Solve $rac{8}{x}+2=rac{x+4}{x-6}$ For X

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Hey everyone! Today, we're diving deep into a juicy mathematics problem that's all about finding the solution set of the equation $\frac{8}{x}+2=\frac{x+4}{x-6}$. Remember, we've got those crucial restrictions: $x \neq \{0,6\}$. These little guys are super important because they tell us which values of $x$ would make any part of our equation undefined, like dividing by zero. So, as we work through this, always keep those values in the back of your mind. Our main goal is to isolate $x$ and find out what values it can actually be. We'll be using a bunch of algebraic tricks, like finding common denominators and cross-multiplying, to get this equation into a simpler form. It's kinda like solving a puzzle, where each step brings us closer to the final answer. Don't worry if it looks a bit intimidating at first; we'll break it down piece by piece, making sure we understand every move we make. By the end of this, you'll have a solid grasp on how to tackle these kinds of rational equations and confidently identify their solution sets. We're going to be doing some serious number crunching and algebraic maneuvering, so grab your favorite thinking cap, maybe a snack, and let's get this math party started! We're aiming to find the specific values of 'x' that make this equation true, and remember, those forbidden values (0 and 6) are off the table. Let's get ready to unravel this equation step-by-step.

Understanding Rational Equations and Our Goal

Alright guys, let's get down to business with our mathematics challenge: solving the equation $\frac{8}{x}+2=\frac{x+4}{x-6}$. This is what we call a rational equation, meaning it has fractions where the variable, our beloved $x$, is in the denominator. Now, the first thing we always gotta do with these is to check for restrictions. The problem kindly tells us that $x \neq \{0,6\}$. This is because if $x$ were 0, the term $\frac{8}{x}$ would be undefined (division by zero, yikes!). And if $x$ were 6, the term $\frac{x+4}{x-6}$ would also be undefined for the same reason. So, any potential solutions we find later must not be 0 or 6. If we find a solution that is 0 or 6, we have to throw it out – it's an extraneous solution. Our main mission here is to find the solution set, which is simply the collection of all valid $x$ values that make the equation true. Think of it as finding the treasure map coordinates for $x$. We're going to use our trusty algebraic skills to simplify this beast. We'll likely need to clear out those denominators to make things way easier to handle. This usually involves finding a common denominator for all the fractions involved. Once we have that, we can multiply the entire equation by it, effectively zapping those denominators away. Then, we'll be left with a more straightforward equation, possibly a quadratic one, that we can solve using standard techniques like factoring or the quadratic formula. It's all about transforming the problem into something we know how to solve. So, keep those restrictions in mind, and let's get ready to simplify!

Step-by-Step Solution: Clearing the Denominators

Okay, team, let's tackle this equation: $\frac{8}{x}+2=\frac{x+4}{x-6}$. Our first major move is to get rid of those pesky denominators. To do this, we need to find the least common denominator (LCD). Looking at the denominators, we have $x$ and $x-6$. The LCD is simply the product of these unique denominators, which is $x(x-6)$. Now, the magic happens when we multiply every single term in the equation by this LCD. This is totally allowed because we're essentially multiplying the equation by 1 (since the LCD over itself equals 1), which doesn't change the equality. So, here we go:

$x(x-6) \times \frac{8}{x} + x(x-6) \times 2 = x(x-6) \times \frac{x+4}{x-6}$

Let's simplify each part:

• For the first term, $\frac{8}{x}$, the $x$ in the denominator cancels out with the $x$ from the LCD. We're left with $(x-6) \times 8$, which simplifies to $8x - 48$.

• For the second term, $x(x-6) \times 2$, we just distribute the 2 across $(x-6)$ and then multiply by $x$. So, it becomes $2x(x-6)$, which expands to $2x^2 - 12x$.

• For the third term, on the right side, the $(x-6)$ in the denominator cancels out with the $(x-6)$ from the LCD. We're left with $x \times (x+4)$, which simplifies to $x^2 + 4x$.

