucial step in solving the equation because it allows us to clear the fractions and work with a simpler, more manageable expression.

Finding the Least Common Multiple (LCM)

Alright, we've factored the denominators, and now we're ready to find the Least Common Multiple (LCM). Think of the LCM as the smallest expression that both denominators can divide into evenly. It's like finding the smallest common ground between the denominators. This is super important because the LCM is exactly what we'll use to multiply both sides of the equation, effectively clearing those fractions. To find the LCM, we look at all the factors that appear in either denominator and take the highest power of each factor. This ensures that the LCM includes all the necessary factors to cancel out each denominator.

Let's recap our factored denominators: we have $(y + 1)(y - 1)$ and $(y + 1)$. What factors do we see? We have $(y + 1)$ and $(y - 1)$. The factor $(y + 1)$ appears in both denominators, but we only need to include it once in the LCM. The factor $(y - 1)$ appears only in the first denominator, so we need to include it as well. Therefore, the LCM is the product of these factors: $(y + 1)(y - 1)$. This expression is the magic key that will unlock the equation and make it much easier to solve. By multiplying both sides of the equation by this LCM, we'll be able to cancel out the denominators and work with a simpler equation without fractions. This is a common technique in algebra, and mastering it will make solving equations like this a breeze.

The Answer and Why

So, what should Janet multiply both sides of the equation by? We've figured out that the LCM of the denominators is $(y + 1)(y - 1)$. But wait! Remember how we factored $y^2 - 1$? We found that $y^2 - 1 = (y + 1)(y - 1)$. This means the LCM is the same as $y^2 - 1$! Looking back at our answer choices, we see that option B, $y^2-1$, is the correct one.

Why does this work? When we multiply both sides of the equation by $y^2 - 1$, we're essentially multiplying each term by $(y + 1)(y - 1)$. Let's see how this clears the fractions: For the term $\frac{y^2-5}{y^2-1}$, when we multiply by $y^2 - 1$, the denominator cancels out completely, leaving us with just $y^2 - 5$. For the term $\frac{y^2+y+2}{y+1}$, when we multiply by $(y + 1)(y - 1)$, the $(y + 1)$ in the denominator cancels with the $(y + 1)$ in our multiplier, leaving us with $(y^2 + y + 2)(y - 1)$. So, multiplying by $y^2 - 1$ neatly eliminates all the fractions, making the equation much easier to solve. This is the power of finding the LCM! It allows us to transform a complex equation with fractions into a simpler equation that we can handle more easily.

Wrapping Up

So there you have it, folks! Janet should multiply both sides of her equation by B. $y^2-1$ to solve it. We walked through the process step by step, from understanding the problem and factoring the denominators to finding the LCM and seeing how it clears the fractions. Remember, the key to solving equations with fractions is to find that LCM and use it to eliminate the denominators. This technique is a fundamental skill in algebra, and mastering it will make you a math whiz in no time! Keep practicing, and you'll be solving equations like a pro.