Solving Rational Equations A Step-by-Step Guide

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In the realm of algebra, solving rational equations is a fundamental skill with wide-ranging applications. However, the process often involves the possibility of encountering extraneous solutions, which are values that satisfy the transformed equation but not the original one. This article delves into the intricacies of solving the rational equation 4x+3+x+4x−3=2xx2−9\frac{4}{x+3}+\frac{x+4}{x-3}=\frac{2 x}{x^2-9} , systematically guiding you through the steps to identify and eliminate extraneous solutions. Understanding these concepts is crucial for success in algebra and related fields.

Understanding Rational Equations

Rational equations are equations that contain one or more rational expressions, where a rational expression is a fraction whose numerator and denominator are polynomials. Solving these equations requires a careful approach to avoid errors and extraneous solutions. Extraneous solutions arise when we perform operations, such as multiplying both sides by an expression that could be zero, which can introduce solutions that do not satisfy the original equation. Therefore, it is essential to check all potential solutions in the original equation to ensure their validity.

Before diving into the solution, let's first understand what rational equations are. Rational equations are equations that contain one or more fractions where the numerator and/or the denominator contain a variable. These equations often appear complex, but they can be solved systematically using algebraic techniques. The key is to eliminate the fractions by finding a common denominator and then solving the resulting polynomial equation. However, a crucial aspect of solving rational equations is identifying and eliminating extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but not the original equation. They typically arise when we multiply both sides of the equation by an expression that can be zero for some value of x. Therefore, it is imperative to check all potential solutions in the original equation to ensure they are valid. In this article, we will tackle the equation 4x+3+x+4x−3=2xx2−9\frac{4}{x+3}+\frac{x+4}{x-3}=\frac{2 x}{x^2-9}, demonstrating each step carefully and highlighting the importance of checking for extraneous solutions. Let's begin by understanding the original equation and identifying any restrictions on the variable x. This involves recognizing that denominators cannot be zero, which will help us determine values of x that must be excluded from the solution set. Once we have the restrictions, we can proceed to clear the fractions and solve the resulting equation, always keeping in mind the need to verify our solutions.

Step-by-Step Solution

1. Identify Restrictions on x

The first step in solving any rational equation is to identify the values of x that would make the denominators equal to zero. These values must be excluded from the solution set because division by zero is undefined. In the given equation, 4x+3+x+4x−3=2xx2−9\frac{4}{x+3}+\frac{x+4}{x-3}=\frac{2 x}{x^2-9}, we have three denominators: x + 3, x - 3, and x² - 9. Setting each of these equal to zero, we find:

  • x + 3 = 0 => x = -3
  • x - 3 = 0 => x = 3
  • x² - 9 = 0 => (x + 3)(x - 3) = 0 => x = -3 or x = 3

Thus, x cannot be -3 or 3. These are the restrictions on x that we must keep in mind when we arrive at potential solutions. If any of our solutions are -3 or 3, they will be extraneous solutions and must be discarded. Identifying these restrictions early on is a crucial step in solving rational equations correctly. It prevents us from accepting invalid solutions and ensures that our final answer is accurate. The process involves careful algebraic manipulation and a thorough understanding of the conditions under which a rational expression is undefined. Next, we will move on to clearing the fractions by multiplying both sides of the equation by the least common denominator (LCD). This will transform the rational equation into a polynomial equation, which is easier to solve. However, we must always remember to check our solutions against the restrictions we have identified to avoid extraneous solutions.

2. Find the Least Common Denominator (LCD)

To solve the rational equation, we need to eliminate the fractions. This is done by multiplying both sides of the equation by the least common denominator (LCD). The LCD is the smallest expression that is divisible by each of the denominators in the equation. In our case, the denominators are x + 3, x - 3, and x² - 9. Notice that x² - 9 can be factored as (x + 3)(x - 3). Therefore, the LCD is (x + 3)(x - 3).

Finding the LCD is a critical step because it allows us to clear the fractions and simplify the equation into a more manageable form. Without the LCD, we would be dealing with complex fractions, making the equation much harder to solve. The LCD ensures that when we multiply both sides of the equation, each denominator will divide evenly into it, eliminating the fractions. This transformation is a key technique in solving rational equations. It converts the equation into a polynomial equation, which we can then solve using standard algebraic methods. However, it is important to remember that multiplying by the LCD can sometimes introduce extraneous solutions, which is why checking our answers against the original equation and the restrictions we identified earlier is so crucial. In this instance, the LCD of (x + 3)(x - 3) simplifies the process significantly. Once we have multiplied both sides by the LCD, we will expand and simplify the resulting equation to find potential solutions. This step requires careful attention to algebraic manipulation and the correct application of the distributive property.

3. Multiply Both Sides by the LCD

Now that we have identified the LCD as (x + 3)(x - 3), we multiply both sides of the equation by this expression. This step will clear the fractions and transform the rational equation into a polynomial equation.