Now, let's put all these simplified terms back into our equation. Our equation transforms from a messy rational one into a much cleaner polynomial equation:

$(8x - 48) + (2x^2 - 12x) = x^2 + 4x$

See? No more fractions! This is a huge win. This step is absolutely crucial for solving rational equations because it turns them into equations we're already familiar with, like linear or quadratic equations. The key here is to be super careful with your distribution and cancellation. A small mistake here can send you down the wrong path. Always double-check your multiplications and cancellations. Remember those initial restrictions ($x \neq 0, 6$)? They still hold true, even though they've disappeared from the equation itself. We need to keep them in mind for when we find our final answers. This is where the transformation really starts paying off, making the rest of the solving process much more manageable. We're on the home stretch to finding our solution set!

Simplifying and Rearranging into a Quadratic Equation

Now that we've cleared those denominators, our equation looks like this: $8x - 48 + 2x^2 - 12x = x^2 + 4x$. Our next mission, should we choose to accept it (and we totally should!), is to simplify this beast and rearrange it into a standard quadratic form, which is $ax^2 + bx + c = 0$. This form is super helpful because it allows us to use well-known methods like factoring or the quadratic formula to find our solutions.

First, let's combine like terms on the left side of the equation. We have $2x^2$ as our only $x^2$ term. For the $x$ terms, we have $8x$ and $-12x$, which combine to give us $-4x$. And our constant term is $-48$. So, the left side simplifies to $2x^2 - 4x - 48$.

Our equation now stands as: $2x^2 - 4x - 48 = x^2 + 4x$.

To get it into the $ax^2 + bx + c = 0$ format, we need to move all terms to one side, usually the left side, to set the equation equal to zero. We can do this by subtracting $x^2$ from both sides and subtracting $4x$ from both sides.

Subtracting $x^2$ from both sides: $(2x^2 - x^2) - 4x - 48 = (x^2 - x^2) + 4x$ $x^2 - 4x - 48 = 4x$

Now, subtracting $4x$ from both sides: $x^2 - 4x - 4x - 48 = 4x - 4x$ $x^2 - 8x - 48 = 0$

Boom! We have successfully transformed our original rational equation into a beautiful, standard quadratic equation: $x^2 - 8x - 48 = 0$. This is a major victory, guys! This form is so much easier to work with. Remember those restrictions we talked about ($x \neq 0, 6$)? They're still lurking in the background, and we'll need them later, but for now, our focus is on solving this quadratic. Finding the roots of this equation will give us the potential solutions to our original problem. It's like we've narrowed down the possibilities significantly. This step requires careful algebra – combining terms, moving them across the equals sign (remembering to change their signs), and keeping everything organized. A tidy workspace and a clear head make all the difference here!

Solving the Quadratic Equation: Factoring and the Quadratic Formula

We've arrived at the quadratic equation $x^2 - 8x - 48 = 0$. Now, we have two main ways to solve this: factoring or using the quadratic formula. Let's explore both!

Method 1: Factoring

Factoring is usually the quickest method if it works. We're looking for two numbers that multiply to $-48$ (the constant term) and add up to $-8$ (the coefficient of the $x$ term). Let's list some pairs of factors for $-48$ and see if any add up to $-8$:

• $1$ and $-48$ (Sum: $-47$)
• $-1$ and $48$ (Sum: $47$)
• $2$ and $-24$ (Sum: $-22$)
• $-2$ and $24$ (Sum: $22$)
• $3$ and $-16$ (Sum: $-13$)
• $-3$ and $16$ (Sum: $13$)
• $4$ and $-12$ (Sum: $-8$)

Bingo! We found our pair: $4$ and $-12$. They multiply to $-48$ and add up to $-8$. So, we can factor our quadratic equation as:

$(x + 4)(x - 12) = 0$

For this product to be zero, at least one of the factors must be zero. So, we set each factor equal to zero and solve for $x$:

• $x + 4 = 0 \implies x = -4$
• $x - 12 = 0 \implies x = 12$

So, our potential solutions from factoring are $x = -4$ and $x = 12$.

