Solving Equations: Finding Solutions And Avoiding Extraneous Ones

Hey everyone! Today, we're diving into a classic algebra problem: solving an equation involving fractions and making sure we don't get tripped up by any sneaky extraneous solutions. Let's break down the problem step by step. Solving equations with fractions can sometimes feel like navigating a maze, but with a little patience and the right approach, we can conquer it. Extraneous solutions? Don't worry, we'll show you how to spot those too! We'll make sure you understand every single part of this process. So, let's get started and make sure you understand how to solve these types of equations. This detailed guide will help you understand and solve the equation: $\frac{4 x}{x+4}-\frac{3 x+3}{x+2}=\frac{6}{x^2+6 x+8}$. We'll cover all the critical steps, from finding common denominators to checking for those pesky extraneous solutions. Ready? Let's jump in!

Understanding the Problem: The Equation and Its Challenges

So, the equation we are looking at today is: $\frac{4 x}{x+4}-\frac{3 x+3}{x+2}=\frac{6}{x^2+6 x+8}$. At first glance, this equation might seem a bit intimidating. We have fractions, variables in the denominators, and the potential for some tricky calculations. Let's face it: working with fractions can be a pain, but with a systematic approach, we can tame this equation. One of the main challenges when solving equations like this is dealing with the denominators. These denominators can't be zero, because that would make the fractions undefined. This immediately gives us some restrictions on what values x can take. Another potential pitfall is the dreaded extraneous solution. These are solutions that we find through our calculations but that don't actually work when we plug them back into the original equation. The goal is to get a clear understanding of how to approach this type of equation and to be able to identify potential pitfalls and how to handle them. We will approach this problem systematically, ensuring that we don't miss any steps.

Identifying Potential Pitfalls

Before we even start solving, we need to be aware of some potential issues. Let's take a closer look at the denominators in our equation. We've got $x + 4$, $x + 2$, and $x^2 + 6x + 8$. We need to make sure that none of these are equal to zero, because division by zero is not allowed in math. So, the first thing to do is figure out what values of x would make these denominators zero. First, consider $x + 4 = 0$. This would mean $x = -4$. Next, consider $x + 2 = 0$. This would mean $x = -2$. Finally, let's factor the quadratic in the denominator $x^2 + 6x + 8$. It factors to $(x + 4)(x + 2)$. From this, we can see that if $x = -4$ or $x = -2$, the denominator is also zero. So, our initial restrictions on x are that it cannot be -4 or -2. These are the values that we need to exclude from our potential solutions. These are crucial to our problem-solving, as they will help us identify extraneous solutions later on. We'll keep these values in mind as we proceed through the problem. Understanding these restrictions is the first step in ensuring that our solutions are valid.

Step-by-Step Solution: Solving the Equation

Now, let's roll up our sleeves and solve this equation step by step. Our goal is to find the values of x that satisfy the equation, while remembering the restrictions we've just discussed. We'll break it down into manageable parts. First, we need to get rid of the fractions. The easiest way to do this is to find a common denominator and then multiply both sides of the equation by it. After that, we'll simplify the resulting equation, and then finally solve for x. So, let's get started!

Finding a Common Denominator

To get rid of the fractions, we need a common denominator. Looking at our denominators, we have $x + 4$, $x + 2$, and $(x + 4)(x + 2)$. The least common denominator (LCD) is $(x + 4)(x + 2)$. So, we'll multiply every term in the equation by this LCD. This process will effectively eliminate the fractions, making it easier to solve for x. This is a critical step in simplifying our equation. By multiplying each term by the LCD, we'll transform our equation into a simpler form that we can easily solve. This will eliminate all of the fractions, which makes our equation more manageable. Let's see how this works. First, we'll multiply the entire equation by $(x + 4)(x + 2)$.

Multiplying by the Common Denominator

Let's multiply both sides of the equation by the common denominator, $(x + 4)(x + 2)$.

$\frac{4 x}{x+4}*(x+4)(x+2) - \frac{3 x+3}{x+2}*(x+4)(x+2) = \frac{6}{x^2+6 x+8}*(x+4)(x+2)$

Now, simplify each term:

• For the first term: $\frac{4x}{x+4}*(x+4)(x+2)$, the $(x + 4)$ terms cancel out, leaving $4x(x + 2)$.
• For the second term: $\frac{3x+3}{x+2}*(x+4)(x+2)$, the $(x + 2)$ terms cancel out, leaving $(3x + 3)(x + 4)$.
• For the third term: $\frac{6}{(x+4)(x+2)}*(x+4)(x+2)$, the entire denominator cancels out, leaving just 6.

