Solving The Equation: A Step-by-Step Guide

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Hey math enthusiasts! Today, we're diving into a classic algebra problem: solving the equation x+23xβˆ’1xβˆ’2=xβˆ’33x\frac{x+2}{3x} - \frac{1}{x-2} = \frac{x-3}{3x}. Don't sweat it, because we're going to break it down step-by-step to make sure everyone understands. We'll navigate through the tricky waters of fractions, common denominators, and, of course, some good old algebraic manipulation. This isn't just about getting an answer; it's about understanding why each step works. So, buckle up, grab your pencils, and let's get started. By the end of this, you'll be handling equations like a pro, able to confidently find the solution to a wide variety of similar problems. Our main focus will be on the strategies, techniques, and the underlying logic that makes these types of problems solvable. We'll start with the initial setup, moving through to simplification, and eventually isolating the variable. Let's make this journey into the world of algebraic equations not just educational, but also fun and engaging. Ready to solve some equations?

Step 1: Combining Fractions and Finding Common Denominators

Alright, guys, our first step in solving the equation is to deal with those pesky fractions. The equation is currently: x+23xβˆ’1xβˆ’2=xβˆ’33x\frac{x+2}{3x} - \frac{1}{x-2} = \frac{x-3}{3x}. Our aim here is to combine these fractions, and to do that, we need a common denominator. Look at the denominators we have: 3x3x and (xβˆ’2)(x-2). A common denominator that includes both of these is going to be 3x(xβˆ’2)3x(x-2). Now, the first fraction, x+23x\frac{x+2}{3x}, is missing the (xβˆ’2)(x-2) part. So, we multiply both the numerator and the denominator by (xβˆ’2)(x-2). This gives us: (x+2)(xβˆ’2)3x(xβˆ’2)\frac{(x+2)(x-2)}{3x(x-2)}. For the second fraction, 1xβˆ’2\frac{1}{x-2}, we multiply both the numerator and denominator by 3x3x, which gets us 3x3x(xβˆ’2)\frac{3x}{3x(x-2)}. Keep in mind, when manipulating these equations, we're essentially multiplying by 1, which doesn't change the value, only the form of the expressions. Now, let's rewrite the equation with these new fractions. We now have: (x+2)(xβˆ’2)3x(xβˆ’2)βˆ’3x3x(xβˆ’2)=xβˆ’33x\frac{(x+2)(x-2)}{3x(x-2)} - \frac{3x}{3x(x-2)} = \frac{x-3}{3x}.

Next, let’s combine those numerators on the left-hand side. Remember, we’re simplifying, not changing the value, so the common denominator stays the same while we subtract the numerators. This simplification is key when finding the solution. The numerator of the first fraction expands to x2βˆ’4x^2 - 4. Our equation now becomes: x2βˆ’4βˆ’3x3x(xβˆ’2)=xβˆ’33x\frac{x^2 - 4 - 3x}{3x(x-2)} = \frac{x-3}{3x}. Keep up the good work; we're making progress. Take a look and ensure that everything is clear. If you have any questions, don’t hesitate to pause and revisit any steps that aren’t crystal clear. Math is about building blocks, so ensure that each layer is strong.

Step 2: Simplifying the Equation and Clearing the Fractions

Okay, team, the next part of solving the equation involves simplifying and getting rid of those fractions. We currently have: x2βˆ’3xβˆ’43x(xβˆ’2)=xβˆ’33x\frac{x^2 - 3x - 4}{3x(x-2)} = \frac{x-3}{3x}. Notice how we’ve already combined fractions. Our goal is to isolate 'x', and getting rid of the fraction helps tremendously. We're going to multiply both sides of the equation by 3x(xβˆ’2)3x(x-2). This cancels the denominator on the left side and simplifies the right side. On the left side, the equation becomes x2βˆ’3xβˆ’4x^2 - 3x - 4, and on the right side, we get (xβˆ’3)(xβˆ’2)(x-3)(x-2). Our equation now looks like this: x2βˆ’3xβˆ’4=(xβˆ’3)(xβˆ’2)x^2 - 3x - 4 = (x-3)(x-2). See how the equation is looking cleaner?

