Solving Rational Equations A Step-by-Step Guide

In this article, we will delve into the process of solving a rational equation. Specifically, we'll tackle the equation:

$\frac{2x}{x+2} + \frac{5}{x-5} = \frac{8}{x^2 - 3x - 10}$

Rational equations, which involve fractions with variables in the denominator, often require a series of algebraic manipulations to arrive at a solution. This guide will provide a step-by-step approach, ensuring clarity and understanding throughout the process. We will cover the critical steps, including identifying restrictions, finding the least common denominator (LCD), clearing fractions, solving the resulting equation, and verifying solutions.

1. Identifying Restrictions

Before we begin solving the equation, it's crucial to identify any values of x that would make the denominators zero. These values are called restrictions because they are not permissible solutions, as division by zero is undefined. Identifying these restrictions at the outset prevents extraneous solutions from being included in the final answer. To find the restrictions, we set each denominator equal to zero and solve for x.

• First Denominator (x + 2):

  • x + 2 = 0
  • x = -2

• Second Denominator (x - 5):

  • x - 5 = 0
  • x = 5

• Third Denominator (x² - 3x - 10):

  • x² - 3x - 10 = 0
  • This quadratic expression can be factored: (x - 5)(x + 2) = 0
  • Setting each factor to zero gives us x = 5 and x = -2, which we already identified.

Therefore, the restrictions are x ≠ -2 and x ≠ 5. These values must be excluded from our final solution set. Identifying restrictions is a foundational step when solving rational equations, preventing division by zero, and ensuring the validity of the solutions. In this context, restrictions are $x \neq -2$ and $x \neq 5$. By understanding the restrictions, we pave the way for accurate resolution of the equation.

2. Finding the Least Common Denominator (LCD)

The next step is to find the least common denominator (LCD) of all the fractions in the equation. The LCD is the smallest expression that is divisible by each of the denominators. This is crucial for clearing fractions and simplifying the equation. In our case, the denominators are (x + 2), (x - 5), and (x² - 3x - 10). We already know that x² - 3x - 10 can be factored into (x + 2)(x - 5). Therefore, the LCD is simply (x + 2)(x - 5).

The LCD represents the smallest expression that all denominators divide into without leaving a remainder. For our equation, denominators include $(x+2)$, $(x-5)$, and $(x^2 - 3x - 10)$. We factor $x^2 - 3x - 10$ into $(x-5)(x+2)$. Consequently, the LCD becomes the product of the distinct factors, which is $(x+2)(x-5)$. This determination is vital as it enables us to eliminate the denominators, simplifying the equation for resolution. The meticulous identification of the LCD is an important step in solving rational equations, leading to an easier manipulation process and minimizing the complexity of the subsequent steps.

3. Clearing Fractions

To clear the fractions, we multiply both sides of the equation by the LCD, which is (x + 2)(x - 5). This step eliminates the denominators, transforming the rational equation into a simpler polynomial equation. This makes the equation easier to solve.

• Original equation:

  $\frac{2x}{x+2} + \frac{5}{x-5} = \frac{8}{(x+2)(x-5)}$

• Multiply both sides by the LCD (x + 2)(x - 5):

  $(x + 2)(x - 5) \cdot \left(\frac{2x}{x+2} + \frac{5}{x-5}\right) = (x + 2)(x - 5) \cdot \frac{8}{(x+2)(x-5)}$

• Distribute the LCD on the left side:

  $(x + 2)(x - 5) \cdot \frac{2x}{x+2} + (x + 2)(x - 5) \cdot \frac{5}{x-5} = (x + 2)(x - 5) \cdot \frac{8}{(x+2)(x-5)}$

• Cancel out common factors:

  $2x(x - 5) + 5(x + 2) = 8$

Clearing fractions involves multiplying both sides of the equation by the LCD, in this instance $(x + 2)(x - 5)$. This action effectively cancels out the denominators, transforming the equation into a simpler, more manageable form. The multiplication ensures each term is correctly factored, leading to a polynomial equation free from fractional components. This pivotal step streamlines the resolution process, enabling straightforward application of algebraic techniques for finding solutions.