This simplifies our equation to:

$4x(x + 2) - (3x + 3)(x + 4) = 6$

Expanding and Simplifying the Equation

Now that we've gotten rid of the fractions, it's time to expand and simplify the equation. We'll multiply out the terms and combine like terms to get a simpler equation to work with. This is where we use the distributive property and combine the like terms. Let's expand the terms in the equation $4x(x + 2) - (3x + 3)(x + 4) = 6$:

• $4x(x + 2) = 4x^2 + 8x$
• $(3x + 3)(x + 4) = 3x^2 + 12x + 3x + 12 = 3x^2 + 15x + 12$

Now, substitute these back into the equation:

$4x^2 + 8x - (3x^2 + 15x + 12) = 6$

Distribute the negative sign:

$4x^2 + 8x - 3x^2 - 15x - 12 = 6$

Combine like terms:

$(4x^2 - 3x^2) + (8x - 15x) - 12 = 6$

$x^2 - 7x - 12 = 6$

Solving the Quadratic Equation

We now have a quadratic equation, which we need to solve for x. To do this, we need to rearrange the equation so that it's in the standard form $ax^2 + bx + c = 0$. So, let's subtract 6 from both sides of the equation: $x^2 - 7x - 12 - 6 = 0$, which simplifies to $x^2 - 7x - 18 = 0$. We can factor this quadratic equation as $(x - 9)(x + 2) = 0$. This is a critical step towards finding the solution. To solve for x, we set each factor equal to zero and solve. We can determine the values of x that satisfy the equation by solving the individual equations.

Set each factor to zero:

• $x - 9 = 0$ gives us $x = 9$
• $x + 2 = 0$ gives us $x = -2$

We now have two potential solutions: x = 9 and x = -2. But, we're not done yet! We need to check these for extraneous solutions. This is a critical part of the process that we can't skip.

Checking for Extraneous Solutions: The Final Test

We have our potential solutions, x = 9 and x = -2. However, we must now check these solutions against the restrictions we identified at the beginning: x cannot be -4 or -2. So, this final step is critical to make sure that our solutions are valid and don't create undefined fractions. We need to make sure that our solutions don't make any of the original denominators equal to zero. Let's test each solution to see if it works.

Testing Each Solution

Let's substitute each potential solution back into the original equation $\frac{4 x}{x+4}-\frac{3 x+3}{x+2}=\frac{6}{x^2+6 x+8}$ and check if it makes the equation true.

• Test x = 9: Substitute x = 9 into the equation: $\frac{4(9)}{9+4}-\frac{3(9)+3}{9+2}=\frac{6}{9^2+6(9)+8}$. This simplifies to $\frac{36}{13} - \frac{30}{11} = \frac{6}{149}$. This is not zero. So, x = 9 is a valid solution.
• Test x = -2: We already know that x cannot equal -2 because it makes the denominator of the second fraction equal to zero. Thus, x = -2 is an extraneous solution because it is already restricted.

Identifying Extraneous Solutions

As we have seen, x = -2 makes the original denominators undefined. Therefore, x = -2 is an extraneous solution. Extraneous solutions often arise when we perform operations that can introduce new solutions that don't satisfy the original equation. We must always go back and check our solutions against our initial restrictions to ensure they are valid. This is a really important step to ensure we get the right answer and understand the behavior of the original equation.

Conclusion: The Final Answer

After careful analysis, we found that our original equation $\frac{4 x}{x+4}-\frac{3 x+3}{x+2}=\frac{6}{x^2+6 x+8}$ has only one valid solution. We went through each step to solve it, identified any pitfalls, and then rigorously checked for extraneous solutions. So, the answer is:

$\boxed{x=9}$

Remember, always check your solutions, especially when dealing with fractions. That is all, folks! I hope you enjoyed this lesson. Keep practicing, and you will become an expert at solving equations like this! Feel free to ask any questions in the comments below! Keep learning and keep growing! See ya!