Solving $\frac{t}{3}+\frac{t+1}{t}=\frac{1}{t}$ A Step-by-Step Guide

1. Identifying the Equation and Its Components

The first step in solving any equation is to carefully identify the equation and its components. In our case, the equation is $\frac{t}{3}+\frac{t+1}{t}=\frac{1}{t}$. This equation involves fractions with the variable 't' in the denominator, making it a rational equation. Recognizing this is crucial because it helps us determine the appropriate methods for solving it. Rational equations often require specific techniques to eliminate the fractions and simplify the equation into a more manageable form. The components of the equation include the terms $\frac{t}{3}$, $\frac{t+1}{t}$, and $\frac{1}{t}$, each involving the variable 't'. Understanding these components allows us to plan our approach, which typically involves finding a common denominator and eliminating it to simplify the equation. Before we dive into solving, it’s also essential to note any restrictions on the variable 't'. Since we have 't' in the denominator, 't' cannot be zero because division by zero is undefined. This restriction is crucial and will be considered when we find the solutions. Ignoring this restriction can lead to incorrect solutions or misinterpretations of the solution set. Identifying the type of equation is an important initial step because different types of equations require different methods. For example, linear equations are solved differently than quadratic equations, and rational equations have their own set of rules. By identifying this as a rational equation, we know we need to focus on eliminating the denominators.

2. Finding the Common Denominator

To solve the equation, we must first find a common denominator for all the fractions. Identifying the common denominator is a crucial step in simplifying rational equations. The given equation is $\frac{t}{3}+\frac{t+1}{t}=\frac{1}{t}$. The denominators are 3 and t. The least common denominator (LCD) for 3 and t is 3t. Finding the LCD is essential because it allows us to combine the fractions into a single fraction, which simplifies the equation. To find the LCD, we look for the smallest expression that each denominator can divide into evenly. In this case, 3t is the smallest expression that both 3 and t can divide into without leaving a remainder. Once we have the LCD, we can rewrite each fraction with the LCD as the new denominator. This involves multiplying the numerator and the denominator of each fraction by the factor that makes the denominator equal to the LCD. By doing this, we ensure that all fractions have the same denominator, making it possible to combine them. This step is vital for the subsequent steps in solving the equation. Failing to find the correct LCD can lead to errors in the solution. The process of finding the LCD might involve factoring the denominators if they are more complex expressions. In such cases, we would identify the prime factors of each denominator and then construct the LCD by including each factor the greatest number of times it appears in any denominator. However, in this case, the denominators are simple, making the LCD straightforward to find. Once the LCD is found, we can proceed to the next step of multiplying each term in the equation by the LCD, which will eliminate the fractions and result in a simpler equation to solve.

3. Multiplying Both Sides by the Common Denominator

Now that we have identified the common denominator as 3t, the next crucial step is to multiply both sides of the equation by this common denominator. This process is essential because it eliminates the fractions, thereby simplifying the equation and making it easier to solve. Our equation is $\frac{t}{3}+\frac{t+1}{t}=\frac{1}{t}$. Multiplying both sides by 3t, we get: 3t * ($\frac{t}{3}+\frac{t+1}{t}$) = 3t * ($\frac{1}{t}$). This step is based on the fundamental algebraic principle that multiplying both sides of an equation by the same non-zero quantity maintains the equality. Distributing 3t on the left side, we multiply 3t by each term inside the parentheses: 3t * ($\frac{t}{3}$) + 3t * ($\frac{t+1}{t}$) = 3t * ($\frac{1}{t}$). Now, we simplify each term: (3t * $\frac{t}{3}$) simplifies to t², (3t * $\frac{t+1}{t}$) simplifies to 3(t+1), and (3t * $\frac{1}{t}$) simplifies to 3. This leads us to the equation t² + 3(t+1) = 3. Multiplying by the common denominator is a powerful technique for solving rational equations because it clears the fractions and transforms the equation into a more familiar form, such as a quadratic equation or a linear equation. Once the fractions are eliminated, the equation becomes much easier to manipulate and solve using standard algebraic methods. This step is crucial because it avoids the complexities of working with fractions, which can often lead to errors. By carefully multiplying each term by the common denominator, we ensure that the equation remains balanced and that we are on the right path to finding the solution set.

4. Simplifying the Equation

After multiplying both sides of the equation by the common denominator, our equation is now t² + 3(t+1) = 3. The next step is to simplify this equation by expanding and combining like terms. Simplifying the equation is a critical step in the process of solving it, as it transforms the equation into a more manageable form. First, we expand the term 3(t+1) by distributing the 3 across the terms inside the parentheses: t² + 3t + 3 = 3. Now, we have a quadratic equation. To further simplify, we need to set the equation equal to zero. Subtracting 3 from both sides of the equation gives us: t² + 3t + 3 - 3 = 3 - 3, which simplifies to t² + 3t = 0. At this point, we have a simplified quadratic equation that is ready to be solved. Simplifying the equation is not just about making it look neater; it’s about transforming it into a standard form that allows us to apply specific solution methods. In this case, we have transformed the equation into a quadratic equation in the form at² + bt = 0, which can be solved by factoring. The process of simplifying can involve various algebraic manipulations, such as distributing terms, combining like terms, and rearranging the equation. Each step in the simplification process should be performed carefully to avoid errors. Simplifying the equation makes it easier to identify the type of equation we are dealing with and the appropriate method for solving it. For instance, in this case, recognizing the equation as a quadratic allows us to use methods such as factoring, completing the square, or using the quadratic formula to find the solutions. By simplifying, we reduce the complexity of the equation and increase the chances of finding the correct solution.

