Solving Linear Equations A Comprehensive Guide To (y-6)/4 - (y-4)/6 = 1 - Y/10

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In the realm of mathematics, solving equations is a fundamental skill. Linear equations, in particular, form the bedrock of many mathematical concepts and real-world applications. This article delves into the step-by-step solution of the linear equation xxv. y−64−y−46=1−y10\frac{y-6}{4}-\frac{y-4}{6}=1-\frac{y}{10}, providing a comprehensive guide to understanding the underlying principles and techniques involved.

Understanding Linear Equations

At its core, a linear equation is an algebraic equation in which the highest power of the variable is one. These equations can be represented in the general form of ax + b = c, where a, b, and c are constants, and x is the variable. Linear equations are characterized by their straight-line graphs when plotted on a coordinate plane, hence the name “linear.” Mastering the art of solving linear equations is crucial for various mathematical disciplines, including algebra, calculus, and differential equations, as well as practical applications in physics, engineering, economics, and computer science. The ability to manipulate these equations, isolate variables, and find solutions forms the foundation for more advanced mathematical problem-solving. Linear equations also serve as building blocks for more complex mathematical models and algorithms, making their understanding indispensable for anyone pursuing a career in STEM fields. In everyday life, linear equations are used to model simple relationships and make predictions, such as calculating costs, determining distances, and budgeting expenses. The simplicity and versatility of linear equations make them an essential tool in both academic and practical settings, underlining the importance of developing proficiency in their solution.

Problem Statement: XXV. (y-6)/4 - (y-4)/6 = 1 - y/10

The specific equation we aim to solve is: y−64−y−46=1−y10\frac{y-6}{4}-\frac{y-4}{6}=1-\frac{y}{10}. This equation involves fractions, which adds a layer of complexity. However, by systematically applying algebraic principles, we can simplify the equation and isolate the variable y to find its value. Before diving into the solution, it’s important to recognize the structure of the equation. We have fractions with y in the numerators and constants in the denominators, as well as constant terms and a term with y on the right side of the equation. Our goal is to eliminate the fractions, combine like terms, and isolate y on one side of the equation. This process involves several steps, including finding the least common multiple (LCM) of the denominators, multiplying both sides of the equation by the LCM, distributing terms, and simplifying the resulting equation. Each step is crucial to ensure the accuracy of the final solution. By breaking down the problem into manageable steps, we can approach the equation methodically and avoid common errors. The equation is a classic example of a linear equation with fractional coefficients, requiring careful application of algebraic techniques to solve. Successfully solving this equation will reinforce your understanding of linear equations and build your confidence in tackling more complex problems. It serves as a valuable exercise in algebraic manipulation and problem-solving skills.

Step 1: Finding the Least Common Multiple (LCM)

The first crucial step in solving this equation is to eliminate the fractions. To do this, we need to find the least common multiple (LCM) of the denominators: 4, 6, and 10. The LCM is the smallest number that is a multiple of all the denominators. Here’s how to find the LCM:

  • List the multiples of each denominator:
    • Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, ...
    • Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ...
    • Multiples of 10: 10, 20, 30, 40, 50, 60, ...
  • Identify the smallest common multiple: The smallest number that appears in all three lists is 60. Therefore, the LCM of 4, 6, and 10 is 60. Finding the LCM is a fundamental skill in algebra and arithmetic, essential for simplifying expressions and solving equations involving fractions. The LCM method ensures that we choose the smallest possible number that will clear the fractions, making the subsequent calculations easier and less prone to errors. In more complex problems, prime factorization may be used to find the LCM, but in this case, listing multiples is a straightforward approach. Once the LCM is determined, we can proceed to the next step, which involves multiplying both sides of the equation by the LCM to eliminate the fractions. This step is crucial in transforming the equation into a more manageable form, paving the way for isolating the variable and finding the solution. The correct calculation of the LCM is paramount, as an incorrect LCM can lead to incorrect solutions.

Step 2: Multiplying Both Sides by the LCM

Now that we have the LCM, which is 60, we multiply both sides of the equation by 60 to eliminate the fractions. This step is crucial because it transforms the equation into a more manageable form without fractions. The original equation is: y−64−y−46=1−y10\frac{y-6}{4}-\frac{y-4}{6}=1-\frac{y}{10}.

Multiplying both sides by 60 gives us:

60∗(y−64−y−46)=60∗(1−y10)60 * (\frac{y-6}{4}-\frac{y-4}{6}) = 60 * (1-\frac{y}{10})

Now, distribute the 60 to each term on both sides of the equation:

60∗y−64−60∗y−46=60∗1−60∗y1060 * \frac{y-6}{4} - 60 * \frac{y-4}{6} = 60 * 1 - 60 * \frac{y}{10}

Next, simplify each term by dividing the numerators by the denominators:

15(y−6)−10(y−4)=60−6y15(y-6) - 10(y-4) = 60 - 6y

This step is critical in simplifying the equation and preparing it for further algebraic manipulations. By multiplying both sides by the LCM, we effectively clear the denominators, making the equation easier to solve. The distribution of 60 across the terms ensures that each term is correctly scaled, maintaining the balance of the equation. The resulting equation, 15(y−6)−10(y−4)=60−6y15(y-6) - 10(y-4) = 60 - 6y, is a linear equation without fractions, which is much simpler to work with. This transformation is a standard technique in solving equations involving fractions and is applicable in various algebraic contexts. The careful application of the distributive property and accurate simplification of the terms are essential to avoid errors and ensure the correct progression towards the solution. The next step will involve expanding the terms and further simplifying the equation.

