Solve The Equation (6x+1)/3 + 1 = (x-3)/6. Solve The Equation 5x - 2(2x-7) = 2(3x-1) + 7/2 And Verify The Result.

Introduction

This article provides a detailed walkthrough on solving two linear equations, Q2 and Q3, offering step-by-step solutions and verification methods. Understanding how to solve linear equations is a fundamental skill in mathematics, crucial for various applications in science, engineering, and everyday problem-solving. We will dissect each equation, explain the algebraic manipulations involved, and demonstrate how to check the accuracy of our solutions. Whether you are a student looking to improve your algebra skills or someone seeking a refresher on the basics, this guide aims to provide clarity and confidence in tackling linear equations.

Q2: Solve ${ \frac{6x+1}{3} + 1 = \frac{x-3}{6} }$

Solving linear equations often requires a strategic approach to isolate the variable. In this section, we will meticulously solve the equation ${ \frac{6x+1}{3} + 1 = \frac{x-3}{6} }$, ensuring every step is clear and understandable. The initial challenge with this equation is the presence of fractions, which can complicate the solving process. Our first goal will be to eliminate these fractions by finding a common denominator and multiplying both sides of the equation by it. This clears the fractions and simplifies the equation, making it easier to manage. Next, we'll distribute any coefficients, combine like terms, and strategically rearrange the equation so that all terms involving x are on one side and constant terms are on the other. This process involves adding or subtracting terms from both sides of the equation, maintaining the balance and equality. Once we have all the x terms on one side and the constants on the other, we can isolate x by dividing both sides by the coefficient of x. This will give us the solution for x. To ensure accuracy, we'll substitute this solution back into the original equation to verify that both sides of the equation are equal. This verification step is crucial as it confirms that our solution is correct and that no errors were made during the solving process. By following this systematic approach, we can confidently solve linear equations of this type. The steps involved, from clearing fractions to isolating the variable and verifying the solution, form the backbone of linear equation solving techniques. Understanding and mastering these steps will equip you with the skills needed to tackle a wide range of algebraic problems.

Step 1: Eliminate Fractions

The first step in solving this linear equation is to eliminate the fractions. To do this, we need to find the least common multiple (LCM) of the denominators, which are 3 and 6. The LCM of 3 and 6 is 6. We then multiply both sides of the equation by 6:

${ 6 \left( \frac{6x+1}{3} + 1 \right) = 6 \left( \frac{x-3}{6} \right) }$

Step 2: Distribute and Simplify

Next, we distribute the 6 on both sides of the equation:

${ 6 \cdot \frac{6x+1}{3} + 6 \cdot 1 = 6 \cdot \frac{x-3}{6} }$

Simplifying this gives us:

${ 2(6x+1) + 6 = x - 3 }$

Further distribution yields:

${ 12x + 2 + 6 = x - 3 }$

Combining like terms on the left side, we get:

${ 12x + 8 = x - 3 }$

Step 3: Isolate the Variable

To isolate the variable x, we first subtract x from both sides:

${ 12x - x + 8 = x - x - 3 }$

${ 11x + 8 = -3 }$

Then, we subtract 8 from both sides:

${ 11x + 8 - 8 = -3 - 8 }$

${ 11x = -11 }$

Step 4: Solve for x

Finally, we divide both sides by 11 to solve for x:

${ \frac{11x}{11} = \frac{-11}{11} }$

${ x = -1 }$

Thus, the solution to the equation is x = -1.

Q3: Solve ${ 5x - 2(2x-7) = 2(3x-1) + \frac{7}{2} }$ and check your results.

Solving this linear equation, ${ 5x - 2(2x-7) = 2(3x-1) + \frac{7}{2} }$, involves several key steps that are crucial for obtaining the correct solution and verifying its accuracy. The initial part of solving this equation requires careful attention to distribution and simplification. We must distribute the coefficients across the parentheses on both sides of the equation. This means multiplying -2 by both terms inside the first set of parentheses and multiplying 2 by both terms inside the second set. Accurate distribution is essential because any error here will propagate through the rest of the solution. Once the distribution is complete, we simplify each side of the equation by combining like terms. This involves adding or subtracting the coefficients of the x terms and the constant terms separately. Simplifying each side makes the equation more manageable and prepares it for the next steps. To further isolate the variable x, we need to move all x terms to one side of the equation and all constant terms to the other side. This is achieved by adding or subtracting terms from both sides of the equation, maintaining the balance. The goal is to have an equation in the form of ax = b, where a and b are constants. Once we have the equation in this form, we can solve for x by dividing both sides by the coefficient a. This gives us the numerical value of x that satisfies the equation. However, our work isn't complete until we verify the solution. Verification is a critical step in solving equations. We substitute the solution we found back into the original equation. If both sides of the equation are equal after the substitution, then our solution is correct. If the sides are not equal, it indicates that an error was made during the solving process, and we need to revisit our steps to find and correct the mistake. By following these steps meticulously, we can confidently solve linear equations and ensure the accuracy of our results. This process reinforces our understanding of algebraic manipulations and enhances our problem-solving skills.