5. Solving the Quadratic Equation by Factoring

Now that we have the simplified quadratic equation t² + 3t = 0, the next step is to solve it. In this case, the equation can be easily solved by factoring. Factoring is a powerful method for solving quadratic equations, especially when the equation can be factored easily. The key to factoring is to find the common factors in the terms of the equation. In our equation, t² + 3t = 0, we can see that 't' is a common factor. Factoring out 't', we get: t(t + 3) = 0. This equation is now in a factored form, which means that the product of two expressions, t and (t + 3), is equal to zero. According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. Therefore, we can set each factor equal to zero and solve for 't': 1. t = 0 2. t + 3 = 0 Solving the second equation for 't', we subtract 3 from both sides: t = -3. Thus, we have found two potential solutions: t = 0 and t = -3. Factoring is an efficient method for solving quadratic equations when it is applicable. It involves rewriting the quadratic expression as a product of two linear expressions. The ability to factor quadratic expressions is a fundamental skill in algebra and is essential for solving a wide range of problems. However, not all quadratic equations can be easily factored. In cases where factoring is difficult or not possible, other methods such as completing the square or using the quadratic formula can be used. Factoring simplifies the process of finding the solutions by breaking down the quadratic equation into simpler linear equations. By setting each factor to zero, we can quickly identify the values of the variable that make the equation true. This method is straightforward and efficient, making it a preferred choice when the quadratic expression can be factored easily.

6. Checking for Extraneous Solutions

After finding the potential solutions, it's crucial to check for extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but not the original equation. In the context of rational equations, this often occurs when a solution makes the denominator of the original equation equal to zero. Our original equation is $\frac{t}{3}+\frac{t+1}{t}=\frac{1}{t}$, and we found two potential solutions: t = 0 and t = -3. Let’s check each solution: 1. For t = 0: If we substitute t = 0 into the original equation, we get $\frac{0}{3}+\frac{0+1}{0}=\frac{1}{0}$. Notice that we have division by zero in the second and third terms, which is undefined. Therefore, t = 0 is an extraneous solution and must be discarded. 2. For t = -3: If we substitute t = -3 into the original equation, we get $\frac{-3}{3}+\frac{-3+1}{-3}=\frac{1}{-3}$. Simplifying, we have -1 + $\frac{-2}{-3}$ = -$\frac{1}{3}$, which becomes -1 + $\frac{2}{3}$ = -$\frac{1}{3}$. Further simplifying, we get -$\frac{1}{3}$ = -$\frac{1}{3}$, which is true. Thus, t = -3 is a valid solution. Checking for extraneous solutions is a vital step in solving rational equations. It ensures that the solutions we find are valid in the context of the original equation and not just artifacts of the algebraic manipulations we performed. The reason we get extraneous solutions in rational equations is because multiplying both sides of the equation by an expression containing the variable can introduce solutions that make the denominator zero, which is not allowed. Therefore, it is always necessary to verify the solutions by substituting them back into the original equation. By doing this, we can confidently identify and discard any extraneous solutions, ensuring that we have the correct solution set.

7. Presenting the Solution Set

After checking for extraneous solutions, we can now confidently present the solution set. We found two potential solutions: t = 0 and t = -3. However, we determined that t = 0 is an extraneous solution because it makes the denominator of the original equation zero. Therefore, the only valid solution is t = -3. The solution set is the set of all values of the variable that satisfy the original equation. In this case, the solution set contains only one element: -3. We can represent the solution set using set notation as {-3}. Presenting the solution set clearly and accurately is the final step in the problem-solving process. It communicates the answer in a concise and understandable manner. The solution set should include all valid solutions and exclude any extraneous solutions. In some cases, an equation may have multiple solutions, while in other cases, it may have no solutions. Understanding how to present the solution set is important for clear communication in mathematics. When presenting the solution set, it is often helpful to provide a brief explanation of why the values are solutions and why any potential extraneous solutions were discarded. This adds clarity and completeness to the answer. In summary, the solution set for the equation $\frac{t}{3}+\frac{t+1}{t}=\frac{1}{t}$ is {-3}. This means that t = -3 is the only value that satisfies the equation without causing any undefined terms. This final step completes the process of solving the equation, from identifying the equation and its components to presenting the final solution set.

In conclusion, solving the equation $\frac{t}{3}+\frac{t+1}{t}=\frac{1}{t}$ involves several key steps, from identifying the equation as a rational equation to presenting the final solution set. We began by finding the common denominator, which was 3t, and then multiplied both sides of the equation by this common denominator to eliminate the fractions. This led us to a simplified quadratic equation: t² + 3t = 0. We then solved this equation by factoring, which gave us two potential solutions: t = 0 and t = -3. However, we recognized the importance of checking for extraneous solutions, especially in rational equations where the denominator cannot be zero. By substituting the potential solutions back into the original equation, we found that t = 0 was an extraneous solution, as it resulted in division by zero. Therefore, we discarded t = 0 as a valid solution. The only remaining solution, t = -3, was verified to be a valid solution, as it satisfied the original equation without causing any undefined terms. Finally, we presented the solution set as {-3}, indicating that -3 is the only value of t that makes the equation true. This comprehensive process highlights the importance of each step in solving equations, from the initial identification and simplification to the final verification and presentation of the solution set. The ability to solve rational equations is a fundamental skill in algebra and is essential for tackling more complex mathematical problems. By following a systematic approach and paying careful attention to potential pitfalls, such as extraneous solutions, one can confidently solve these types of equations and arrive at the correct answer. This detailed guide provides a clear and thorough understanding of how to solve rational equations, equipping you with the knowledge and skills necessary to tackle similar problems in the future.