Solving (2/(x-2))-(4/(x^2-2x)) = 5 A Step-by-Step Guide

In this comprehensive guide, we will delve into the process of solving the equation $\frac{2}{x-2}-\frac{4}{x^2-2 x}=5$. This equation, which involves rational expressions, requires a systematic approach to ensure accurate solutions. We will break down each step, providing clear explanations and justifications along the way. Our aim is to not only find the solution set but also to equip you with the understanding and skills to tackle similar problems in the future. We'll begin by identifying the restrictions on the variable x, then proceed to clear the fractions by finding a common denominator, and finally solve the resulting quadratic equation. By the end of this guide, you will have a solid grasp of how to solve rational equations effectively.

Understanding the Equation and Identifying Restrictions

The first crucial step in solving any equation, especially one involving rational expressions, is to understand the equation's structure and identify any restrictions on the variable. In our case, the equation is $\frac{2}{x-2}-\frac{4}{x^2-2 x}=5$. We observe that the variable x appears in the denominators of the fractions. This immediately alerts us to potential restrictions because division by zero is undefined. Therefore, we must determine the values of x that would make the denominators equal to zero and exclude them from our solution set.

Let's examine the denominators individually. The first denominator is x - 2. Setting this equal to zero, we get x - 2 = 0, which implies x = 2. Thus, x cannot be equal to 2. The second denominator is x² - 2x. We can factor this expression as x(x - 2). Setting this equal to zero, we get x(x - 2) = 0, which gives us two possible solutions: x = 0 or x - 2 = 0. The latter equation, x - 2 = 0, again yields x = 2. So, we have two restrictions: x cannot be 0 and x cannot be 2. These restrictions are critical because any solution we find must satisfy these conditions. Failing to account for these restrictions can lead to extraneous solutions, which are values that satisfy the transformed equation but not the original equation. Therefore, before we proceed with solving the equation, it's essential to explicitly state that x ≠ 0 and x ≠ 2. These values will be excluded from our final solution set. Understanding these restrictions from the outset ensures that our solution process remains valid and accurate. With the restrictions clearly identified, we can now move on to the next step: clearing the fractions.

Clearing Fractions and Simplifying the Equation

With the restrictions on x clearly defined, the next step in solving the equation $\frac{2}{x-2}-\frac{4}{x^2-2 x}=5$ is to eliminate the fractions. This is achieved by multiplying both sides of the equation by the least common denominator (LCD) of the fractions. Identifying the LCD is crucial for simplifying the equation effectively. In our case, the denominators are x - 2 and x² - 2x. As we noted earlier, x² - 2x can be factored as x(x - 2). Therefore, the LCD of the fractions is x(x - 2). Multiplying both sides of the equation by this LCD will clear the fractions and transform the equation into a more manageable form.

Let's perform the multiplication. We have:

x(x - 2) * [$\frac{2}{x-2}-\frac{4}{x^2-2 x}$] = 5 * x(x - 2)

Distributing the LCD on the left side, we get:

x(x - 2) * $\frac{2}{x-2}$ - x(x - 2) * $\frac{4}{x^2-2 x}$ = 5x(x - 2)

Now, we simplify each term. In the first term, the (x - 2) terms cancel out, leaving us with 2x. In the second term, x(x - 2) is exactly the denominator, so the entire fraction simplifies to -4. Thus, the left side of the equation becomes 2x - 4. On the right side, we distribute the 5x to get 5x² - 10x. Now our equation looks like this:

2x - 4 = 5x² - 10x

This simplified equation is a quadratic equation, which we can solve using various methods, such as factoring, completing the square, or the quadratic formula. Before we proceed to solving the quadratic equation, it's essential to ensure that all terms are on one side, setting the equation equal to zero. This will put the equation in standard quadratic form, which is necessary for applying the quadratic formula or factoring techniques. In the next section, we will rearrange the equation and prepare it for solving.