Step 1: Distribute

The first step is to distribute the constants outside the parentheses:

${ 5x - 2(2x-7) = 2(3x-1) + \frac{7}{2} }$

${ 5x - 4x + 14 = 6x - 2 + \frac{7}{2} }$

Step 2: Combine Like Terms

Next, combine like terms on both sides of the equation:

${ x + 14 = 6x - 2 + \frac{7}{2} }$

To combine the constants on the right side, we need a common denominator. Convert -2 to a fraction with a denominator of 2:

${ -2 = \frac{-4}{2} }$

Now we can combine the constants:

${ x + 14 = 6x + \frac{-4}{2} + \frac{7}{2} }$

${ x + 14 = 6x + \frac{3}{2} }$

Step 3: Isolate the Variable

To isolate the variable x, subtract x from both sides:

${ x - x + 14 = 6x - x + \frac{3}{2} }$

${ 14 = 5x + \frac{3}{2} }$

Now, subtract ${ \frac{3}{2} }$ from both sides:

${ 14 - \frac{3}{2} = 5x + \frac{3}{2} - \frac{3}{2} }$

Convert 14 to a fraction with a denominator of 2:

${ 14 = \frac{28}{2} }$

So the equation becomes:

${ \frac{28}{2} - \frac{3}{2} = 5x }$

${ \frac{25}{2} = 5x }$

Step 4: Solve for x

To solve for x, divide both sides by 5:

${ \frac{\frac{25}{2}}{5} = \frac{5x}{5} }$

${ \frac{25}{2} \cdot \frac{1}{5} = x }$

${ x = \frac{5}{2} }$

Thus, the solution to the equation is x = 5/2.

Step 5: Check the Solution

To check the solution, substitute x = 5/2 back into the original equation:

${ 5\left(\frac{5}{2}\right) - 2\left(2\left(\frac{5}{2}\right)-7\right) = 2\left(3\left(\frac{5}{2}\right)-1\right) + \frac{7}{2} }$

${ \frac{25}{2} - 2(5-7) = 2\left(\frac{15}{2}-1\right) + \frac{7}{2} }$

${ \frac{25}{2} - 2(-2) = 2\left(\frac{15}{2}-\frac{2}{2}\right) + \frac{7}{2} }$

${ \frac{25}{2} + 4 = 2\left(\frac{13}{2}\right) + \frac{7}{2} }$

Convert 4 to a fraction with a denominator of 2:

${ 4 = \frac{8}{2} }$

So the equation becomes:

${ \frac{25}{2} + \frac{8}{2} = 13 + \frac{7}{2} }$

${ \frac{33}{2} = \frac{26}{2} + \frac{7}{2} }$

${ \frac{33}{2} = \frac{33}{2} }$

Since both sides of the equation are equal, the solution x = 5/2 is correct.

Conclusion

In this article, we have thoroughly explored the solutions to two linear equations, Q2 and Q3. For Q2, we meticulously solved ${ \frac{6x+1}{3} + 1 = \frac{x-3}{6} }$ and determined the solution to be x = -1. Our step-by-step approach involved eliminating fractions, distributing constants, combining like terms, isolating the variable, and finally, solving for x. Each of these steps is crucial in solving linear equations accurately. Similarly, for Q3, we tackled the equation ${ 5x - 2(2x-7) = 2(3x-1) + \frac{7}{2} }$. The solution we found was x = 5/2. To ensure the correctness of our solution, we performed a verification step, substituting x = 5/2 back into the original equation and confirming that both sides of the equation were equal. This verification process is a vital practice in algebra, helping to catch any potential errors made during the solving process. By mastering the techniques demonstrated in this article, such as clearing fractions, distributing and simplifying expressions, and isolating variables, you can build a strong foundation in algebra. These skills are not only essential for academic success in mathematics but also applicable in various real-world scenarios. Whether you are balancing a budget, calculating measurements, or solving scientific problems, the ability to confidently solve linear equations is a valuable asset. We encourage you to practice these methods with other linear equations to further solidify your understanding and skills. Remember, the key to mastering algebra is consistent practice and a systematic approach to problem-solving. Linear equations are just the beginning, but they form a critical stepping stone to more advanced mathematical concepts. Keep practicing, and you will continue to improve your algebraic abilities.