Solving Rational Equations A Step By Step Guide
In mathematics, rational equations play a crucial role in various fields, ranging from algebra to calculus. Understanding how to solve rational equations is essential for students and professionals alike. This article aims to provide a comprehensive guide on solving rational equations, focusing on two specific examples. We will delve into the step-by-step process, ensuring clarity and accuracy in our solutions. Rational equations, at their core, involve fractions where the numerator and/or the denominator contain variables. These equations can often seem daunting, but with a systematic approach, they become manageable and even elegant to solve. The beauty of rational equations lies in their ability to model real-world scenarios, such as rates, work problems, and mixture problems. Therefore, mastering the techniques to solve them opens doors to a deeper understanding of mathematical applications in various domains. The examples we will explore in this article are designed to illustrate common challenges and effective strategies for tackling them. By breaking down each step and providing detailed explanations, we aim to empower you with the knowledge and confidence to solve a wide range of rational equations. Whether you are a student preparing for an exam or a professional seeking to refresh your skills, this guide will serve as a valuable resource. So, let's embark on this journey of mathematical exploration and unravel the intricacies of solving rational equations together.
g) Solving the Rational Equation: 5/x² + 4/(3x) = (15x + 4x²)/(3x²)
Step 1: Identify the Equation
The first step in solving any equation is to clearly identify what you are working with. In this case, we have the rational equation:
5/x² + 4/(3x) = (15x + 4x²)/(3x²)
This equation involves fractions with variables in the denominators, making it a rational equation. Our goal is to find the value(s) of x that satisfy this equation. To begin, it's crucial to recognize the structure of the equation and the operations involved. We have two fractions on the left-hand side that are being added together, and a single fraction on the right-hand side. The presence of x² and x in the denominators suggests that we might need to consider potential restrictions on the values of x, as division by zero is undefined. Before we proceed with any algebraic manipulations, let's take a moment to appreciate the elegance of this equation. It represents a relationship between different terms involving x, and by solving it, we are essentially uncovering the hidden values of x that make this relationship true. This process is akin to solving a puzzle, where each step brings us closer to the final solution. Now, let's move on to the next step, which involves clearing the fractions to simplify the equation.
Step 2: Find the Least Common Denominator (LCD)
The next step is to find the least common denominator (LCD) of the fractions in the equation. The denominators are x², 3x, and 3x². The LCD is the smallest expression that is divisible by all the denominators. To find the LCD, we need to consider the prime factors of each denominator. The factors of x² are x * x, the factors of 3x are 3 * x, and the factors of 3x² are 3 * x * x. The LCD must include each unique factor raised to the highest power it appears in any of the denominators. In this case, the LCD is 3x². Understanding the concept of the LCD is crucial because it allows us to eliminate the fractions from the equation, making it easier to solve. By multiplying both sides of the equation by the LCD, we essentially clear the denominators and transform the equation into a more manageable form. This is a common technique used in solving rational equations and is a fundamental step in the overall process. The LCD acts as a bridge, connecting the different fractions and allowing us to combine them into a single expression. Once we have the LCD, we can proceed to the next step, which involves multiplying both sides of the equation by it. This will effectively remove the fractions and pave the way for solving for x.
Step 3: Multiply Both Sides by the LCD
To eliminate the fractions, multiply both sides of the equation by the LCD, which is 3x². This gives us:
3x² * (5/x² + 4/(3x)) = 3x² * ((15x + 4x²)/(3x²))
Distribute 3x² on the left side:
3x² * (5/x²) + 3x² * (4/(3x)) = 3x² * ((15x + 4x²)/(3x²))
Simplify each term:
15 + 4x = 15x + 4x²
Multiplying by the LCD is a pivotal step in solving rational equations because it transforms the equation into a more familiar form, typically a polynomial equation. This step eliminates the complexity introduced by the fractions, allowing us to apply standard algebraic techniques to solve for the variable. The distribution of the LCD across the terms on the left-hand side ensures that each fraction is properly accounted for. The simplification process involves canceling out common factors between the LCD and the denominators of the fractions. This cancellation is what effectively removes the fractions from the equation. After multiplying by the LCD and simplifying, we are left with a polynomial equation that we can solve using techniques such as factoring, completing the square, or the quadratic formula. This step demonstrates the power of algebraic manipulation in transforming complex equations into simpler forms. Now that we have a polynomial equation, let's move on to the next step, which involves rearranging the terms and solving for x.