Transforming to Standard Quadratic Form and Solving

Having cleared the fractions and simplified the equation to 2x - 4 = 5x² - 10x, the next step is to transform this equation into the standard quadratic form, which is ax² + bx + c = 0. This form is essential for applying methods such as factoring or the quadratic formula to find the solutions. To achieve this, we need to move all terms to one side of the equation, leaving zero on the other side. Let's rearrange the equation by subtracting 2x and adding 4 to both sides:

0 = 5x² - 10x - 2x + 4

Combining like terms, we get:

0 = 5x² - 12x + 4

Now the equation is in standard quadratic form, with a = 5, b = -12, and c = 4. We can now proceed to solve this quadratic equation. There are several methods we could use, such as factoring, completing the square, or the quadratic formula. In this case, let's try factoring first. We are looking for two numbers that multiply to a * c (5 * 4 = 20) and add up to b (-12). The numbers -2 and -10 satisfy these conditions. So, we can rewrite the middle term as -10x - 2x:

0 = 5x² - 10x - 2x + 4

Now, we factor by grouping:

0 = 5x(x - 2) - 2(x - 2)

We can factor out the common term (x - 2):

0 = (x - 2)(5x - 2)

Now we have two factors, and for the product to be zero, at least one of the factors must be zero. So, we set each factor equal to zero and solve for x:

x - 2 = 0 or 5x - 2 = 0

Solving the first equation, x - 2 = 0, we get x = 2. Solving the second equation, 5x - 2 = 0, we get 5x = 2, which gives us x = $\frac{2}{5}$. However, we must remember the restrictions we identified earlier: x cannot be 0 or 2. The solution x = 2 violates this restriction, so it is an extraneous solution and must be discarded. Therefore, the only valid solution is x = $\frac{2}{5}$. In the next section, we will summarize the solution set and discuss the importance of verifying solutions.

Verifying the Solution and Final Answer

After solving the quadratic equation and obtaining potential solutions, it is crucial to verify these solutions against the original equation and the restrictions identified earlier. This step ensures that the solutions are valid and not extraneous. In our case, we found two potential solutions: x = 2 and x = $\frac{2}{5}$. However, we know that x cannot be 2 because it would make the denominators in the original equation equal to zero, violating the restrictions we established at the beginning of our solution process. Therefore, x = 2 is an extraneous solution and must be discarded.

The remaining potential solution is x = $\frac{2}{5}$. To verify this solution, we substitute it back into the original equation: $\frac{2}{x-2}-\frac{4}{x^2-2 x}=5$. Substituting x = $\frac{2}{5}$, we get:

$\frac{2}{\frac{2}{5}-2}-\frac{4}{(\frac{2}{5})^2-2(\frac{2}{5})}=5$

Let's simplify this expression step by step. First, we simplify the denominators:

$\frac{2}{-\frac{8}{5}}-\frac{4}{\frac{4}{25}-\frac{4}{5}}=5$

Next, we simplify the fractions within the denominators:

$\frac{2}{-\frac{8}{5}}-\frac{4}{\frac{4-20}{25}}=5$

$\frac{2}{-\frac{8}{5}}-\frac{4}{-\frac{16}{25}}=5$

Now, we divide by the fractions, which is the same as multiplying by their reciprocals:

2 * (-\frac{5}{8}) - 4 * (-\frac{25}{16}) = 5

Simplifying further:

-\frac{5}{4} + \frac{25}{4} = 5

$\frac{20}{4} = 5$

5 = 5

Since the equation holds true when x = $\frac{2}{5}$, this solution is valid. Therefore, the solution set for the equation $\frac{2}{x-2}-\frac{4}{x^2-2 x}=5$ is {$\frac{2}{5}$}. By systematically solving the equation, identifying restrictions, and verifying the solution, we have ensured that our answer is accurate and complete. This thorough approach is essential for solving rational equations and avoiding common pitfalls such as extraneous solutions.

Final Answer: The solution set is {$\frac{2}{5}$}.