Original equation: 4x+3+x+4x−3=2xx2−9\frac{4}{x+3}+\frac{x+4}{x-3}=\frac{2 x}{x^2-9}

Multiplying both sides by (x + 3)(x - 3):

(x + 3)(x - 3) * [4x+3+x+4x−3\frac{4}{x+3}+\frac{x+4}{x-3}] = (x + 3)(x - 3) * 2xx2−9\frac{2 x}{x^2-9}

Distribute the LCD to each term:

(x + 3)(x - 3) * 4x+3\frac{4}{x+3} + (x + 3)(x - 3) * x+4x−3\frac{x+4}{x-3} = (x + 3)(x - 3) * 2x(x+3)(x−3)\frac{2 x}{(x+3)(x-3)}

Simplify by canceling out common factors:

4(x - 3) + (x + 4)(x + 3) = 2x

This step is crucial in solving rational equations as it eliminates the fractions, making the equation easier to manipulate and solve. By multiplying by the LCD, we are essentially multiplying each term by a form of 1 that eliminates the denominators. This transforms the equation into a more familiar form, a polynomial equation. However, this is also the step where extraneous solutions can be introduced, which is why the earlier step of identifying restrictions on x is so important. We must remember that any solution we find must be checked against these restrictions. The resulting equation, 4(x - 3) + (x + 4)(x + 3) = 2x, is now a quadratic equation, which we can solve using standard methods such as factoring, completing the square, or using the quadratic formula. The next step will involve expanding and simplifying this equation to bring it into a standard quadratic form.

4. Simplify and Solve the Equation

Now, we expand and simplify the equation we obtained in the previous step:

4(x - 3) + (x + 4)(x + 3) = 2x

Expand the terms:

4x - 12 + (x² + 3x + 4x + 12) = 2x

Combine like terms:

4x - 12 + x² + 7x + 12 = 2x

x² + 11x - 12 + 12 = 2x

x² + 11x = 2x

Move all terms to one side to set the equation to zero:

x² + 11x - 2x = 0

x² + 9x = 0

Factor out the common factor x:

x(x + 9) = 0

Set each factor equal to zero and solve for x:

  • x = 0
  • x + 9 = 0 => x = -9

So, the potential solutions are x = 0 and x = -9. Simplifying and solving the equation is a crucial step in finding potential solutions. This process involves careful algebraic manipulation, including expanding products, combining like terms, and factoring. The goal is to transform the equation into a form where we can easily identify the values of x that satisfy it. In this case, we have successfully transformed the rational equation into a quadratic equation, which we then factored to find the potential solutions. Each step in this process must be performed accurately to ensure we arrive at the correct potential solutions. However, the job is not done yet. We must now check these solutions against the restrictions we identified earlier to eliminate any extraneous solutions. This is the final, critical step in solving rational equations, and it ensures that our answers are valid in the context of the original equation.

5. Check for Extraneous Solutions

Finally, we need to check our potential solutions, x = 0 and x = -9, in the original equation to ensure they are not extraneous solutions. Recall that we identified x = -3 and x = 3 as restrictions, meaning these values cannot be solutions.

Check x = 0:

Substitute x = 0 into the original equation:

40+3+0+40−3=2(0)02−9\frac{4}{0+3}+\frac{0+4}{0-3}=\frac{2 (0)}{0^2-9}

43+4−3=0−9\frac{4}{3}+\frac{4}{-3}=\frac{0}{-9}

43−43=0\frac{4}{3}-\frac{4}{3}=0

0 = 0

This is true, so x = 0 is a valid solution.

Check x = -9:

Substitute x = -9 into the original equation:

4−9+3+−9+4−9−3=2(−9)(−9)2−9\frac{4}{-9+3}+\frac{-9+4}{-9-3}=\frac{2 (-9)}{(-9)^2-9}

4−6+−5−12=−1881−9\frac{4}{-6}+\frac{-5}{-12}=\frac{-18}{81-9}

−23+512=−1872\frac{-2}{3}+\frac{5}{12}=\frac{-18}{72}

−812+512=−14\frac{-8}{12}+\frac{5}{12}=\frac{-1}{4}

−312=−14\frac{-3}{12}=\frac{-1}{4}

−14=−14\frac{-1}{4}=\frac{-1}{4}

This is also true, so x = -9 is a valid solution.

Checking for extraneous solutions is a vital step in solving rational equations. It ensures that the solutions we have found are valid in the original equation and not introduced by the process of clearing fractions. This step involves substituting each potential solution back into the original equation and verifying that the equation holds true. If a potential solution makes the equation false or results in division by zero, it is an extraneous solution and must be discarded. In our case, we checked both x = 0 and x = -9 in the original equation and found that both solutions satisfy the equation. Since neither of these values violates our initial restrictions (x ≠ -3 and x ≠ 3), we can confidently conclude that they are both valid solutions. This careful verification process is what distinguishes solving rational equations from solving other types of equations and is essential for obtaining accurate results.

Final Answer

The solutions to the equation 4x+3+x+4x−3=2xx2−9\frac{4}{x+3}+\frac{x+4}{x-3}=\frac{2 x}{x^2-9} are x = -9 and x = 0.

Answer: x = -9, 0

Summary

Solving rational equations requires a systematic approach that includes identifying restrictions, finding the LCD, clearing fractions, solving the resulting equation, and, most importantly, checking for extraneous solutions. The equation 4x+3+x+4x−3=2xx2−9\frac{4}{x+3}+\frac{x+4}{x-3}=\frac{2 x}{x^2-9} serves as a comprehensive example of this process. By following each step carefully, we can confidently arrive at the correct solutions and avoid the pitfalls of extraneous solutions. Understanding and applying these techniques is crucial for mastering algebra and tackling more complex mathematical problems. Remember to always verify your solutions in the original equation to ensure their validity.

Keywords

Solving rational equations, extraneous solutions, least common denominator (LCD), restrictions on x, algebraic manipulation, quadratic equation, checking solutions.