Solving The Rational Equation: X/(x+2) + 1/x = 1

Dec 10, 2025 by Andrew McMorgan 49 views

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the wild world of mathematics to tackle a tricky rational equation: $\frac{x}{x+2}+\frac{1}{x}=1$ . Now, you might be looking at this and thinking, "Whoa, what's going on here?" But don't worry, we're going to break it down step by step. Our main goal is to figure out what kind of solutions this equation spits out – specifically, whether we're dealing with valid solutions or those sneaky extraneous solutions. You know, the ones that look like they work but actually break the original equation? We'll be on the lookout for those!

First things first, let's talk about rational equations and why they can be a bit of a headache. These are equations that involve fractions where the variables are in the denominator. The big problem with denominators is that they can't be zero. If we end up with a solution that makes any denominator zero in the original equation, then that solution is no good – it's extraneous. So, for our equation $\frac{x}{x+2}+\frac{1}{x}=1$ , we immediately know that $x$ cannot be 0 (because of the $\frac{1}{x}$ term) and $x+2$ cannot be 0, which means $x$ cannot be -2. Keep these restrictions in mind, because they're super important!

To solve this equation, we need to get rid of those pesky fractions. The standard move here is to multiply everything by the least common denominator (LCD). The denominators are $x+2$ and $x$ . So, our LCD is simply $x(x+2)$ . Let's multiply both sides of the equation by this LCD:

$x(x+2) \left( \frac{x}{x+2} + \frac{1}{x} \right) = x(x+2) (1)$

Now, we distribute the LCD to each term on the left side:

$\frac{x(x+2)x}{x+2} + \frac{x(x+2)(1)}{x} = x(x+2)$

See how the denominators start canceling out? This is the magic of using the LCD!

On the first term, the $(x+2)$ cancels out, leaving us with $x \cdot x$ , which is $x^2$ .

On the second term, the $x$ cancels out, leaving us with $(x+2) \cdot 1$ , which is simply $x+2$ .

And on the right side, we just have $x(x+2)$ , which we can expand to $x^2 + 2x$ .

So, after multiplying by the LCD and simplifying, our equation transforms into:

$x^2 + (x+2) = x^2 + 2x$

Looking pretty good, right? We've turned a complex rational equation into a much simpler polynomial equation. Now, let's keep simplifying and try to isolate $x$ . We can start by combining like terms on the left side:

$x^2 + x + 2 = x^2 + 2x$

Now, we want to get all the $x$ terms on one side and the constants on the other. Let's subtract $x^2$ from both sides. Notice that the $x^2$ terms cancel out completely!

$x + 2 = 2x$

This is awesome! We've gone from a quadratic-looking equation to a simple linear equation. Now, let's get the $x$ terms together. Subtract $x$ from both sides:

$2 = 2x - x$

$2 = x$

So, we found a potential solution: $x=2$ . But wait! Remember those restrictions we talked about at the beginning? We said that $x$ cannot be 0 and $x$ cannot be -2.

Does our solution $x=2$ violate either of these restrictions? Nope! $2 \neq 0$ and $2 \neq -2$ . This means that $x=2$ is a valid solution to the original equation. It doesn't make any denominators zero, and it satisfies the equation we derived.

Now, let's revisit the options given:

A. The equation has two valid solutions and no extraneous solutions. B. The equation has no valid solutions and two extraneous solutions. C. The equation has one valid solution and no extraneous solutions.

Based on our work, we found exactly one solution, $x=2$ , and it's valid because it doesn't violate our initial restrictions. This means option C is the correct description. We didn't find any other solutions, and the one we found isn't extraneous. Pretty neat, huh? Keep practicing these steps, guys, and you'll be a rational equation master in no time!

Understanding Extraneous Solutions in Rational Equations

So, let's really dig into this idea of extraneous solutions, because they're the trickiest part of solving equations with variables in the denominator. When we're dealing with a rational equation like $\frac{x}{x+2}+\frac{1}{x}=1$ , the main pitfall is encountering a denominator that equals zero. In this specific problem, we identified early on that $x \neq 0$ and $x \neq -2$ . These are our domain restrictions. They tell us which values of $x$ are not allowed in the original equation. Any number that violates these restrictions cannot be a solution, even if it pops out of our algebraic manipulations.

Why do these extraneous solutions even show up? Well, it's usually a consequence of the method we use to eliminate the fractions, primarily by multiplying by the least common denominator (LCD). When we multiply both sides of an equation by an expression involving the variable (like $x(x+2)$ in our case), we are essentially assuming that this expression is non-zero. If, for a particular value of $x$ , the LCD is zero, then multiplying by it doesn't maintain the equivalence of the equation. This can lead to introducing solutions that weren't there originally.