Step 4: Rearrange the Equation
Now, let's rearrange the equation into a standard quadratic form. Subtract 15x and 4x² from both sides:
0 = 4x² + 15x - 4x - 15
0 = 4x² + 11x - 15
Rearranging the equation into a standard form is crucial because it allows us to easily identify the coefficients and apply appropriate solving methods. In this case, we have a quadratic equation, which is a polynomial equation of degree two. The standard form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants. By rearranging the terms, we can clearly see the values of a, b, and c, which are essential for applying methods such as factoring or the quadratic formula. The process of rearranging involves moving all the terms to one side of the equation, leaving zero on the other side. This ensures that we have a clear and concise representation of the equation, making it easier to work with. In this specific example, we subtracted 15x and 4x² from both sides to achieve the standard quadratic form. This step demonstrates the importance of algebraic manipulation in transforming equations into a format that is conducive to solving. Now that we have the equation in standard form, let's move on to the next step, which involves solving the quadratic equation for x.
Step 5: Solve the Quadratic Equation
To solve the quadratic equation, we can use factoring, the quadratic formula, or completing the square. Let's try factoring:
0 = 4x² + 16x - 5x - 15
0 = 4x(x + 4) - 5(x + 3)
0 = (4x - 5)(x + 3)
Setting each factor to zero gives us:
4x - 5 = 0 or x + 3 = 0
Solving for x:
x = 5/4 or x = -3
Solving the quadratic equation is a critical step in finding the solutions to the original rational equation. There are several methods available for solving quadratic equations, each with its own advantages and disadvantages. Factoring is a method that involves breaking down the quadratic expression into two linear factors. This method is efficient when the quadratic expression can be easily factored. The quadratic formula is a general method that can be used to solve any quadratic equation, regardless of whether it can be factored or not. Completing the square is another method that can be used to solve quadratic equations, and it is particularly useful when the quadratic formula is not easily applicable. In this case, we attempted to solve the quadratic equation by factoring. This involved finding two binomials that, when multiplied together, give the original quadratic expression. By setting each factor equal to zero, we can find the values of x that make the equation true. These values are the solutions to the quadratic equation and, therefore, potential solutions to the original rational equation. Now that we have found the potential solutions, let's move on to the next step, which involves checking these solutions in the original equation.
Step 6: Check for Extraneous Solutions
It is crucial to check the solutions in the original equation to ensure they are valid. Substitute x = 5/4:
5/(5/4)² + 4/(3*(5/4)) = (15*(5/4) + 4*(5/4)²)/(3*(5/4)²)
5/(25/16) + 4/(15/4) = (75/4 + 4*(25/16))/(3*(25/16))
(516)/25 + (44)/15 = (75/4 + 25/4)/(75/16)
16/5 + 16/15 = (100/4)/(75/16)
(48 + 16)/15 = (25)*(16/75)
64/15 = 16/3
This solution is valid.
Now substitute x = -3:
5/(-3)² + 4/(3*(-3)) = (15*(-3) + 4*(-3)²)/(3*(-3)²)
5/9 + 4/(-9) = (-45 + 49)/(39)
5/9 - 4/9 = (-45 + 36)/27
1/9 = -9/27
1/9 = -1/3
This solution is extraneous.
Checking for extraneous solutions is a vital step in solving rational equations because it ensures that the solutions we obtain are valid and do not lead to undefined expressions in the original equation. Extraneous solutions are solutions that satisfy the transformed equation but do not satisfy the original equation. These solutions typically arise when we perform operations that can introduce new solutions, such as squaring both sides or multiplying by an expression that contains the variable. In the context of rational equations, extraneous solutions often occur when a solution makes one of the denominators in the original equation equal to zero. Since division by zero is undefined, these solutions are not valid. To check for extraneous solutions, we substitute each potential solution back into the original equation and verify that it satisfies the equation. If a solution leads to a contradiction or an undefined expression, it is an extraneous solution and must be discarded. This step highlights the importance of rigor in mathematical problem-solving. It is not enough to simply find a solution; we must also ensure that it is a valid solution within the context of the original problem. Now that we have checked for extraneous solutions, we can confidently state the valid solutions to the equation.
Step 7: State the Solution
The solution to the equation is x = 5/4.
