Solving Rational Equations A Step-by-Step Guide To 2/(x-9) = 1/(x-3) - 1/(x-4)

In this article, we will delve into the process of solving the equation 2/(x-9) = 1/(x-3) - 1/(x-4). This equation involves rational expressions, which require careful manipulation to arrive at a solution. We'll explore the steps involved, from finding a common denominator to solving the resulting quadratic equation. Our goal is to provide a comprehensive, step-by-step guide that will help you understand the underlying principles and techniques for solving similar equations. This exploration is crucial for anyone studying algebra or preparing for more advanced mathematical concepts. By the end of this article, you should be confident in your ability to tackle equations involving rational expressions and understand the importance of verifying solutions to avoid extraneous roots.

1. Finding a Common Denominator

The first crucial step in solving this equation is to combine the fractions on the right-hand side. To do this, we need to find a common denominator for the terms 1/(x-3) and 1/(x-4). The least common denominator (LCD) in this case is the product of the two denominators, which is (x-3)(x-4). This is because the denominators (x-3) and (x-4) do not share any common factors, so their product is the smallest expression that both denominators can divide into evenly. Finding the common denominator is essential as it allows us to combine the fractions into a single term, simplifying the equation and making it easier to solve. Without this step, we cannot effectively compare and manipulate the terms to isolate the variable x.

To combine the fractions on the right-hand side, we rewrite each fraction with the common denominator. For the first fraction, 1/(x-3), we multiply both the numerator and the denominator by (x-4). This gives us (1 * (x-4)) / ((x-3) * (x-4)), which simplifies to (x-4) / (x-3)(x-4). Similarly, for the second fraction, 1/(x-4), we multiply both the numerator and the denominator by (x-3). This yields (1 * (x-3)) / ((x-4) * (x-3)), which simplifies to (x-3) / (x-3)(x-4). Now that both fractions have the same denominator, we can subtract them. This process of finding a common denominator and rewriting fractions is a fundamental technique in algebra and is used extensively when dealing with rational expressions and equations.

2. Combining Fractions

After finding the common denominator, we can combine the fractions on the right-hand side of the equation. We now have the equation 2/(x-9) = (x-4)/(x-3)(x-4) - (x-3)/(x-3)(x-4). Since both fractions on the right-hand side have the same denominator, we can subtract the numerators. Subtracting the numerators, we get (x-4) - (x-3). Remember to distribute the negative sign correctly when subtracting polynomials. This step is crucial because a mistake in distributing the negative sign can lead to an incorrect numerator and, ultimately, an incorrect solution for the equation. Accuracy in this step is paramount to the overall success in solving the problem. This combined fraction makes the equation simpler and sets the stage for further simplification and solving.

Subtracting the numerators (x-4) - (x-3), we get x - 4 - x + 3. Simplifying this expression, the x terms cancel each other out (x - x = 0), and we are left with -4 + 3, which equals -1. So, the right-hand side of the equation simplifies to -1 / (x-3)(x-4). This simplification is a critical step as it transforms the right-hand side into a single fraction, making it easier to manipulate and compare with the left-hand side of the equation. At this point, the equation becomes 2/(x-9) = -1 / (x-3)(x-4). This form of the equation is now ready for the next step: cross-multiplication.

3. Cross-Multiplication

With the equation now in the form 2/(x-9) = -1 / (x-3)(x-4), we can proceed with cross-multiplication. Cross-multiplication is a technique used to eliminate fractions in an equation by multiplying the numerator of one side by the denominator of the other side and setting them equal. In this case, we will multiply 2 by (x-3)(x-4) and -1 by (x-9). This method is based on the principle that if two fractions are equal, then their cross-products are also equal. Cross-multiplication is a powerful tool for solving equations involving fractions, as it transforms the equation into a simpler, more manageable form, often a polynomial equation. It is an essential technique in algebra and is used in various mathematical contexts.

Performing the cross-multiplication, we get 2 * (x-3)(x-4) = -1 * (x-9). This step is crucial as it removes the fractions from the equation, making it easier to solve. Now, we have a more standard algebraic equation to work with. The next step involves expanding the expressions on both sides of the equation. Expanding these expressions is necessary to simplify the equation further and combine like terms. This process sets the stage for solving for x, and any errors in this expansion can lead to an incorrect solution. Therefore, careful attention to detail is required during this step to ensure the accurate progression of the solution.

4. Expanding and Simplifying

After cross-multiplication, our equation is 2(x-3)(x-4) = -1(x-9). Now, we need to expand both sides of the equation to simplify it further. First, let's expand the left-hand side. We have 2(x-3)(x-4). We'll first multiply (x-3) and (x-4). Multiplying these two binomials, we get x * x - 4x - 3x + 12, which simplifies to x² - 7x + 12. Now, we multiply this result by 2, giving us 2(x² - 7x + 12) = 2x² - 14x + 24. Expanding expressions like this is a fundamental skill in algebra, and it's crucial to ensure that each term is correctly multiplied and that signs are handled properly. The correct expansion is essential for arriving at the correct final solution.

Next, we expand the right-hand side of the equation, which is -1(x-9). Distributing the -1, we get -x + 9. Now, our equation looks like this: 2x² - 14x + 24 = -x + 9. The next step is to move all terms to one side of the equation to set it equal to zero. This is a standard technique when dealing with quadratic equations because setting the equation to zero allows us to use methods like factoring or the quadratic formula to find the solutions for x. Bringing all terms to one side is a critical step in preparing the equation for solving.

