Solve For X: Algebra Equation Explained

Hey guys! Today, we're diving deep into a classic algebra problem that often pops up in math classes: solving for x. We'll be tackling the equation $7 + \frac{5}{(x - 5)} = \frac{2x}{(x - 5)}$. Don't let those fractions scare you; by the end of this, you'll be a pro at handling these types of equations. We'll break it down step-by-step, making sure you understand every move. So, grab your notebooks, and let's get started on this algebraic adventure!

Understanding the Equation and Key Concepts

Alright, let's first get comfortable with the equation we're working with: $7 + \frac{5}{(x - 5)} = \frac{2x}{(x - 5)}$. The main thing to notice here, my math enthusiasts, is that we have a variable, 'x', in the denominator of our fractions. This is super important because it tells us there's a restriction on the possible values of x. Specifically, the denominator, $(x - 5)$, can never be equal to zero. If it were, we'd be dividing by zero, which is a big no-no in mathematics – it leads to undefined results. So, right off the bat, we know that x cannot be 5. Keep this little nugget in mind; it's crucial for checking our final answers later on. Understanding these restrictions is a fundamental part of solving rational equations, and it often trips people up if they forget. We're essentially looking for values of 'x' that make the left side of the equation perfectly equal to the right side, but we must also ensure that these values don't break the rules of arithmetic by making any denominator zero. This initial check is your first line of defense against incorrect solutions.

Step-by-Step Solution

Now, let's get down to business and solve this bad boy. The goal is to isolate 'x'. Since we have fractions with the same denominator, $(x - 5)$, the first smart move is to eliminate the denominators. We can do this by multiplying both sides of the equation by $(x - 5)$. This is a valid operation as long as $(x - 5)$ is not zero, which we've already established (x ≠ 5).

Let's see how that plays out:

$$(x - 5) \times \left( 7 + \frac{5}{(x - 5)} \right) = (x - 5) \times \left( \frac{2x}{(x - 5)} \right)$$

On the left side, we distribute $(x - 5)$:

$$(x - 5) \times 7 + (x - 5) \times \frac{5}{(x - 5)}$$

This simplifies to:

$$7(x - 5) + 5$$

On the right side, the $(x - 5)$ terms cancel out neatly:

$$2x$$

So, our equation now looks much friendlier:

$$7(x - 5) + 5 = 2x$$

This is a huge step! We've transformed a rational equation into a simple linear equation. Now, let's continue simplifying the left side by distributing the 7:

$$7x - 35 + 5 = 2x$$

Combine the constant terms on the left:

$$7x - 30 = 2x$$

The next step is to gather all the 'x' terms on one side and the constant terms on the other. Let's subtract $2x$ from both sides:

$$7x - 2x - 30 = 2x - 2x$$

$$5x - 30 = 0$$

Now, add 30 to both sides to isolate the 'x' term:

$$5x - 30 + 30 = 0 + 30$$

$$5x = 30$$

Finally, to get 'x' by itself, divide both sides by 5:

$$\frac{5x}{5} = \frac{30}{5}$$

$$x = 6$$

There we have it! Our potential solution is $x = 6$. But hold up, we're not done yet. Remember that initial restriction?

Checking for Extraneous Solutions

This is arguably the most critical step when dealing with rational equations, guys. We found a potential solution, $x = 6$. Now we need to check if this value violates our initial restriction, which was x ≠ 5. Since 6 is definitely not equal to 5, our solution is valid! It doesn't make any denominator zero, and it satisfies the original equation. If our solution had turned out to be 5, we would have had to discard it, and the answer would be 'no solution'. This process of checking is called identifying extraneous solutions. An extraneous solution is a value that we get through our algebraic manipulations but that doesn't actually work in the original equation. It's like finding a cool key that looks like it fits a lock, but when you try it, it just won't turn. We always need to make sure our keys (our solutions) actually work in the original lock (the original equation). So, for our equation $7 + \frac{5}{(x - 5)} = \frac{2x}{(x - 5)}$, the value $x = 6$ is indeed the correct and only solution.

Conclusion: The Solution for X

So, after carefully navigating through the algebraic steps and, most importantly, checking for any pesky extraneous solutions, we've arrived at our final answer. The equation $7 + \frac{5}{(x - 5)} = \frac{2x}{(x - 5)}$ has one unique solution. That solution is x = 6. Remember, the key takeaways here are to always identify denominator restrictions early on and to always check your potential solutions against those restrictions in the original equation. This method works not just for this specific problem but for a whole class of equations involving fractions with variables in the denominator. Keep practicing these steps, and you'll become an algebra whiz in no time! Happy problem-solving!