Stating the solution clearly and concisely is the final step in the problem-solving process. It is important to present the solution in a way that is easy to understand and unambiguous. In this case, we have found that the only valid solution to the rational equation is x = 5/4. This means that this is the only value of x that, when substituted into the original equation, will make the equation true. It is also important to note that we had an extraneous solution, x = -3, which we discarded after checking it in the original equation. This highlights the importance of the step of checking for extraneous solutions, as failing to do so could lead to an incorrect answer. By stating the solution clearly, we are providing a definitive answer to the problem and completing the solution process. This step also allows us to reflect on the entire process and ensure that we have addressed all aspects of the problem. Now that we have solved the first rational equation, let's move on to the second example and apply the same systematic approach to find its solution.
h) Solving the Rational Equation: 3/(x+2) - 2/(x-1)
Step 1: Identify the Equation
The first step is to identify the equation:
3/(x+2) - 2/(x-1) = 0
This is a rational equation because it involves fractions with expressions containing x in the denominators. The equation presents a slightly different challenge compared to the previous one, as it involves subtraction of two fractions and the goal is to find the value(s) of x that make the expression equal to zero. Before we proceed, it's crucial to recognize the potential restrictions on the values of x. The denominators (x+2) and (x-1) cannot be equal to zero, as this would result in division by zero, which is undefined. Therefore, we must keep in mind that x cannot be equal to -2 or 1. This understanding will be important when we check for extraneous solutions later in the process. The structure of the equation suggests that we will need to combine the fractions by finding a common denominator. This will allow us to simplify the equation and eventually solve for x. The process of solving this rational equation is akin to piecing together a puzzle. Each step we take brings us closer to the solution, and with careful attention to detail, we can successfully unravel the value(s) of x that satisfy the equation. Now, let's move on to the next step, which involves finding the least common denominator (LCD) of the fractions.
Step 2: Find the Least Common Denominator (LCD)
The denominators are (x+2) and (x-1). The LCD is the product of these two expressions, as they have no common factors. Therefore, the LCD is (x+2)(x-1).
Finding the least common denominator (LCD) is a fundamental step in solving rational equations, as it allows us to combine fractions and simplify the equation. The LCD is the smallest expression that is divisible by all the denominators in the equation. In this case, the denominators are (x+2) and (x-1). Since these expressions have no common factors, the LCD is simply their product, which is (x+2)(x-1). Understanding the concept of the LCD is crucial because it enables us to eliminate the fractions from the equation, making it easier to solve. By multiplying both sides of the equation by the LCD, we effectively clear the denominators and transform the equation into a more manageable form. This is a common technique used in solving rational equations and is a cornerstone of the overall process. The LCD acts as a bridge, connecting the different fractions and allowing us to combine them into a single expression. Once we have the LCD, we can proceed to the next step, which involves multiplying both sides of the equation by it. This will effectively remove the fractions and pave the way for solving for x. Now, let's move on to the next step, which involves multiplying both sides of the equation by the LCD.
Step 3: Multiply Both Sides by the LCD
Multiply both sides of the equation by the LCD, (x+2)(x-1):
(x+2)(x-1) * [3/(x+2) - 2/(x-1)] = 0 * (x+2)(x-1)
Distribute (x+2)(x-1) on the left side:
(x+2)(x-1) * [3/(x+2)] - (x+2)(x-1) * [2/(x-1)] = 0
Simplify each term:
3(x-1) - 2(x+2) = 0
Multiplying by the LCD is a pivotal step in solving rational equations because it transforms the equation into a more familiar form, typically a polynomial equation. This step eliminates the complexity introduced by the fractions, allowing us to apply standard algebraic techniques to solve for the variable. The distribution of the LCD across the terms on the left-hand side ensures that each fraction is properly accounted for. The simplification process involves canceling out common factors between the LCD and the denominators of the fractions. This cancellation is what effectively removes the fractions from the equation. After multiplying by the LCD and simplifying, we are left with a linear equation that we can solve using basic algebraic techniques. This step demonstrates the power of algebraic manipulation in transforming complex equations into simpler forms. Now that we have a linear equation, let's move on to the next step, which involves expanding the terms and simplifying the equation further.
