Solving For V In The Equation -7/4v - 1/3 = 2/3v - 3

In this article, we will walk through the step-by-step process of solving for v in the given linear equation:

$-\frac{7}{4}v - \frac{1}{3} = \frac{2}{3}v - 3$

Understanding how to solve such equations is a fundamental skill in algebra, essential for various mathematical and real-world applications. We'll break down each step to make the solution clear and easy to follow. This comprehensive guide ensures that anyone, whether a student learning algebra or someone brushing up on their math skills, can confidently tackle similar problems. We aim to simplify each stage of the process, providing a detailed explanation that aids in comprehension and application. By the end of this guide, you will not only know how to solve this specific equation but also understand the principles behind solving linear equations in general.

Step 1: Eliminate Fractions

To make the equation easier to work with, the first step is to eliminate the fractions. We achieve this by multiplying every term in the equation by the least common multiple (LCM) of the denominators, which are 4 and 3. The LCM of 4 and 3 is 12. Multiplying each term by 12 will clear the fractions and give us a more straightforward equation to solve. This approach is a standard technique in algebra to simplify equations involving fractions. It allows us to deal with whole numbers, which are often easier to manipulate. This step is crucial for maintaining the balance of the equation while transforming it into a simpler form. The elimination of fractions is not just about making the numbers look cleaner; it’s about setting up the equation for the subsequent steps in the solving process.

Multiplying both sides of the equation by 12, we get:

$12(-\frac{7}{4}v - \frac{1}{3}) = 12(\frac{2}{3}v - 3)$

Step 2: Distribute the Multiplication

Next, we distribute the multiplication across the terms inside the parentheses on both sides of the equation. This involves multiplying 12 by each term inside the parentheses. Distributing the multiplication is a key step in expanding the equation and preparing it for further simplification. It ensures that every term is correctly accounted for and that the balance of the equation is maintained. This process transforms the equation from a condensed form into a more expanded form, making it easier to combine like terms and isolate the variable. The careful application of the distributive property is vital for achieving an accurate solution. Missteps in this stage can lead to incorrect results, underscoring the importance of precision in algebraic manipulation.

This gives us:

$12(-\frac{7}{4}v) - 12(\frac{1}{3}) = 12(\frac{2}{3}v) - 12(3)$

Simplifying each term:

$-21v - 4 = 8v - 36$

Step 3: Collect Like Terms

The next step involves collecting like terms. This means moving all the terms containing v to one side of the equation and all the constant terms to the other side. This is a fundamental technique in solving algebraic equations, as it isolates the variable we are trying to find. To do this, we add or subtract terms from both sides of the equation, ensuring that the equation remains balanced. The goal is to consolidate the variable terms and the constant terms, which simplifies the equation and brings us closer to the solution. Collecting like terms is not just about grouping similar elements; it’s about reorganizing the equation in a way that makes the variable's value more apparent.

Let’s add 21v to both sides:

$-21v - 4 + 21v = 8v - 36 + 21v$

$-4 = 29v - 36$

Now, add 36 to both sides:

$-4 + 36 = 29v - 36 + 36$

$32 = 29v$

Step 4: Isolate v

To isolate v, we need to divide both sides of the equation by the coefficient of v, which is 29. Isolating the variable is the ultimate goal in solving an equation, as it directly reveals the variable's value. Division is the inverse operation of multiplication, and by dividing both sides by the same number, we maintain the balance of the equation. This step is crucial for finding the specific value that satisfies the original equation. It is the culmination of the previous steps, where we simplified and rearranged the equation to make this final isolation possible. The act of isolating the variable is not just a mechanical step; it’s the point where the solution becomes clear.

Dividing both sides by 29:

$\frac{32}{29} = \frac{29v}{29}$

$v = \frac{32}{29}$

Step 5: Simplify the Answer

The fraction $\frac{32}{29}$ is already in its simplest form since 32 and 29 have no common factors other than 1. Therefore, the solution is: Simplifying the answer is the final touch in the problem-solving process. It involves reducing the solution to its most basic form, making it easier to understand and use. In the case of fractions, this means ensuring that the numerator and denominator have no common factors. Simplification is not just about aesthetics; it’s about presenting the solution in the clearest and most concise manner. A simplified answer is easier to compare with other results and less prone to errors in subsequent calculations. This final step ensures that the solution is both accurate and practical.

$v = \frac{32}{29}$

So, the solution to the equation $-\frac{7}{4}v - \frac{1}{3} = \frac{2}{3}v - 3$ is $v = \frac{32}{29}$. This is the final, simplified answer.

Conclusion

In this article, we have demonstrated a detailed, step-by-step solution for finding v in the equation $-\frac{7}{4}v - \frac{1}{3} = \frac{2}{3}v - 3$. We began by eliminating fractions to simplify the equation, followed by distributing multiplication, collecting like terms, and finally, isolating v to find its value. The process of solving linear equations involves several key algebraic techniques, including the use of the distributive property, combining like terms, and performing inverse operations to isolate the variable. Each step is crucial for maintaining the balance of the equation and arriving at the correct solution. Understanding these techniques is essential for success in algebra and related fields. The ability to solve linear equations is a fundamental skill that underpins more advanced mathematical concepts and has practical applications in various disciplines. By mastering this skill, one can confidently tackle a wide range of problems in mathematics and beyond.

The final solution, $v = \frac{32}{29}$, represents the value of v that satisfies the original equation. This comprehensive approach ensures clarity and understanding for anyone learning or reviewing algebra.