Solving Rational Equations 8/(y-2) - 4/y = (6y+2)/(y^2-4)

Hey guys! Let's dive into solving a rational equation today. Rational equations might look intimidating at first glance with their fractions and variables in the denominator, but don't worry, we'll break it down step by step. Specifically, we're going to tackle this equation:

$$\frac{8}{y-2} - \frac{4}{y} = \frac{6y+2}{y^2-4}$$

I know, I know, it looks like a beast, but trust me, we can tame it. Our goal here is not just to find the solution, but to understand the process of solving rational equations. We'll cover everything from identifying restrictions on the variable to checking our final answers. So, grab a pencil and some paper, and let's get started!

1. Identifying Restrictions: The Ground Rules

Before we even think about manipulating the equation, we need to lay down some ground rules. In the world of fractions, there's one cardinal sin: thou shalt not divide by zero! Seriously, it breaks the universe (or at least your calculator). So, we need to identify any values of y that would make the denominator of any of our fractions equal to zero. These values are called restrictions, and we need to exclude them from our possible solutions.

Let's look at our equation again:

$$\frac{8}{y-2} - \frac{4}{y} = \frac{6y+2}{y^2-4}$$

We have three denominators to worry about: y - 2, y, and y² - 4. Let's analyze each one:

• y - 2: If y - 2 = 0, then y = 2. So, y cannot be 2.
• y: If y = 0, then we have division by zero. So, y cannot be 0.
• y² - 4: This one is a bit trickier, but we can factor it! y² - 4 is a difference of squares, which factors into (y - 2)(y + 2). If (y - 2)(y + 2) = 0, then either y - 2 = 0 or y + 2 = 0. This gives us y = 2 or y = -2. We already knew y couldn't be 2, but now we also know that y cannot be -2.

So, our restrictions are y ≠ 2, y ≠ 0, and y ≠ -2. Write these down somewhere – they're important! We'll need to check our final answer against these restrictions later.

Why are restrictions so important? Imagine we solve the equation and get y = 2 as a solution. If we plug that back into the original equation, we'd be dividing by zero, which is undefined. That means y = 2 isn't a real solution, even if our algebra looks right. Restrictions protect us from these false solutions.

2. Clearing the Fractions: Multiplying by the LCD

Now that we've established the ground rules, let's get rid of those pesky fractions! The key to doing this is to multiply both sides of the equation by the least common denominator (LCD) of all the fractions. The LCD is the smallest expression that all the denominators will divide into evenly.

Looking at our denominators – y - 2, y, and y² - 4 (which we know factors into (y - 2)(y + 2)) – what's the LCD? Well, it needs to include all the factors present in each denominator. So, the LCD is y(y - 2)(y + 2). Notice how it includes y, (y - 2), and (y + 2), covering all our bases.

Now, we multiply both sides of the equation by this LCD:

$$y(y-2)(y+2) \left( \frac{8}{y-2} - \frac{4}{y} \right) = y(y-2)(y+2) \left( \frac{6y+2}{y^2-4} \right)$$

This looks messy, I know, but the magic is about to happen! We distribute the LCD to each term on the left side:

$$y(y-2)(y+2) \cdot \frac{8}{y-2} - y(y-2)(y+2) \cdot \frac{4}{y} = y(y-2)(y+2) \cdot \frac{6y+2}{(y-2)(y+2)}$$

Now comes the fun part: canceling! Notice how terms in the numerators and denominators start to cancel each other out. This is the whole point of multiplying by the LCD – to eliminate the fractions.

• In the first term, (y - 2) cancels out, leaving us with 8y(y + 2).
• In the second term, y cancels out, leaving us with 4(y - 2)(y + 2).
• On the right side, (y - 2)(y + 2) cancels out, leaving us with y(6y + 2).

After canceling, our equation looks much simpler:

$$8y(y+2) - 4(y-2)(y+2) = y(6y+2)$$

We've successfully cleared the fractions! Give yourself a pat on the back. The hardest part is often getting rid of those denominators. Now, we're left with a good ol' polynomial equation, which we know how to handle.

3. Simplifying and Solving: The Algebra Gym

Okay, now it's time to flex our algebraic muscles! We need to simplify both sides of the equation and then solve for y. This involves distributing, combining like terms, and possibly factoring.

Let's start by distributing:

$$8y^2 + 16y - 4(y^2 - 4) = 6y^2 + 2y$$

Notice that we multiplied (y - 2)(y + 2) to get (y² - 4) directly, since it's a difference of squares. Now, distribute the -4:

$$8y^2 + 16y - 4y^2 + 16 = 6y^2 + 2y$$

Combine like terms on the left side:

$$4y^2 + 16y + 16 = 6y^2 + 2y$$

Now, let's get all the terms on one side to set the equation equal to zero. We'll subtract 4y² from both sides, subtract 16y from both sides, and subtract 16 from both sides:

$$0 = 2y^2 - 14y - 16$$

We now have a quadratic equation! Before we jump into factoring or using the quadratic formula, let's see if we can simplify it. Notice that all the coefficients are even, so we can divide both sides by 2:

$$0 = y^2 - 7y - 8$$

Ah, much better! Now, let's try to factor this quadratic. We're looking for two numbers that multiply to -8 and add up to -7. Those numbers are -8 and 1. So, we can factor the quadratic as:

$$0 = (y - 8)(y + 1)$$

Now, we use the zero product property, which says that if the product of two factors is zero, then at least one of the factors must be zero. So:

• y - 8 = 0 => y = 8
• y + 1 = 0 => y = -1

So, we have two potential solutions: y = 8 and y = -1.

4. Checking for Extraneous Solutions: The Reality Check

We're almost there! But we're not done until we've done the most crucial step: checking for extraneous solutions. Remember those restrictions we identified way back in step 1? This is where they come into play. We need to make sure that neither of our potential solutions violates those restrictions.

Our restrictions were y ≠ 2, y ≠ 0, and y ≠ -2. Let's check our solutions:

• y = 8: 8 is not equal to 2, 0, or -2. So, y = 8 is a valid solution.
• y = -1: -1 is not equal to 2, 0, or -2. So, y = -1 is also a valid solution.

Both of our solutions passed the reality check! This means they are both actual solutions to the original equation.

What are extraneous solutions? Extraneous solutions are values that we get when solving an equation that appear to be solutions, but when we plug them back into the original equation, they don't work. They usually arise when we perform operations that can introduce false solutions, like squaring both sides of an equation or, in this case, multiplying by an expression containing a variable (our LCD). That's why checking is so important!

5. The Grand Finale: Stating the Solution

We did it! We successfully solved the rational equation. Our solutions are y = 8 and y = -1. We can write this in set notation as {-1, 8}.

Final Answer: The solution(s) is/are -1, 8.

Key Takeaways: Mastering Rational Equations

Solving rational equations might seem daunting at first, but by following these steps, you can conquer them:

1. Identify Restrictions: Find any values of the variable that would make a denominator equal to zero. These values are not allowed solutions.
2. Clear the Fractions: Multiply both sides of the equation by the least common denominator (LCD) of all the fractions.
3. Simplify and Solve: Distribute, combine like terms, and solve the resulting polynomial equation. This might involve factoring, using the quadratic formula, or other algebraic techniques.
4. Check for Extraneous Solutions: Plug each potential solution back into the original equation to make sure it doesn't violate any restrictions or create undefined terms. Discard any extraneous solutions.
5. State the Solution: Write down the valid solutions you found.

By practicing these steps, you'll become a rational equation-solving pro! Remember to always be careful with your algebra and pay close attention to those restrictions. Good luck, and happy solving!