Now, let's expand the right side. Multiplying (xβˆ’3)(x-3) and (xβˆ’2)(x-2), we get x2βˆ’5x+6x^2 - 5x + 6. So, the equation is now: x2βˆ’3xβˆ’4=x2βˆ’5x+6x^2 - 3x - 4 = x^2 - 5x + 6. We're getting closer to isolating 'x'. Always ensure that you're taking your time, and checking your math. One small mistake can lead to the wrong answer. Take advantage of all the available resources and never hesitate to double-check.

Then, let’s move all terms to one side. We're aiming to simplify the equation to the point where we can easily solve for xx. We can subtract x2x^2 from both sides, which simplifies things quite a bit. This leaves us with: βˆ’3xβˆ’4=βˆ’5x+6-3x - 4 = -5x + 6. It’s looking simpler already. What do you think about solving this problem so far? Have you seen other similar equations? The more you practice, the more familiar these algebraic maneuvers will become. Now, our next steps will involve solving the equation, which includes bringing like terms together to isolate the variable, 'x'.

Step 3: Isolating x and Solving for the Variable

Alright, folks, in this step, our goal is straightforward: solving the equation and getting 'x' all by itself. We currently have: βˆ’3xβˆ’4=βˆ’5x+6-3x - 4 = -5x + 6. To isolate 'x', let's start by adding 5x5x to both sides. This gives us: 2xβˆ’4=62x - 4 = 6. The equation looks a lot cleaner now, right? It's much simpler to work with. Next, we’ll add 4 to both sides of the equation to isolate the term with 'x'. Doing that, we get: 2x=102x = 10. We are so close to the finish line!

Now, the last step is super easy. To solve for 'x', we simply divide both sides by 2. This isolates 'x' and gives us the solution. So, we divide both sides by 2. Therefore, x=5x = 5. We’ve successfully found the solution to our equation. Isn’t that amazing? We started with a complex-looking equation with fractions, and with each step, we made it simpler and simpler until we were able to isolate the variable. We started by handling those fractions, ensuring we used a common denominator to combine and simplify them. Then, we cleared the fractions by multiplying both sides by the common denominator. This step-by-step method shows how we successfully tackled the equation and found our solution.

Step 4: Verification and Final Thoughts

Okay, guys, we've found our answer, but we're not quite done yet. We always want to verify our solution to solving the equation to make sure it's correct. Verification helps us to build confidence and ensures we catch any errors. Let's substitute x=5x = 5 back into the original equation: x+23xβˆ’1xβˆ’2=xβˆ’33x\frac{x+2}{3x} - \frac{1}{x-2} = \frac{x-3}{3x}. Plugging in x=5x = 5, we get: 5+23βˆ—5βˆ’15βˆ’2=5βˆ’33βˆ—5\frac{5+2}{3*5} - \frac{1}{5-2} = \frac{5-3}{3*5}. Simplifying this, we get: 715βˆ’13=215\frac{7}{15} - \frac{1}{3} = \frac{2}{15}. Let's simplify the left side: 715βˆ’515=215\frac{7}{15} - \frac{5}{15} = \frac{2}{15}. Therefore, 215=215\frac{2}{15} = \frac{2}{15}.

The equation holds true! This means that our solution, x=5x = 5, is correct. Yay! This is a great example of solving the equation and making sure that all steps are done correctly. Remember, the journey of solving an algebraic equation is just as important as the destination. Knowing how you got the answer helps you apply these methods to other problems and builds a stronger foundation in math. Always, always check your work; it saves time and builds confidence. And most importantly, keep practicing! The more you work through different types of problems, the better you’ll get at recognizing patterns and applying the right steps. Math can be fun; believe it, and you will achieve it. Great job today, everyone!