Solving The Equation: $1+\frac{1}{n^2-2 N}=\frac{n^2+4 N+3}{3 N^2-6 N}$

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Hey Plastik Magazine readers! Today, we're diving deep into the fascinating world of mathematics to tackle a rather intriguing equation: $1+\frac{1}{n^2-2 n}=\frac{n^2+4 n+3}{3 n^2-6 n}$. If you're a math enthusiast or just looking to sharpen your problem-solving skills, you're in for a treat. Let's break down this equation step by step and unravel its solution. So, grab your calculators and let's get started!

Understanding the Problem

Before we jump into solving, let's take a moment to truly understand what we're dealing with. This equation involves fractions, variables, and a mix of algebraic expressions. At first glance, it might seem a bit daunting, but don't worry, guys! We're going to approach it methodically. The key is to recognize the different components and how they interact with each other. We have a variable, 'n', that we need to find the value(s) for, which will make the equation true. We've got fractions with polynomials in the denominators, and our goal is to simplify and solve for 'n'. The beauty of mathematics lies in its precision, so let's aim for accuracy and clarity in our solution.

Initial Observations

First off, let’s identify the domain restrictions. Notice that we have $n^2 - 2n$ and $3n^2 - 6n$ in the denominators of our fractions. We need to make sure these denominators don't equal zero, because division by zero is a big no-no in math. So, we need to find the values of 'n' that would make these expressions zero and exclude them from our possible solutions. This is a crucial first step because it helps us avoid mathematical errors and ensures our solution is valid.

To find these values, we can set each denominator equal to zero and solve for 'n':

1. $n^2 - 2n = 0$ can be factored as $n(n - 2) = 0$. This gives us $n = 0$ and $n = 2$ as restricted values.
2. $3n^2 - 6n = 0$ can be factored as $3n(n - 2) = 0$. This also gives us $n = 0$ and $n = 2$ as restricted values.

So, we know that $n$ cannot be 0 or 2. Keep this in mind as we proceed with solving the equation. Any solution we find must not be 0 or 2, otherwise, it's not a valid solution. This step is super important because it lays the groundwork for a correct solution. We’re essentially setting the boundaries within which we can operate, ensuring our final answer makes sense within the mathematical context.

Simplifying the Equation

Now that we've identified our domain restrictions, let's dive into simplifying the equation. Simplification is often the key to making complex equations more manageable. Our goal here is to eliminate the fractions and consolidate terms so we can work with a more straightforward expression. To do this, we'll find a common denominator for the fractions and then combine them. This is a standard technique in algebra, and mastering it will help you tackle many types of equations.

Finding a Common Denominator

Looking at the equation $1+\frac{1}{n^2-2 n}=\frac{n^2+4 n+3}{3 n^2-6 n}$, we need to find a common denominator for the fractions. Let's factor the denominators to make this easier:

• $n^2 - 2n$ can be factored as $n(n - 2)$.
• $3n^2 - 6n$ can be factored as $3n(n - 2)$.

The least common denominator (LCD) is the smallest expression that is divisible by both denominators. In this case, the LCD is $3n(n - 2)$. Now we'll rewrite each term in the equation with this common denominator. Remember, whatever we do to the denominator, we must also do to the numerator to maintain the equation's balance. This is like keeping a scale even – we can add or subtract the same amount on both sides, or in this case, multiply both the numerator and denominator by the same factor.

Rewriting with the Common Denominator

Let's rewrite each term with the LCD of $3n(n - 2)$:

1. The first term, 1, can be rewritten as $\frac{3n(n - 2)}{3n(n - 2)}$.
2. The second term, $\frac{1}{n^2 - 2n}$, can be rewritten as $\frac{1}{n(n - 2)}$. To get the LCD, we multiply the numerator and denominator by 3, resulting in $\frac{3}{3n(n - 2)}$.
3. The third term, $\frac{n^2 + 4n + 3}{3n^2 - 6n}$, is already over the LCD, so we leave it as $\frac{n^2 + 4n + 3}{3n(n - 2)}$.

