Solving Rational Equations: A Step-by-Step Guide

Have you ever stumbled upon an equation that looks like a fraction party, with variables hanging out in the denominators? These are called rational equations, and while they might seem intimidating at first, they're actually quite manageable with the right approach. In this guide, we'll break down the process of solving the rational equation $\frac{3}{y+5}-\frac{9}{y-5}=\frac{6}{y^2-25}$ step by step. So, grab your algebraic toolkit, and let's dive in!

Understanding Rational Equations

Before we get our hands dirty with the equation, let's understand rational equations better. A rational equation is simply an equation that contains one or more rational expressions. A rational expression, in turn, is a fraction where the numerator and/or the denominator are polynomials. Think of it as a fancy way of saying equations with fractions where variables can pop up in the bottom part. For instance, $\frac{x+1}{x-2}$, $\frac{5}{x}$, and even plain old polynomials like $x^2 + 3$ (which can be thought of as $\frac{x^2+3}{1}$) can be part of a rational equation.

Solving rational equations is crucial in many areas of mathematics and its applications. They frequently appear in problems involving rates, proportions, and inverse relationships. Mastering the techniques to solve these equations opens doors to tackling more complex mathematical models and real-world problems. For instance, if you're dealing with mixing solutions in chemistry, figuring out travel times with varying speeds, or even modeling electrical circuits, you might encounter rational equations.

When facing these equations, the key is to eliminate the fractions. We do this by finding a common denominator, multiplying both sides of the equation, and then solving the resulting equation, which is hopefully simpler. However, there's a crucial step: always check your solutions! Because we're dealing with fractions, there might be values that make the denominator zero, leading to undefined expressions. We call these extraneous solutions, and we need to discard them.

Identifying the Challenge

In this particular case, our equation is $\frac{3}{y+5}-\frac{9}{y-5}=\frac{6}{y^2-25}$. The presence of variables in the denominators ($y+5$, $y-5$, and $y^2-25$) is what makes this a rational equation. Our goal is to find the value(s) of y that make this equation true. But before we jump into solving, it's wise to take a quick peek at the denominators. We need to be mindful of any values of y that would make them zero, as division by zero is a big no-no in mathematics. These values will be our potential "problem spots" or extraneous solutions that we'll need to watch out for later.

Step 1: Factoring and Finding the Least Common Denominator (LCD)

Our first strategic move in tackling this rational equation is to identify the least common denominator (LCD). The LCD is the smallest expression that all the denominators can divide into evenly. Think of it as the common ground where all our fractions can meet. To find it, we first need to factor each denominator completely. This is like breaking down the denominators into their prime "ingredients."

Looking at our equation, $\frac{3}{y+5}-\frac{9}{y-5}=\frac{6}{y^2-25}$, we have three denominators: $y+5$, $y-5$, and $y^2-25$. The first two, $y+5$ and $y-5$, are already in their simplest form – they can't be factored any further. However, the third denominator, $y^2-25$, looks familiar. It's a difference of squares, a classic factoring pattern!

Remember the difference of squares pattern? It states that $a^2 - b^2$ can be factored into $(a + b)(a - b)$. Applying this to our denominator, $y^2 - 25$ can be factored into $(y + 5)(y - 5)$. Now we have all our denominators in their factored form: $y+5$, $y-5$, and $(y + 5)(y - 5)$.

With the denominators factored, finding the LCD becomes much clearer. We need an expression that contains all the factors present in any of the denominators. We have $(y + 5)$ and $(y - 5)$ as factors. Therefore, the LCD is simply the product of these factors: $(y + 5)(y - 5)$. This expression is divisible by each of the individual denominators.

Identifying the LCD is a critical step because it allows us to clear the fractions from the equation. It’s like finding the right tool for the job – in this case, the tool that will make our equation much easier to handle.

Step 2: Multiplying Both Sides by the LCD

Now that we've identified the LCD as $(y + 5)(y - 5)$, the next step is to multiply both sides of our equation by this LCD. This is a crucial step because it will eliminate the fractions, transforming our rational equation into a more manageable linear equation. Think of it as using a powerful "fraction-busting" tool.

Our equation is $\frac{3}{y+5}-\frac{9}{y-5}=\frac{6}{(y+5)(y-5)}$. We'll multiply both the left-hand side and the right-hand side of this equation by $(y + 5)(y - 5)$. This gives us:

$(y + 5)(y - 5) * [\frac{3}{y+5}-\frac{9}{y-5}] = (y + 5)(y - 5) * \frac{6}{(y+5)(y-5)}$

Now, we distribute the LCD on the left-hand side. This means we multiply each term inside the brackets by $(y + 5)(y - 5)$:

$(y + 5)(y - 5) * \frac{3}{y+5} - (y + 5)(y - 5) * \frac{9}{y-5} = (y + 5)(y - 5) * \frac{6}{(y+5)(y-5)}$

This is where the magic happens! Notice how terms start to cancel out. In the first term on the left, $(y + 5)$ in the numerator and denominator cancel out. In the second term, $(y - 5)$ cancels out. On the right-hand side, the entire $(y + 5)(y - 5)$ term cancels out. This is the power of multiplying by the LCD – it gets rid of those pesky fractions!