5. Forming a Quadratic Equation

After expanding and simplifying both sides, our equation is 2x² - 14x + 24 = -x + 9. To solve for x, we need to set the equation to zero. We can do this by adding x to both sides and subtracting 9 from both sides. Adding x to both sides gives us 2x² - 13x + 24 = 9. Subtracting 9 from both sides gives us 2x² - 13x + 15 = 0. This is now a quadratic equation in the standard form ax² + bx + c = 0, where a = 2, b = -13, and c = 15. Forming the quadratic equation is a critical step because it puts the equation into a form that can be solved using standard methods, such as factoring, completing the square, or the quadratic formula.

Now that we have the quadratic equation 2x² - 13x + 15 = 0, we need to solve it for x. There are several methods to solve quadratic equations, but in this case, we will use factoring. Factoring involves finding two binomials that, when multiplied together, give us the quadratic equation. If we can factor the quadratic equation, we can then set each factor equal to zero and solve for x. Factoring is often the quickest method for solving quadratic equations when it is possible, and it relies on recognizing patterns and relationships between the coefficients of the quadratic equation. This is an important skill in algebra and is used in various mathematical contexts.

6. Solving the Quadratic Equation

We now have the quadratic equation 2x² - 13x + 15 = 0. To solve this equation, we will attempt to factor it. We are looking for two binomials that multiply to give 2x² - 13x + 15. The general form of these binomials will be (ax + b)(cx + d), where a, b, c, and d are constants. To factor this quadratic, we need to find two numbers that multiply to 2 * 15 = 30 and add up to -13. These numbers are -10 and -3. Using these numbers, we can rewrite the middle term of the quadratic equation.

We rewrite -13x as -10x - 3x, so the equation becomes 2x² - 10x - 3x + 15 = 0. Now, we can factor by grouping. We group the first two terms and the last two terms: (2x² - 10x) + (-3x + 15) = 0. We factor out the greatest common factor from each group. From the first group, we can factor out 2x, and from the second group, we can factor out -3. This gives us 2x(x - 5) - 3(x - 5) = 0. Notice that we now have a common factor of (x - 5). We factor out (x - 5), and we are left with (2x - 3)(x - 5) = 0. This factorization is a crucial step in solving the quadratic equation, and it allows us to find the values of x that make the equation true.

Now that we have factored the quadratic equation as (2x - 3)(x - 5) = 0, we can use the zero-product property to find the solutions for x. The zero-product property states that if the product of two factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for x. Setting the first factor (2x - 3) equal to zero gives us 2x - 3 = 0. Adding 3 to both sides gives us 2x = 3. Dividing both sides by 2 gives us x = 3/2 or x = 1.5. This is one of the possible solutions for x.

Next, we set the second factor (x - 5) equal to zero, which gives us x - 5 = 0. Adding 5 to both sides gives us x = 5. This is the second possible solution for x. So, we have two potential solutions for the equation: x = 1.5 and x = 5. However, it's crucial to remember that when solving rational equations, we need to check these solutions to make sure they are not extraneous. Extraneous solutions are values that satisfy the transformed equation but not the original equation, and they often arise when we perform operations that can introduce new solutions, such as squaring both sides or multiplying by an expression that could be zero.

7. Checking for Extraneous Solutions

We have found two potential solutions for the equation 2/(x-9) = 1/(x-3) - 1/(x-4): x = 1.5 and x = 5. Now, it's essential to check these solutions in the original equation to ensure they are valid and not extraneous. Extraneous solutions can arise when we perform operations that change the domain of the equation, such as multiplying by an expression that contains x. To check our solutions, we will substitute each value of x back into the original equation and see if it holds true. This step is crucial because it ensures that we only accept solutions that are valid for the original equation, maintaining the integrity of our solution process.

First, let's check x = 1.5. Substituting x = 1.5 into the original equation 2/(x-9) = 1/(x-3) - 1/(x-4), we get 2/(1.5-9) = 1/(1.5-3) - 1/(1.5-4). Simplifying the denominators, we have 2/(-7.5) = 1/(-1.5) - 1/(-2.5). Converting the decimals to fractions, we get 2/(-15/2) = 1/(-3/2) - 1/(-5/2). Simplifying further, we have -4/15 = -2/3 + 2/5. To add the fractions on the right-hand side, we need a common denominator, which is 15. So, we rewrite the fractions as -4/15 = -10/15 + 6/15. This simplifies to -4/15 = -4/15, which is true. Therefore, x = 1.5 is a valid solution.

Now, let's check x = 5. Substituting x = 5 into the original equation 2/(x-9) = 1/(x-3) - 1/(x-4), we get 2/(5-9) = 1/(5-3) - 1/(5-4). Simplifying the denominators, we have 2/(-4) = 1/2 - 1/1, which simplifies to -1/2 = 1/2 - 1. Further simplifying the right-hand side, we get -1/2 = 1/2 - 2/2, which is -1/2 = -1/2. This is also true, so x = 5 is a valid solution. Since both potential solutions check out in the original equation, we can confidently conclude that they are both valid solutions.

In conclusion, we have successfully solved the equation 2/(x-9) = 1/(x-3) - 1/(x-4). The solutions we found are x = 1.5 and x = 5. We arrived at these solutions by following a step-by-step process that included finding a common denominator, combining fractions, cross-multiplication, expanding and simplifying, forming a quadratic equation, solving the quadratic equation, and, most importantly, checking for extraneous solutions. This process highlights the importance of each step in solving rational equations. Checking for extraneous solutions is particularly crucial, as it ensures that the solutions we find are valid for the original equation and not artifacts of the algebraic manipulations.

Solving rational equations like this requires a solid understanding of algebraic principles and techniques. Each step, from finding the common denominator to checking for extraneous solutions, plays a critical role in obtaining the correct result. By mastering these techniques, you will be well-equipped to tackle a wide range of algebraic problems and gain a deeper appreciation for the elegance and power of mathematics. Remember, practice is key to proficiency, so continue to work through similar problems to reinforce your understanding and skills.