Solving For V A Step By Step Guide
Introduction
In this article, we will delve into the process of solving for the variable v in the given equation: $ rac{4v-5}{7} = 8 - rac{v+8}{6}$. This is a typical algebraic problem that requires us to manipulate the equation to isolate v on one side. We will go through each step meticulously, explaining the logic and the algebraic principles involved. Our goal is not just to find the solution but also to provide a clear and understandable guide that can be applied to similar problems. This detailed explanation aims to enhance your understanding of equation solving, making you more confident in tackling algebraic challenges.
Understanding the Equation
Before we begin solving, let's take a closer look at the equation itself. We have a linear equation with fractions, which might seem daunting at first. However, by breaking it down into smaller, manageable steps, we can simplify it. The equation $ rac{4v-5}{7} = 8 - rac{v+8}{6}$ involves fractions with expressions containing v in the numerators and constants in the denominators. To solve for v, we need to eliminate these fractions and rearrange the terms. The key to solving such equations lies in understanding the order of operations and the properties of equality. We'll start by clearing the fractions, which is a crucial step in simplifying the equation. This involves multiplying both sides of the equation by the least common multiple (LCM) of the denominators, a technique we'll discuss in detail in the next section.
Clearing the Fractions: A Step-by-Step Approach
The presence of fractions in an equation can often complicate the solving process. To eliminate these fractions, we employ a common strategy: multiplying both sides of the equation by the least common multiple (LCM) of the denominators. In our equation, $ rac4v-5}{7} = 8 - rac{v+8}{6}$, the denominators are 7 and 6. The LCM of 7 and 6 is 42. This means we will multiply both sides of the equation by 42. Multiplying by the LCM ensures that each fraction's denominator will divide evenly into 42, effectively eliminating the fractions. The process looks like this7} = 42 * (8 - rac{v+8}{6})$. Now, we distribute the 42 on both sides. On the left side, 42 divided by 7 is 6, so we have $6(4v - 5)$. On the right side, we distribute the 42 to both terms{6}$. Simplifying further, we get $336 - 7(v+8)$. This step is crucial because it transforms the equation from one with fractions to a simpler equation without fractions, making it easier to solve for v.
Distributing and Simplifying: Expanding the Equation
After clearing the fractions, our equation looks like this: $6(4v - 5) = 336 - 7(v+8)$. The next step is to distribute the numbers outside the parentheses to the terms inside. On the left side, we distribute the 6 across $4v$ and $-5$, resulting in $24v - 30$. On the right side, we distribute the -7 across v and 8, which gives us $-7v - 56$. The equation now becomes $24v - 30 = 336 - 7v - 56$. This step is vital because it expands the equation, removing the parentheses and making the terms more accessible. Next, we simplify the right side by combining the constant terms, 336 and -56. This gives us $280$. So, the equation is further simplified to $24v - 30 = 280 - 7v$. This simplified form is much easier to work with and brings us closer to isolating v.
Isolating the Variable: Bringing v Terms Together
Now that we have the simplified equation $24v - 30 = 280 - 7v$, our next goal is to isolate the variable v. This means we need to bring all the terms containing v to one side of the equation and all the constant terms to the other side. To do this, we can add $7v$ to both sides of the equation. This eliminates the $-7v$ term on the right side and adds it to the left side. The equation becomes $24v + 7v - 30 = 280$. Combining the v terms on the left side, we get $31v - 30 = 280$. This step is crucial in the process of isolating v, as it groups all the v terms together. Now, we need to move the constant term from the left side to the right side. We'll do this in the next step.
Isolating the Variable: Moving Constant Terms
Continuing our quest to isolate v, we now have the equation $31v - 30 = 280$. Our next step is to move the constant term, -30, from the left side to the right side. To do this, we add 30 to both sides of the equation. This eliminates the -30 on the left side and adds it to the constant term on the right side. The equation becomes $31v = 280 + 30$. Simplifying the right side, we get $31v = 310$. This step is a critical part of isolating v, as it separates the variable term from the constant terms. Now, we have a simple equation where the variable term is on one side and a constant on the other. The final step to solve for v is to divide both sides by the coefficient of v, which we will do in the next section.
Solving for v: The Final Step
We've reached the final stage of solving for v. Our equation is now in the form $31v = 310$. To find the value of v, we need to divide both sides of the equation by the coefficient of v, which is 31. This will isolate v on one side and give us its value. Dividing both sides by 31, we get $ rac{31v}{31} = rac{310}{31}$. Simplifying, we find that $v = 10$. This is the solution to our equation. We have successfully isolated v and determined its value. It's always a good practice to check our solution by plugging it back into the original equation, which we will do in the next section.
Checking the Solution: Ensuring Accuracy
To ensure the accuracy of our solution, it's crucial to check it by substituting the value we found for v back into the original equation. Our original equation was $ rac{4v-5}{7} = 8 - rac{v+8}{6}$, and we found that $v = 10$. Substituting v with 10, we get $ rac{4(10)-5}{7} = 8 - rac{10+8}{6}$. Now, we simplify both sides of the equation. On the left side, $4(10) - 5$ is $40 - 5$, which equals 35. So, the left side becomes $ rac{35}{7}$, which simplifies to 5. On the right side, $10 + 8$ is 18. So, we have $8 - rac{18}{6}$. $ rac{18}{6}$ simplifies to 3, so the right side becomes $8 - 3$, which equals 5. Thus, we have $5 = 5$, which is a true statement. This confirms that our solution, $v = 10$, is correct. Checking our solution is a vital step in problem-solving, as it helps us catch any potential errors and ensures that we have the correct answer.
Conclusion: Mastering Algebraic Equations
In conclusion, we have successfully solved for v in the equation $ rac{4v-5}{7} = 8 - rac{v+8}{6}$. We systematically worked through the steps, starting with clearing the fractions by multiplying both sides by the LCM of the denominators. We then distributed and simplified the equation, isolated the variable terms, and finally solved for v. We found that $v = 10$. To ensure the accuracy of our solution, we checked it by substituting it back into the original equation and confirmed that both sides were equal. This process demonstrates the importance of methodical problem-solving in algebra. By understanding each step and the underlying algebraic principles, you can confidently tackle similar equations. Remember, practice is key to mastering these skills. Keep practicing, and you'll become more proficient in solving algebraic equations. This comprehensive guide should serve as a valuable resource for students and anyone looking to enhance their understanding of algebra.