Solving Equations With Fractions What To Multiply By

Hey everyone! Let's dive into a math problem that Janet's tackling. She's trying to solve the equation:

$$y+\frac{y^2-5}{y^2-1}=\frac{y^2+y+2}{y+1}$$

And the question is, what should she multiply both sides of the equation by to get rid of those pesky fractions? We've got a few options: A. y, B. y²-1, C. y+1, and D. y²+y+2. Let's break this down step by step so we can figure out the right move for Janet.

Understanding the Equation

Before we jump into possible solutions, let's really look at this equation. We need to identify what's causing the trouble – and in this case, it's the fractions. Fractions can be a bit of a headache when solving equations, so our main goal is to eliminate them. To do this, we need to find a common expression that we can multiply throughout the equation. This common expression should clear out the denominators in each fraction. When you're staring down fractions in an equation, think about the denominators. The denominators are the key to unlocking the next step. In Janet's equation, we have two denominators: y²-1 and y+1. We need to figure out a single expression that both of these will divide into evenly.

Now, let's take a closer look at those denominators. The first one, y²-1, might look familiar. It's a difference of squares! Remember that cool trick? We can factor y²-1 into (y-1)(y+1). Ah-ha! That's a crucial piece of the puzzle. The second denominator is y+1, which is already in its simplest form. So, now we know our denominators are actually (y-1)(y+1) and (y+1). This factorization is super important because it reveals a common factor, which will guide us to the solution. Factoring helps simplify things and makes the problem much more manageable. By recognizing the difference of squares pattern, we've made a big step forward in figuring out what to multiply by. Always keep an eye out for these patterns, as they can be real lifesavers in algebra!

Finding the Common Denominator

Okay, so we've identified the denominators as (y-1)(y+1) and (y+1). Now, how do we find the expression that both of these divide into? We're essentially looking for the least common denominator (LCD). Think of it like this: we need a single expression that, when we multiply both sides of the equation by it, will cancel out all the denominators. To find the LCD, we need to consider all the unique factors present in our denominators. We have (y-1) and (y+1). The denominator (y-1)(y+1) already includes both of these, and (y+1) is just a part of that. Therefore, the least common denominator is (y-1)(y+1). This is because it contains all the factors from both denominators, ensuring that when we multiply by it, everything will cancel out nicely. So, the expression (y-1)(y+1), which is also y²-1, is our key to eliminating the fractions and making the equation easier to solve. This is a classic technique in algebra: find the LCD, multiply through, and simplify. It turns a messy-looking equation into something much more manageable.

Choosing the Correct Multiplier

Now that we've found the common denominator, y²-1, we know what Janet needs to multiply both sides of the equation by. Let's revisit our options: A. y, B. y²-1, C. y+1, and D. y²+y+2. We've already determined that y²-1 is the key to clearing the fractions, so the correct answer is B. But let's quickly see why the other options don't quite work. Multiplying by just y (option A) wouldn't eliminate the fractions. It would only change the numerators, leaving us with the same problem. Multiplying by y+1 (option C) would get rid of the denominator on the right side of the equation, but it would still leave the fraction with y²-1 on the left side. Option D, y²+y+2, doesn't directly relate to the denominators we're trying to eliminate. It might complicate things further rather than simplifying them. So, by systematically breaking down the problem and identifying the common denominator, we've confidently arrived at the correct multiplier: y²-1. This is a great illustration of how understanding the underlying principles of algebra can lead you to the right answer.

Therefore, Janet should multiply both sides of the equation by B. y²-1.

Why This Works: A Deeper Dive

Let's really understand why multiplying by y²-1 works. It's not just a magic trick; it's based on solid mathematical principles. When we multiply both sides of an equation by the same expression, we maintain the equality. This is a fundamental rule of algebra. The goal here is to strategically choose an expression that will eliminate the denominators. When we multiply the left side of Janet's equation by y²-1, we get:

$$(y^2-1) * \left(y+\frac{y^2-5}{y^2-1}\right)$$

This distributes to:

$$y(y^2-1) + (y^2-1)*\frac{y^2-5}{y^2-1}$$

Notice what happens in the second term? The (y²-1) in the numerator and denominator cancel each other out, leaving us with just (y²-5). That's the magic of the common denominator at work! On the right side of the equation, we have:

$$(y^2-1) * \frac{y^2+y+2}{y+1}$$

Remember that y²-1 can be factored into (y-1)(y+1). So, we have:

$$\frac{(y-1)(y+1)(y^2+y+2)}{y+1}$$

Here, the (y+1) terms cancel out, leaving us with (y-1)(y²+y+2). Again, the common denominator has worked its magic, eliminating the fraction. So, by multiplying both sides by y²-1, we've transformed the original equation with fractions into an equation without fractions, which is much easier to solve. This technique is a cornerstone of algebraic manipulation, and understanding why it works will make you a more confident problem solver. Always remember the power of the common denominator to simplify and conquer!

Key Takeaways for Solving Equations with Fractions

Alright, guys, let's wrap this up with some key takeaways that you can use whenever you encounter equations with fractions. These steps will help you approach these problems systematically and confidently. First and foremost, identify the denominators. This is your starting point. Look closely at each fraction and note what's in the denominator. These are the culprits we need to eliminate. Next, factor the denominators. This is a crucial step, especially if you see expressions like differences of squares or other factorable polynomials. Factoring will help you identify common factors and simplify the process of finding the least common denominator. Then, find the least common denominator (LCD). The LCD is the expression that all the denominators will divide into evenly. It's the key to clearing those fractions. To find it, consider all the unique factors present in the denominators and take the highest power of each. After finding the LCD, multiply both sides of the equation by the LCD. This is where the magic happens. By multiplying every term in the equation by the LCD, you'll eliminate the denominators and transform the equation into a more manageable form. Finally, simplify and solve. Once you've cleared the fractions, you'll be left with a simpler equation that you can solve using standard algebraic techniques. Remember to combine like terms, isolate the variable, and check your solution. By following these steps, you'll be well-equipped to tackle any equation with fractions that comes your way. Keep practicing, and you'll become a fraction-busting pro in no time!

Janet's Next Steps

So, now that we know Janet should multiply both sides of the equation by y²-1, what should she do next? Well, the next step is to actually do the multiplication! She needs to carefully distribute y²-1 to each term on both sides of the equation. This will eliminate the fractions, as we discussed earlier. Once she's done that, she'll have a new equation without any fractions, which should be much easier to work with. The next task is to simplify the equation. This means combining like terms, expanding any products, and rearranging the terms so that the equation is in a standard form (like a polynomial equation set equal to zero). After simplifying, Janet might need to factor or use other algebraic techniques to solve for y. Depending on the complexity of the equation, this could involve factoring a quadratic, using the quadratic formula, or employing other methods. Finally, and this is super important, Janet needs to check her solutions. When you solve equations, especially those involving fractions, it's crucial to plug your answers back into the original equation to make sure they work. Sometimes, you might get solutions that don't actually satisfy the original equation (these are called extraneous solutions). By checking her solutions, Janet can ensure that she has the correct answers. So, Janet's journey isn't over yet, but she's off to a great start by identifying the correct multiplier! And remember, guys, solving math problems is like building a puzzle – each step leads you closer to the final solution.