Now our equation looks like this:

$\frac{3n(n - 2)}{3n(n - 2)} + \frac{3}{3n(n - 2)} = \frac{n^2 + 4n + 3}{3n(n - 2)}$

Combining Terms and Simplifying Further

With all terms now sharing a common denominator, we can combine the fractions on the left side of the equation. Combining terms is a fundamental step in simplifying algebraic expressions, and it allows us to consolidate our equation into a more manageable form. By adding the numerators together, we reduce the number of individual fractions and move closer to isolating the variable 'n'.

Combining Fractions

Let's combine the fractions on the left side of the equation:

$\frac{3n(n - 2) + 3}{3n(n - 2)} = \frac{n^2 + 4n + 3}{3n(n - 2)}$

Now, let’s expand the numerator on the left side:

$\frac{3n^2 - 6n + 3}{3n(n - 2)} = \frac{n^2 + 4n + 3}{3n(n - 2)}$

Eliminating the Denominator

Since both sides of the equation have the same denominator, we can eliminate it by multiplying both sides by $3n(n - 2)$. However, we need to remember our earlier restriction: $n$ cannot be 0 or 2. By eliminating the denominator, we're essentially focusing on the numerators, which will lead us to a simpler equation to solve. This step is a common technique in solving equations involving fractions, as it transforms the equation into a more familiar form.

So, multiplying both sides by $3n(n - 2)$, we get:

$3n^2 - 6n + 3 = n^2 + 4n + 3$

Solving the Quadratic Equation

Now we're left with a quadratic equation. Solving quadratic equations is a crucial skill in algebra, and there are several methods we can use, such as factoring, completing the square, or using the quadratic formula. In this case, we'll rearrange the equation to bring all terms to one side and then attempt to factor it. Factoring, when possible, is often the quickest way to find the solutions, but if factoring proves difficult, we can always resort to the quadratic formula. The key is to manipulate the equation into a standard form that allows us to apply these techniques effectively.

Rearranging the Equation

Let's rearrange the equation to get all terms on one side:

$3n^2 - 6n + 3 - (n^2 + 4n + 3) = 0$

Simplifying, we get:

$2n^2 - 10n = 0$

Factoring the Quadratic

Now, let’s factor out the common factor, which is $2n$:

$2n(n - 5) = 0$

Finding the Solutions

To find the solutions, we set each factor equal to zero:

1. $2n = 0$ gives us $n = 0$.
2. $n - 5 = 0$ gives us $n = 5$.

Checking for Extraneous Solutions

We've found two potential solutions: $n = 0$ and $n = 5$. However, remember our initial domain restrictions? We determined that $n$ cannot be 0 or 2 because these values would make the denominators in the original equation equal to zero. Checking for extraneous solutions is a critical step in solving equations, especially those involving fractions or radicals. It ensures that the solutions we've found are valid within the context of the original equation.

Applying the Restrictions

We found that $n = 0$ is one of our solutions, but we also know that $n$ cannot be 0 because it would make the denominators in the original equation zero. Therefore, $n = 0$ is an extraneous solution, and we must discard it. This highlights the importance of checking our solutions against any restrictions we identified at the beginning of the problem.

Validating the Solution

The other solution we found was $n = 5$. This value is not in our restricted domain, so it's a potential valid solution. To be absolutely sure, let's plug $n = 5$ back into the original equation to see if it holds true.

Original equation: $1+\frac{1}{n^2-2 n}=\frac{n^2+4 n+3}{3 n^2-6 n}$

Substituting $n = 5$:

$1 + \frac{1}{5^2 - 2(5)} = \frac{5^2 + 4(5) + 3}{3(5^2) - 6(5)}$

$1 + \frac{1}{25 - 10} = \frac{25 + 20 + 3}{3(25) - 30}$

$1 + \frac{1}{15} = \frac{48}{75 - 30}$

$1 + \frac{1}{15} = \frac{48}{45}$

$\frac{16}{15} = \frac{16}{15}$

The equation holds true! So, $n = 5$ is indeed a valid solution.

Conclusion

Alright, guys! We've successfully solved the equation $1+\frac{1}{n^2-2 n}=\frac{n^2+4 n+3}{3 n^2-6 n}$. It was quite the journey, but we tackled it step by step, from understanding the problem and simplifying the equation to solving the quadratic and checking for extraneous solutions. The final solution is $n = 5$. Remember, the key to mastering math is practice and perseverance. Keep challenging yourselves, and you'll be amazed at what you can achieve. Until next time, keep those mathematical gears turning!