4. Solving the Resulting Equation

After clearing the fractions, we are left with a polynomial equation. Now, we simplify and solve for x. In our case, the equation is:

$2x(x - 5) + 5(x + 2) = 8$

• Expand the terms: $2x^2 - 10x + 5x + 10 = 8$

• Combine like terms: $2x^2 - 5x + 10 = 8$

• Move all terms to one side to set the equation to zero: $2x^2 - 5x + 10 - 8 = 0$ $2x^2 - 5x + 2 = 0$

• Now, we solve the quadratic equation. This can be done by factoring, completing the square, or using the quadratic formula. Let's try factoring: $(2x - 1)(x - 2) = 0$

• Set each factor equal to zero and solve for x:

  • $2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$
  • $x - 2 = 0 \implies x = 2$

Thus, we have two potential solutions: $x = \frac{1}{2}$ and $x = 2$.

Solving the resulting equation after clearing fractions involves simplifying and isolating the variable. We've transformed our rational equation into a quadratic equation: $2x^2 - 5x + 2 = 0$. This quadratic can be solved through factoring, completing the square, or applying the quadratic formula. In this instance, factoring leads to $(2x - 1)(x - 2) = 0$. By setting each factor to zero, we derive the potential solutions $x = \frac{1}{2}$ and $x = 2$. These values are critical candidates for our final solution set, pending verification against the initial restrictions.

5. Verifying Solutions

It is essential to verify the solutions we found against the restrictions we identified earlier. Recall that our restrictions were x ≠ -2 and x ≠ 5. Both of our potential solutions, $x = \frac{1}{2}$ and $x = 2$, are different from these restricted values. Therefore, they are valid solutions.

To further ensure accuracy, we can substitute each solution back into the original equation to check if it holds true.

• Checking x = ½:

  $\frac{2(\frac{1}{2})}{\frac{1}{2}+2} + \frac{5}{\frac{1}{2}-5} = \frac{8}{(\frac{1}{2})^2-3(\frac{1}{2})-10}$

  $\frac{1}{\frac{5}{2}} + \frac{5}{-\frac{9}{2}} = \frac{8}{\frac{1}{4}-\frac{3}{2}-10}$

  $\frac{2}{5} - \frac{10}{9} = \frac{8}{\frac{1-6-40}{4}}$

  $\frac{18 - 50}{45} = \frac{8}{\frac{-45}{4}}$

  $\frac{-32}{45} = \frac{-32}{45}$

  This solution checks out.

• Checking x = 2:

  $\frac{2(2)}{2+2} + \frac{5}{2-5} = \frac{8}{2^2-3(2)-10}$

  $\frac{4}{4} + \frac{5}{-3} = \frac{8}{4-6-10}$

  $1 - \frac{5}{3} = \frac{8}{-12}$

  $\frac{3-5}{3} = -\frac{2}{3}$

  $\frac{-2}{3} = -\frac{2}{3}$

  This solution also checks out.

Since both solutions satisfy the original equation and do not violate our restrictions, they are valid.

Verifying solutions is a critical step to ensure the accuracy of results. We must confirm that our potential solutions, $x = \frac{1}{2}$ and $x = 2$, do not coincide with our previously identified restrictions, which are $x \neq -2$ and $x \neq 5$. Both solutions pass this initial check. To further validate, we substitute each solution back into the original equation. This substitution confirms that both $x = \frac{1}{2}$ and $x = 2$ indeed satisfy the equation, thus affirming their validity as solutions.

Conclusion

By following these steps, we have successfully solved the rational equation:

$\frac{2x}{x+2} + \frac{5}{x-5} = \frac{8}{x^2-3x-10}$

The solutions are $x = \frac{1}{2}$ and $x = 2$. This process highlights the importance of identifying restrictions, finding the LCD, clearing fractions, solving the resulting equation, and verifying solutions. These steps are crucial for solving any rational equation accurately and efficiently. Understanding each step thoroughly allows for the confident resolution of even complex rational equations, making this methodology an invaluable tool in algebraic problem-solving.