Step 4: Expand and Simplify
Expand the terms:
3x - 3 - 2x - 4 = 0
Combine like terms:
x - 7 = 0
Expanding and simplifying the equation is a crucial step in solving for the variable. It involves applying the distributive property to remove parentheses and then combining like terms to reduce the equation to its simplest form. In this case, we expanded the terms 3(x-1) and -2(x+2) by multiplying the constants with the terms inside the parentheses. This gave us 3x - 3 and -2x - 4, respectively. Next, we combined the like terms, which are the terms that have the same variable raised to the same power. In this case, we combined the terms 3x and -2x to get x, and we combined the constants -3 and -4 to get -7. This resulted in the simplified equation x - 7 = 0. Simplifying the equation makes it easier to isolate the variable and solve for its value. This step demonstrates the importance of algebraic manipulation in transforming equations into a format that is conducive to solving. Now that we have a simplified equation, let's move on to the next step, which involves isolating the variable and finding its value.
Step 5: Solve for x
To solve for x, add 7 to both sides of the equation:
x = 7
Solving for x is the culmination of the algebraic manipulations we have performed. It involves isolating the variable on one side of the equation to determine its value. In this case, we have the simplified equation x - 7 = 0. To isolate x, we need to undo the operation that is being performed on it, which is subtraction by 7. We can do this by adding 7 to both sides of the equation. This gives us x = 7. This value represents the solution to the equation, meaning that when we substitute x = 7 back into the original equation, it will make the equation true. It is important to remember that we should always check our solution in the original equation to ensure that it is valid and does not lead to any contradictions or undefined expressions. Now that we have found a potential solution, let's move on to the next step, which involves checking this solution in the original equation.
Step 6: Check for Extraneous Solutions
Substitute x = 7 into the original equation:
3/(7+2) - 2/(7-1) = 0
3/9 - 2/6 = 0
1/3 - 1/3 = 0
0 = 0
The solution is valid.
Checking for extraneous solutions is a vital step in solving rational equations because it ensures that the solution we have obtained is valid and does not lead to undefined expressions in the original equation. Extraneous solutions are solutions that satisfy the transformed equation but do not satisfy the original equation. These solutions typically arise when we perform operations that can introduce new solutions, such as squaring both sides or multiplying by an expression that contains the variable. In the context of rational equations, extraneous solutions often occur when a solution makes one of the denominators in the original equation equal to zero. Since division by zero is undefined, these solutions are not valid. To check for extraneous solutions, we substitute the potential solution back into the original equation and verify that it satisfies the equation. In this case, we substituted x = 7 into the original equation and found that it does indeed satisfy the equation, as it leads to a true statement (0 = 0). This means that x = 7 is a valid solution and not an extraneous solution. This step highlights the importance of rigor in mathematical problem-solving. It is not enough to simply find a solution; we must also ensure that it is a valid solution within the context of the original problem. Now that we have checked for extraneous solutions, we can confidently state the solution to the equation.
Step 7: State the Solution
The solution to the equation is x = 7.
Stating the solution clearly and concisely is the final step in the problem-solving process. It is important to present the solution in a way that is easy to understand and unambiguous. In this case, we have found that the solution to the rational equation is x = 7. This means that when x is equal to 7, the original equation is true. It is also important to note that we checked our solution and found it to be valid, meaning that it does not lead to any contradictions or undefined expressions in the original equation. By stating the solution clearly, we are providing a definitive answer to the problem and completing the solution process. This step also allows us to reflect on the entire process and ensure that we have addressed all aspects of the problem. Now that we have solved both rational equations, let's summarize the key steps and techniques involved in solving these types of equations.
In conclusion, solving rational equations involves a systematic approach that includes identifying the equation, finding the least common denominator (LCD), multiplying both sides by the LCD, simplifying the equation, solving for the variable, and checking for extraneous solutions. Mastering these steps is essential for success in algebra and beyond. Rational equations are a fundamental concept in mathematics, and the ability to solve them is crucial for various applications in science, engineering, and economics. The examples we have explored in this article illustrate the common techniques and challenges involved in solving rational equations. By understanding the underlying principles and practicing these techniques, you can develop the skills and confidence to tackle a wide range of rational equations. Remember, the key to success is to approach each equation methodically, paying close attention to detail and always checking for extraneous solutions. With practice and perseverance, you can master the art of solving rational equations and unlock their power in various mathematical contexts. The journey of mathematical exploration is a rewarding one, and the ability to solve rational equations is a valuable tool in your mathematical toolkit. So, keep practicing, keep exploring, and keep pushing the boundaries of your mathematical knowledge. We hope this article has provided you with a comprehensive guide to solving rational equations and has empowered you to approach these equations with confidence and skill.