After canceling, we're left with:

$3(y - 5) - 9(y + 5) = 6$

We've successfully transformed our rational equation into a linear equation! This equation looks much friendlier and easier to solve. The fractions are gone, and we're left with a simple algebraic expression.

Multiplying by the LCD is a cornerstone technique in solving rational equations. It clears the path, allowing us to work with a cleaner, more straightforward equation. But remember, it's crucial to perform this multiplication on both sides of the equation to maintain the balance and ensure we're solving for the correct values.

Step 3: Simplifying and Solving the Equation

With the fractions cleared, we're now facing a much simpler equation: $3(y - 5) - 9(y + 5) = 6$. This is a linear equation, and we have a standard set of tools to tackle it. Our goal is to isolate y on one side of the equation, revealing its value.

The first step in simplifying this equation is to distribute the constants. This means multiplying the 3 by both terms inside the first parentheses and the -9 by both terms inside the second parentheses:

$3 * y - 3 * 5 - 9 * y - 9 * 5 = 6$

This simplifies to:

$3y - 15 - 9y - 45 = 6$

Now, we combine like terms on the left-hand side. We have two terms with y ($3y$ and $-9y$) and two constant terms ($-15$ and $-45$). Combining them gives us:

$-6y - 60 = 6$

Our next step is to isolate the term with y. To do this, we add 60 to both sides of the equation. This will cancel out the -60 on the left-hand side:

$-6y - 60 + 60 = 6 + 60$

Which simplifies to:

$-6y = 66$

Finally, to solve for y, we divide both sides of the equation by -6:

$\frac{-6y}{-6} = \frac{66}{-6}$

This gives us our solution:

$y = -11$

We've found a potential solution for y! But before we celebrate, we have one crucial step remaining: checking for extraneous solutions.

Simplifying and solving the equation after clearing fractions is a fundamental algebraic process. It involves distribution, combining like terms, and isolating the variable – skills that are essential not just for rational equations but for a wide range of mathematical problems. By carefully applying these steps, we can transform complex-looking equations into simple solutions.

Step 4: Checking for Extraneous Solutions

We've arrived at a potential solution: $y = -11$. However, in the world of rational equations, things aren't always as they seem. Because we've manipulated fractions with variables in the denominators, there's a chance that our solution might be an extraneous solution. An extraneous solution is a value that satisfies the transformed equation (the linear equation we solved) but doesn't work in the original rational equation.

Why does this happen? Extraneous solutions arise when a value makes one of the denominators in the original equation equal to zero. Division by zero is undefined in mathematics, so any such value cannot be a true solution to the equation. It's like a mirage – it looks like a solution, but it disappears when you get closer.

To check for extraneous solutions, we need to go back to our original equation, $\frac{3}{y+5}-\frac{9}{y-5}=\frac{6}{y^2-25}$, and substitute our potential solution, $y = -11$, into the denominators. If any of the denominators become zero, then $y = -11$ is an extraneous solution, and we must discard it.

Let's check each denominator:

• $y + 5 = -11 + 5 = -6$ (Not zero)
• $y - 5 = -11 - 5 = -16$ (Not zero)
• $y^2 - 25 = (-11)^2 - 25 = 121 - 25 = 96$ (Not zero)

None of the denominators are zero when $y = -11$. This means that $y = -11$ is a valid solution to our original rational equation!

Checking for extraneous solutions is a non-negotiable step when solving rational equations. It's the final quality control check that ensures our solution is not just a mathematical illusion. By substituting our solution back into the original equation's denominators, we can confidently identify and eliminate any extraneous solutions.

Solution

After navigating the twists and turns of solving a rational equation, we've arrived at our destination! We successfully solved the equation $\frac{3}{y+5}-\frac{9}{y-5}=\frac{6}{y^2-25}$ and found that the solution is $y = -11$. We factored denominators, found the LCD, cleared fractions, simplified the equation, and, most importantly, checked for extraneous solutions. This thorough approach ensures that our answer is not only mathematically correct but also valid within the context of the original equation.

The solution $y = -11$ is the value that makes the original equation true. If we substitute -11 for y in the equation, both sides will be equal. This is the essence of solving an equation – finding the value(s) that satisfy the given relationship.

Rational equations might seem complex at first glance, but by breaking them down into manageable steps and understanding the underlying principles, we can confidently tackle them. Remember the key steps: factor, find the LCD, multiply by the LCD, simplify, solve, and, crucially, check for extraneous solutions. With practice and a solid understanding of these techniques, you'll be well-equipped to solve a wide range of rational equations.

This journey through solving rational equations highlights the power of algebraic techniques in unraveling complex mathematical problems. By mastering these skills, you gain a valuable toolset for tackling not just equations but also a variety of real-world applications where rational relationships come into play.

For further information and more examples on solving rational equations, you can visit Khan Academy.