Solving The Equation X/3 + 2x/9 = 5x/6 - 1/18

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In this article, we will walk through the steps to solve the equation x3+2x9=5x6−118{ \frac{x}{3} + \frac{2x}{9} = \frac{5x}{6} - \frac{1}{18} }. This involves simplifying the equation by finding a common denominator, combining like terms, and isolating the variable x to find its value. Understanding how to solve such equations is crucial for various mathematical and real-world applications. Let's dive into the step-by-step solution.

Step 1: Finding the Least Common Denominator (LCD)

To effectively solve the equation, our initial step involves identifying the least common denominator (LCD) of the fractions present. The given equation is: x3+2x9=5x6−118{ \frac{x}{3} + \frac{2x}{9} = \frac{5x}{6} - \frac{1}{18} } We observe the denominators: 3, 9, 6, and 18. To find the LCD, we need to determine the smallest multiple that all these numbers can divide into. Let's list the multiples of each denominator:

  • Multiples of 3: 3, 6, 9, 12, 15, 18, 21, ...
  • Multiples of 9: 9, 18, 27, 36, ...
  • Multiples of 6: 6, 12, 18, 24, 30, ...
  • Multiples of 18: 18, 36, 54, ...

From the lists above, it's clear that the least common multiple (LCM) of 3, 9, 6, and 18 is 18. Therefore, the LCD for this equation is 18. Identifying the LCD is a critical step as it allows us to eliminate the fractions by multiplying each term in the equation by this common denominator. This simplifies the equation, making it easier to manipulate and solve. By converting each fraction to an equivalent fraction with the LCD as the denominator, we can combine like terms more efficiently and proceed toward isolating the variable x. The LCD acts as a bridge, allowing us to perform arithmetic operations uniformly across all terms, thus ensuring the equation remains balanced and the solution accurate. This process not only simplifies the equation at hand but also enhances our understanding of fractional arithmetic, a fundamental concept in algebra. Once we've established the LCD, we're well-prepared to move on to the next phase: clearing the fractions by multiplying each term by the LCD. This action will transform our equation into a more manageable form, setting the stage for further simplification and the ultimate solution. Understanding this foundational step is essential for anyone looking to master algebraic problem-solving.

Step 2: Clearing the Fractions

After identifying the least common denominator (LCD) as 18, the next crucial step is to eliminate the fractions from the equation. This is achieved by multiplying every term in the equation by the LCD. Our equation is: x3+2x9=5x6−118{ \frac{x}{3} + \frac{2x}{9} = \frac{5x}{6} - \frac{1}{18} } Multiplying each term by 18 gives us: 18⋅x3+18⋅2x9=18⋅5x6−18⋅118{ 18 \cdot \frac{x}{3} + 18 \cdot \frac{2x}{9} = 18 \cdot \frac{5x}{6} - 18 \cdot \frac{1}{18} } Now, we simplify each term:

  • 18â‹…x3=6x{ 18 \cdot \frac{x}{3} = 6x }
  • 18â‹…2x9=4x{ 18 \cdot \frac{2x}{9} = 4x }
  • 18â‹…5x6=15x{ 18 \cdot \frac{5x}{6} = 15x }
  • 18â‹…118=1{ 18 \cdot \frac{1}{18} = 1 }

Substituting these simplified terms back into the equation, we get: 6x+4x=15x−1{ 6x + 4x = 15x - 1 } This transformation is a pivotal moment in solving the equation because it converts a fractional equation into a simpler linear equation. The absence of fractions makes the equation much easier to work with, as we can now focus solely on combining like terms and isolating the variable x. By clearing the fractions, we've effectively removed a significant obstacle, paving the way for a more straightforward solution process. This step is not just about simplifying the equation; it's about making it accessible and manageable. It showcases the power of algebraic manipulation in transforming complex expressions into simpler forms, a fundamental skill in mathematics. Furthermore, this technique is widely applicable across various mathematical problems involving fractions, highlighting its importance in a broader context. With the fractions now cleared, we have a clear path forward. The next step involves combining like terms on each side of the equation, bringing us closer to isolating x and finding its value. The simplification achieved here underscores the elegance and efficiency of mathematical methods in solving problems.

Step 3: Combining Like Terms

Having cleared the fractions, our equation now stands as: 6x+4x=15x−1{ 6x + 4x = 15x - 1 } The next step involves combining like terms on each side of the equation to further simplify it. On the left side, we have two terms with the variable x: 6x and 4x. Combining these gives us: 6x+4x=10x{ 6x + 4x = 10x } So, the left side of the equation simplifies to 10x. The right side of the equation, 15x - 1, already has its terms simplified, with 15x being the variable term and -1 being the constant term. Thus, the equation now looks like this: 10x=15x−1{ 10x = 15x - 1 } This simplification is crucial because it reduces the number of terms we need to work with, making the equation more manageable. By combining like terms, we consolidate the variable terms and constant terms, which brings us closer to isolating x. This process is a fundamental aspect of algebraic manipulation, allowing us to rearrange and simplify equations in a systematic way. The goal here is to collect similar terms together, which helps in seeing the equation's structure more clearly. In this case, we've successfully combined the x terms on the left side, which sets the stage for the next step: moving all the x terms to one side of the equation. This step is a building block in the overall strategy of solving for x, highlighting the importance of each simplification we make. By reducing the complexity of the equation, we make it easier to apply further operations and ultimately find the value of x. Understanding the concept of like terms and how to combine them is essential for solving algebraic equations effectively. It's a skill that forms the basis for more advanced algebraic techniques and problem-solving strategies. With the equation simplified to 10x = 15x - 1, we are now well-positioned to isolate the variable x and determine its value.

