Solving $x+\frac{3}{x-1}=\frac{3 X}{x-1}$: A Step-by-Step Guide

Hey guys! Today, we're going to dive into solving the equation $x+\frac{3}{x-1}=\frac{3 x}{x-1}$. This is a fun little problem that involves fractions and some basic algebra. We'll break it down step by step so it’s super easy to follow. So, grab your pencils and let’s get started!

Understanding the Equation

First off, let's make sure we understand what we're dealing with. We have the equation $x+\frac{3}{x-1}=\frac{3 x}{x-1}$. Our goal is to find the value(s) of x that make this equation true. Equations like these often involve a bit of algebraic manipulation to isolate x. The key here is to approach it systematically, one step at a time. We need to be mindful of the fraction, especially the denominator $(x-1)$, as it cannot be zero (we can’t divide by zero, right?). So, before we even start solving, we know that x cannot be 1.

When dealing with equations involving fractions, a common strategy is to eliminate the fractions early on. This makes the equation simpler and easier to work with. To do this, we'll need to find a common denominator. In this case, the common denominator is simply $(x-1)$. This is because both fractions already have this denominator. Multiplying both sides of the equation by this common denominator will clear the fractions. This step is crucial because it transforms the equation into a more manageable form, typically a polynomial equation, which we can then solve using standard techniques. Remember, it's essential to ensure that any solutions we find are valid by checking them against the original equation, particularly considering the restriction on x.

Also, bear in mind that this equation might have multiple solutions, or it might have no solutions at all. It all depends on how the algebra plays out. Sometimes you’ll find yourself with a quadratic equation that has two solutions, and sometimes the equation will simplify to a linear equation with just one solution. And every so often, you might end up with a situation where the variable cancels out completely, leaving you with a statement that's either always true (meaning any x would work) or always false (meaning there are no solutions). The beauty of algebra is in figuring out which scenario you’re in!

Step-by-Step Solution

1. Identify the Restriction

Before we do anything, let's identify the restriction on x. The denominator in the fractions is $(x-1)$. We can't divide by zero, so:

$x - 1 \neq 0$

$x \neq 1$

So, x cannot be 1. This is super important to remember, because if we get x = 1 as a solution later on, we'll know it's an extraneous solution and we'll have to throw it out. Always good to keep these things in mind right from the start!

2. Multiply Both Sides by $(x-1)$

To get rid of the fractions, we'll multiply both sides of the equation by $(x-1)$:

$(x-1) \left(x+\frac{3}{x-1}\right) = (x-1) \left(\frac{3 x}{x-1}\right)$

Now, distribute $(x-1)$ on the left side:

$x(x-1) + (x-1)\frac{3}{x-1} = (x-1) \frac{3 x}{x-1}$

This simplifies to:

$x(x-1) + 3 = 3x$

Clearing fractions like this is a classic move in algebra. It transforms the equation from something that looks intimidating into something much more manageable. When you multiply each term by the common denominator, the denominators neatly cancel out, leaving you with a cleaner equation. This step is not only practical but also insightful, as it reveals the underlying structure of the equation. Suddenly, what seemed complex becomes a polynomial equation, often a quadratic or linear one, which we have well-established methods for solving. It’s like magic, but it’s just math!

3. Expand and Simplify

Next, we expand the terms and simplify the equation:

$x^2 - x + 3 = 3x$

Now, let's move all terms to one side to set the equation to zero:

$x^2 - x - 3x + 3 = 0$

$x^2 - 4x + 3 = 0$

At this point, we've transformed the equation into a standard quadratic form. This is a pivotal moment in the solving process. Quadratic equations are a cornerstone of algebra, and there are several ways to tackle them. You might choose to factor the quadratic, use the quadratic formula, or even complete the square. The best method often depends on the specific equation at hand. Factoring is usually the quickest route if you can spot the factors easily. The quadratic formula is a foolproof method that works every time, but it can be a bit more computationally intensive. Completing the square is more of a technique to have in your back pocket, useful in certain theoretical contexts and for understanding the structure of quadratic equations.

4. Factor the Quadratic

We can factor the quadratic equation:

$(x - 3)(x - 1) = 0$

Factoring a quadratic equation is like unlocking a puzzle. It involves finding two binomials that, when multiplied together, give you the quadratic expression you started with. The beauty of factoring is that it directly leads you to the solutions (also called roots or zeros) of the equation. Each factor gives you a potential solution by setting it equal to zero. For instance, if you have $(x - a)(x - b) = 0$, then either $(x - a)$ must be zero, or $(x - b)$ must be zero, leading to solutions $x = a$ and $x = b$. Factoring is a powerful technique, but it relies on recognizing patterns and being comfortable with manipulating algebraic expressions. Practice makes perfect when it comes to factoring, and the more you do it, the quicker you'll become at spotting those key factors!

5. Solve for $x$

Now, set each factor equal to zero:

$x - 3 = 0$ or $x - 1 = 0$

Solving these gives us:

$x = 3$ or $x = 1$

6. Check for Extraneous Solutions

Remember our restriction? x cannot be 1. So, $x = 1$ is an extraneous solution.

Therefore, the only valid solution is:

$x = 3$

Checking for extraneous solutions is a crucial step, especially when dealing with equations involving fractions or radicals. Extraneous solutions are values that you get when solving the equation algebraically, but they don't actually satisfy the original equation. This often happens because certain operations, like squaring both sides or multiplying by a variable expression, can introduce solutions that weren't there to begin with. In our case, we had to exclude $x = 1$ because it made the denominator zero in the original equation, which is a big no-no in math. Always plug your solutions back into the original equation to make sure they work. It's a bit like verifying your answer in a science experiment – you want to be sure your results are valid!

Final Answer

The solution to the equation $x+\frac{3}{x-1}=\frac{3 x}{x-1}$ is $x = 3$.

Wrapping up, solving equations like this one is all about taking it one step at a time. We identified the restriction, cleared the fractions, simplified, factored, and checked for extraneous solutions. Each step is a building block, and together they lead us to the correct answer. Keep practicing, guys, and you'll become equation-solving pros in no time! Math can be challenging, but it's also incredibly rewarding when you crack a tough problem. Keep up the great work!