Method 2: The Quadratic Formula

If factoring wasn't obvious, or if the numbers were messier, the quadratic formula is our reliable backup. For an equation $ax^2 + bx + c = 0$, the formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $x^2 - 8x - 48 = 0$, we have $a=1$, $b=-8$, and $c=-48$.

Let's plug these values into the formula:

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-48)}}{2(1)}$

$x = \frac{8 \pm \sqrt{64 + 192}}{2}$

$x = \frac{8 \pm \sqrt{256}}{2}$

Now, we need to find the square root of 256. We know $15^2 = 225$ and $16^2 = 256$. So, $\sqrt{256} = 16$.

$x = \frac{8 \pm 16}{2}$

This gives us two possible solutions:

• $x_1 = \frac{8 + 16}{2} = \frac{24}{2} = 12$
• $x_2 = \frac{8 - 16}{2} = \frac{-8}{2} = -4$

Both methods give us the same potential solutions: $x = 12$ and $x = -4$. Pretty neat how they line up, right? This confirms our calculations are on the right track. The quadratic formula is a lifesaver when factoring feels impossible or when dealing with non-integer roots.

Checking for Extraneous Solutions and Finalizing the Solution Set

We've done the heavy lifting, guys! We found two potential solutions for our equation: $x = -4$ and $x = 12$. But remember that crucial first step we talked about? The restrictions! We were told that $x \neq \{0,6\}$. This is super important because sometimes, when we manipulate equations, we can introduce solutions that don't actually work in the original equation. These are called extraneous solutions. We absolutely must check our potential solutions against these restrictions.

Let's check $x = -4$: Is $-4$ equal to $0$? No. Is $-4$ equal to $6$? No. So, $-4$ is not an extraneous solution. It seems like a valid solution.

Let's check $x = 12$: Is $12$ equal to $0$? No. Is $12$ equal to $6$? No. So, $12$ is not an extraneous solution either. It also looks like a valid solution.

Since neither of our potential solutions ($x = -4$ and $x = 12$) violates the restrictions ($x \neq 0$ and $x \neq 6$), both are valid solutions to the original equation $\frac{8}{x}+2=\frac{x+4}{x-6}$.

Therefore, the solution set of the equation is the collection of these two values. We write this set using curly braces.

The solution set is $\{-4, 12\}$.

This is our final answer! We successfully navigated the complexities of a rational equation, simplified it down to a quadratic, solved the quadratic, and most importantly, verified our solutions against the initial restrictions. This process is fundamental in algebra, and mastering it means you're well-equipped to handle a wide range of equation-solving challenges. Always remember to check those restrictions – they are the gatekeepers of valid solutions!

Conclusion: Mastering Rational Equations

So there you have it, folks! We've successfully conquered the equation $\frac{8}{x}+2=\frac{x+4}{x-6}$ and found its solution set to be $\{-4, 12\}$. The journey involved understanding rational equations, identifying restrictions, clearing denominators by multiplying by the LCD, simplifying to a quadratic equation, and finally, solving that quadratic using either factoring or the quadratic formula. The absolute key takeaway here is the importance of those initial restrictions ($x \neq 0, 6$). Failing to check for extraneous solutions can lead to incorrect answers. Remember, every step in solving these equations is like a building block; get one wrong, and the whole structure might tumble. But with careful algebraic manipulation and a constant eye on those restrictions, you can confidently tackle any rational equation thrown your way.

Working through problems like this isn't just about getting the right answer; it's about developing critical thinking and problem-solving skills. You learn to break down complex problems into smaller, manageable steps, which is a valuable skill not just in mathematics, but in life! Keep practicing, keep experimenting with different methods, and don't be afraid to ask questions. The more you practice, the more intuitive these steps will become. You'll start spotting patterns and potential pitfalls more easily. So, pat yourselves on the back for making it through this one! You've deepened your understanding of algebraic techniques and reinforced the importance of rigorous checking. Keep that mathematical curiosity alive, and you'll continue to unlock new levels of understanding!