Step 4: Isolating the Variable x

After combining like terms, our equation is: 10x=15x−1{ 10x = 15x - 1 } To isolate the variable x, we need to get all the x terms on one side of the equation and the constant terms on the other side. A common approach is to subtract 15x from both sides of the equation. This will move the x term from the right side to the left side: 10x−15x=15x−1−15x{ 10x - 15x = 15x - 1 - 15x } Simplifying both sides, we get: −5x=−1{ -5x = -1 } Now, to solve for x, we need to divide both sides of the equation by the coefficient of x, which is -5: −5x−5=−1−5{ \frac{-5x}{-5} = \frac{-1}{-5} } This simplifies to: x=15{ x = \frac{1}{5} } Isolating the variable is a critical step in solving any algebraic equation. It involves performing operations on both sides of the equation to get the variable by itself. In this case, we used subtraction and division to achieve this. The process of isolating x is a clear demonstration of the principle of maintaining balance in an equation; whatever operation we perform on one side, we must also perform on the other side to keep the equation true. This step-by-step approach ensures that we are systematically working towards the solution without altering the fundamental relationship expressed by the equation. The act of dividing by the coefficient of x is the final piece of the puzzle, allowing us to reveal the value of x. This process is not just about finding a number; it's about understanding the structure of the equation and how to manipulate it to uncover the unknown. The result, x = 1/5, is the solution to our original equation, and it represents the value that, when substituted back into the equation, will make both sides equal. This step solidifies the algebraic methodology we've employed and highlights the precision required in mathematical problem-solving. With x now isolated and its value determined, we've successfully solved the equation.

Step 5: Verifying the Solution

After finding a solution, it's always a good practice to verify it by substituting the value back into the original equation. This ensures that our solution is correct and that we haven't made any errors in our calculations. Our original equation was: x3+2x9=5x6−118{ \frac{x}{3} + \frac{2x}{9} = \frac{5x}{6} - \frac{1}{18} } We found that x=15{ x = \frac{1}{5} }. Now, let's substitute this value into the equation: 153+2(15)9=5(15)6−118{ \frac{\frac{1}{5}}{3} + \frac{2(\frac{1}{5})}{9} = \frac{5(\frac{1}{5})}{6} - \frac{1}{18} } Simplifying each term, we get: 115+245=16−118{ \frac{1}{15} + \frac{2}{45} = \frac{1}{6} - \frac{1}{18} } To compare both sides, we need to find a common denominator, which is 90. Converting each fraction to an equivalent fraction with a denominator of 90, we have: 690+490=1590−590{ \frac{6}{90} + \frac{4}{90} = \frac{15}{90} - \frac{5}{90} } Adding and subtracting the fractions, we get: 1090=1090{ \frac{10}{90} = \frac{10}{90} } Since both sides of the equation are equal, our solution x=15{ x = \frac{1}{5} } is verified to be correct.

Verification is a crucial step in the problem-solving process because it provides a check on our work and ensures the accuracy of our solution. By substituting the solution back into the original equation, we are essentially reversing the steps we took to solve it, which helps us catch any potential errors. This process not only confirms the solution but also reinforces our understanding of the equation and the algebraic manipulations we performed. In this case, the verification step clearly demonstrates that x=15{ x = \frac{1}{5} } satisfies the original equation, giving us confidence in our answer. Moreover, verification is a fundamental aspect of mathematical rigor and critical thinking. It teaches us the importance of not just finding an answer but also ensuring its validity. This practice is invaluable in mathematics and other fields, where accuracy and precision are paramount. The successful verification of our solution underscores the effectiveness of our problem-solving approach and highlights the importance of each step in the process, from identifying the LCD to isolating the variable. With the solution verified, we can confidently conclude that we have solved the equation correctly.

Conclusion

In conclusion, we have successfully solved the equation x3+2x9=5x6−118{ \frac{x}{3} + \frac{2x}{9} = \frac{5x}{6} - \frac{1}{18} } by systematically simplifying the equation, clearing fractions, combining like terms, isolating the variable x, and verifying the solution. The solution we found is x=15{ x = \frac{1}{5} }. This process demonstrates the importance of following a structured approach in solving algebraic equations. Each step, from finding the LCD to verifying the solution, plays a critical role in ensuring accuracy and understanding. Mastering these techniques is essential for success in algebra and other areas of mathematics. The ability to solve equations like this is not just a mathematical skill; it's a problem-solving skill that can be applied in various real-world situations. The process we've outlined here provides a framework for tackling similar problems and reinforces the fundamental principles of algebraic manipulation. By understanding these principles, we can confidently approach more complex equations and mathematical challenges. The journey from the initial equation to the final solution highlights the power of mathematical reasoning and the elegance of algebraic methods. The successful resolution of this problem is a testament to the effectiveness of a methodical and step-by-step approach, underscoring the value of precision and attention to detail in mathematics. With the solution verified and the process understood, we can apply these skills to future problems with confidence and competence.