Consider this: if we multiply an equation $A=B$ by $C$ , we get $AC=BC$ . This step is valid as long as $C \neq 0$ . If $C=0$ for some value of $x$ , then $AC=0$ and $BC=0$ , so $AC=BC$ will always be true for that $x$ , regardless of whether $A=B$ was true. This is how a solution that makes the multiplier (our LCD) zero can be introduced.

In our equation $\frac{x}{x+2}+\frac{1}{x}=1$ , the LCD is $x(x+2)$ . This LCD is zero if $x=0$ or $x=-2$ . These are exactly the values we excluded from our domain. If our solving process had yielded $x=0$ or $x=-2$ as a potential solution, we would have had to discard it immediately because it would make the original denominators zero. For example, if we tried to plug $x=0$ into the original equation, we'd have $\frac{0}{0+2} + \frac{1}{0} = 1$ . The term $\frac{1}{0}$ is undefined, so $x=0$ is definitely not a solution.

Similarly, if we tried $x=-2$ , we'd get $\frac{-2}{-2+2} + \frac{1}{-2} = 1$ , which simplifies to $\frac{-2}{0} - \frac{1}{2} = 1$ . Again, we have a division by zero ( $\frac{-2}{0}$ ), making $x=-2$ an invalid value for the original equation.

So, the process of solving rational equations typically involves these key steps:

Identify Domain Restrictions: Find all values of the variable that make any denominator equal to zero. These values are excluded from the possible solutions.
Clear Fractions: Multiply both sides of the equation by the LCD to eliminate all denominators.
Solve the Resulting Equation: Simplify and solve the new equation (which is usually a polynomial equation).
Check for Extraneous Solutions: Compare the solutions found in step 3 with the domain restrictions identified in step 1. Any solution that matches a domain restriction must be discarded as extraneous. The remaining solutions are the valid solutions to the original equation.

In our specific case, we followed these steps. We found the restrictions $x \neq 0$ and $x \neq -2$ . After clearing fractions, we simplified the equation to $x=2$ . When we checked this solution against our restrictions, we found that $2$ is not equal to $0$ or $-2$ . Therefore, $x=2$ is a valid solution, and there are no extraneous solutions. This neatly leads us to the conclusion that the equation has one valid solution and no extraneous solutions, aligning perfectly with option C.

Understanding this distinction is crucial. It’s not enough to just find a number that makes your simplified equation true; you must ensure it works in the original equation, which means it can't cause any division by zero. This is a fundamental concept when working with rational expressions and equations in algebra, guys, so make sure you’ve got a firm grip on it!

The Journey from Complex Equation to Simple Linear Solution

Let's take a moment to appreciate the transformation our equation $\frac{x}{x+2}+\frac{1}{x}=1$ underwent. It started off looking like a formidable challenge, a rational equation that could intimidate even seasoned math enthusiasts. The presence of variables in the denominators ( $x+2$ and $x$ ) immediately flagged it as an equation requiring careful handling, particularly concerning potential extraneous solutions. These are solutions that arise during the solving process but do not satisfy the original equation, typically because they cause a zero in a denominator. Our initial step, as discussed, was to pinpoint these potential pitfalls: $x \neq 0$ and $x \neq -2$ . These exclusions are our guardians, protecting us from invalid answers.

The core strategy to tame such an equation is to eliminate the denominators. This is most effectively done by multiplying every term on both sides of the equation by the least common denominator (LCD). For our equation, the denominators are $(x+2)$ and $x$ . Their LCD is the product of these two distinct factors: $x(x+2)$ . Multiplying both sides by $x(x+2)$ is a pivotal move. It transforms the equation from a fraction-filled beast into something much more manageable, ideally a polynomial.

Let’s retrace that algebraic dance:

x(x+2) \left( \frac{x}{x+2} \right) + x(x+2) \left( \frac{1}{x} \right) = x(x+2) (1)

Observe the cancellations: In the first term, $(x+2)$ cancels, leaving $x \cdot x = x^2$ . In the second term, $x$ cancels, leaving $(x+2) \cdot 1 = x+2$ . The right side remains $x(x+2)$ , which expands to $x^2 + 2x$ . Thus, the equation morphs into:

x^2 + (x+2) = x^2 + 2x

This is a significant simplification! We've gone from dealing with fractions to a polynomial equation. The next logical step is to gather like terms and isolate the variable $x$ . Notice the $x^2$ term appears on both sides. Subtracting $x^2$ from both sides is a brilliant simplification that changes the nature of the equation:

x^2 + x + 2 - x^2 = x^2 + 2x - x^2

x + 2 = 2x

Here's where the equation sheds its quadratic guise entirely and becomes a linear equation. Linear equations are generally straightforward to solve. To isolate $x$ , we can subtract $x$ from both sides:

x + 2 - x = 2x - x

2 = x

And there we have it – a single, clear potential solution: $x=2$ . The beauty of this transformation lies in its systematic nature. By applying algebraic rules correctly, we systematically reduced the complexity of the problem. The initial equation, which required understanding domain restrictions and common denominators, has been elegantly resolved into a simple statement $x=2$ .