Step 3: Distributing and Simplifying

After multiplying both sides of the equation by the LCM, we have: 15(y−6)−10(y−4)=60−6y15(y-6) - 10(y-4) = 60 - 6y. The next step is to distribute the numbers outside the parentheses to the terms inside the parentheses. This process involves multiplying 15 by both y and -6, and -10 by both y and -4.

Distributing the terms, we get:

15y−90−10y+40=60−6y15y - 90 - 10y + 40 = 60 - 6y

Now, we combine like terms on the left side of the equation. Like terms are terms that have the same variable raised to the same power. In this case, we combine the y terms (15y and -10y) and the constant terms (-90 and +40):

(15y−10y)+(−90+40)=60−6y(15y - 10y) + (-90 + 40) = 60 - 6y

Simplifying the terms, we have:

5y−50=60−6y5y - 50 = 60 - 6y

This step is essential for simplifying the equation and bringing it closer to the standard form of a linear equation. By distributing the terms and combining like terms, we reduce the complexity of the equation and make it easier to isolate the variable y. The accuracy in distributing and combining like terms is crucial to avoid errors and maintain the balance of the equation. The process involves basic arithmetic operations, such as multiplication and addition/subtraction, and a careful attention to signs. The resulting equation, 5y−50=60−6y5y - 50 = 60 - 6y, is a simplified linear equation that is ready for the next step, which will involve moving the variable terms to one side and the constant terms to the other side. This simplification is a fundamental technique in solving algebraic equations and is widely applicable in various mathematical contexts.

Step 4: Isolating the Variable

Now that we have the simplified equation 5y−50=60−6y5y - 50 = 60 - 6y, our goal is to isolate the variable y on one side of the equation. This involves moving all the terms containing y to one side and all the constant terms to the other side. Let's start by adding 6y to both sides of the equation:

5y−50+6y=60−6y+6y5y - 50 + 6y = 60 - 6y + 6y

This simplifies to:

11y−50=6011y - 50 = 60

Next, we add 50 to both sides of the equation to move the constant term to the right side:

11y−50+50=60+5011y - 50 + 50 = 60 + 50

This simplifies to:

11y=11011y = 110

This step is a critical part of solving linear equations, as it brings us closer to finding the value of the variable. By performing the same operations on both sides of the equation, we maintain the balance and ensure that the equality holds. Adding 6y to both sides eliminates the -6y term on the right side, and adding 50 to both sides eliminates the -50 term on the left side. The resulting equation, 11y=11011y = 110, is a simple linear equation with the variable y isolated on one side. This form is easy to solve by dividing both sides by the coefficient of y. The process of isolating the variable is a fundamental technique in algebra and is used extensively in solving various types of equations. The key is to perform the operations in a systematic manner, ensuring that the balance of the equation is maintained at each step. The next step will be to divide both sides by the coefficient of y to find the value of y.

Step 5: Solving for y

After isolating the variable, we have the equation 11y=11011y = 110. To solve for y, we need to divide both sides of the equation by the coefficient of y, which is 11.

Dividing both sides by 11, we get:

11y11=11011\frac{11y}{11} = \frac{110}{11}

Simplifying, we find:

y=10y = 10

Therefore, the solution to the equation y−64−y−46=1−y10\frac{y-6}{4}-\frac{y-4}{6}=1-\frac{y}{10} is y = 10. This step is the final step in solving the equation and provides the value of the unknown variable. Dividing both sides by the coefficient of y isolates y and gives its numerical value. The simplicity of this step underscores the importance of the preceding steps, where the equation was systematically simplified and rearranged to reach this point. The result, y = 10, is the solution that satisfies the original equation. It is good practice to check the solution by substituting it back into the original equation to verify its correctness. This final step completes the process of solving the linear equation and demonstrates the power of algebraic manipulation in finding solutions to mathematical problems. The value of y = 10 can be used in various applications where this equation may arise, highlighting the practical significance of solving linear equations.

Conclusion

In conclusion, the solution to the equation y−64−y−46=1−y10\frac{y-6}{4}-\frac{y-4}{6}=1-\frac{y}{10} is y = 10. This article has provided a detailed, step-by-step guide to solving this linear equation, covering essential techniques such as finding the least common multiple (LCM), multiplying both sides by the LCM, distributing and simplifying terms, isolating the variable, and finally, solving for the variable. Each step is crucial in the process, and understanding the underlying principles ensures accurate and efficient problem-solving. Linear equations are a fundamental part of mathematics, and mastering their solution is essential for various mathematical disciplines and real-world applications. The skills acquired in solving this equation can be applied to a wide range of problems in algebra, calculus, physics, engineering, and other fields. The systematic approach presented in this article emphasizes the importance of careful and methodical problem-solving, which is a valuable skill in any academic or professional pursuit. By breaking down the problem into manageable steps and applying the appropriate algebraic techniques, complex equations can be solved effectively. This comprehensive guide serves as a valuable resource for students and anyone looking to enhance their understanding of linear equations and algebraic problem-solving. The process of solving equations not only yields solutions but also enhances analytical and logical reasoning skills, which are crucial for success in various domains.