However, the journey isn't quite complete until we perform the crucial final check. We must confirm that our derived solution, $x=2$ , does not conflict with the domain restrictions we identified at the very beginning: $x \neq 0$ and $x \neq -2$ . Since $2$ is neither $0$ nor $-2$ , it successfully passes this test. This confirms that $x=2$ is a valid solution. It is the only solution we found, and it is not extraneous.

This process highlights why understanding each step is so vital. The simplification from a rational form to a linear form is powerful, but it hinges on the correct application of rules, especially the multiplication by the LCD, which is only valid if the LCD is not zero. Because our solution $x=2$ does not make the LCD $x(x+2)$ equal to zero, the simplification steps were all valid for this solution.

Therefore, the statement that accurately describes the solutions is that the equation has exactly one valid solution ( $x=2$ ) and no extraneous solutions. This outcome aligns perfectly with option C provided in the initial problem description. It’s a great example of how, with the right techniques and careful checking, even complex-looking algebraic problems can yield straightforward answers. Keep practicing, and you'll find these transformations become second nature, guys!

Analyzing the Options: Why C is the Champion

Alright, we've done the heavy lifting, solved the equation $\frac{x}{x+2}+\frac{1}{x}=1$ , and found our potential solution $x=2$ . Now, it's time to make sure we select the correct description of the solution set from the given options. This is where understanding the nuances of rational equations and extraneous solutions really pays off. Let's break down why option C is the winner and why the others don't quite hit the mark.

Our options were:

First, let's revisit our work. We started with $\frac{x}{x+2}+\frac{1}{x}=1$ . We identified our domain restrictions immediately: $x \neq 0$ and $x \neq -2$ . These are the values $x$ cannot be if it's to be a valid solution to the original equation, as they would cause division by zero.

We then multiplied by the LCD, $x(x+2)$ , to clear the denominators. This led us to the simplified equation: $x^2 + (x+2) = x^2 + 2x$ . After further simplification, we found that the $x^2$ terms canceled out, leaving us with a linear equation: $x+2 = 2x$ . Solving this linear equation yielded a single potential solution: $x=2$ .

Now, the crucial step: checking this potential solution against our domain restrictions. Is $x=2$ equal to $0$ ? No. Is $x=2$ equal to $-2$ ? No. Since $x=2$ does not violate any of our initial restrictions, it is a valid solution. We can plug it back into the original equation to confirm:

$\frac{2}{2+2} + \frac{1}{2} = \frac{2}{4} + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1$ .

This confirms that $x=2$ is indeed a correct and valid solution.

So, we have found one solution, and it is valid. This immediately points towards option C. But let's quickly consider why the other options are incorrect.

Why Option A is Incorrect: Option A claims there are two valid solutions. Our algebraic process, after simplification, resulted in a linear equation ( $x+2=2x$ ), which can have at most one solution. There was no point in our simplification where we introduced the possibility of two distinct solutions that would then be confirmed as valid. If the equation had simplified to something like $x^2 = 4$ , then we might have had two potential solutions ( $x=2, x=-2$ ) that we'd then check against restrictions. But that wasn't the case here; we arrived at a single solution.

Why Option B is Incorrect: Option B states that the equation has no valid solutions and two extraneous solutions. If we had ended up with, say, $x=0$ and $x=-2$ as our potential solutions from the simplified equation, then both would have been extraneous because they violate our domain restrictions. However, we found $x=2$ , which is valid. Furthermore, the simplification process yielded only one potential solution, not two that would need to be classified as extraneous. So, this option is definitively wrong.

Why Option C is Correct: Option C states, "The equation has one valid solution and no extraneous solutions." This perfectly matches our findings. We discovered exactly one solution, $x=2$ . We verified that this solution is valid by checking it against the domain restrictions ( $x \neq 0, x \neq -2$ ). Since our only solution is valid, there are, by definition, no extraneous solutions. The equation doesn't have any other solutions that we missed; the simplification process led us directly and solely to $x=2$ .

It's this careful checking and understanding of what constitutes a valid versus an extraneous solution that makes solving these types of problems rewarding. Always remember to state your restrictions upfront and verify your final answers against them. That's the golden rule for tackling rational equations, guys. Keep this checklist in mind, and you'll navigate these problems like pros!