Solving Linear Equations Step-by-Step Guide

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In the realm of mathematics, solving linear equations is a fundamental skill. Linear equations, characterized by a variable raised to the first power, appear in various real-world applications, from calculating finances to understanding physics. This article provides a comprehensive guide to solving the linear equation 0.13−0.01(x+1)=−0.03(4−x)0.13 - 0.01(x + 1) = -0.03(4 - x). We will dissect each step, ensuring clarity and understanding. Our goal is to not only find the solution but also to equip you with the knowledge to tackle similar equations with confidence. Remember, practice makes perfect when it comes to mastering mathematical concepts. So, let's embark on this journey of solving linear equations together. Understanding the underlying principles will empower you to navigate the world of algebra with ease and precision.

1. The Equation: 0.13 - 0.01(x + 1) = -0.03(4 - x)

To begin, let's restate the given equation: 0.13−0.01(x+1)=−0.03(4−x)0.13 - 0.01(x + 1) = -0.03(4 - x). This equation is a linear equation because the highest power of the variable 'x' is 1. Our mission is to isolate 'x' on one side of the equation to determine its value. To achieve this, we'll employ a series of algebraic manipulations, ensuring that each step maintains the equality. This process involves distributing, combining like terms, and using inverse operations. The beauty of algebra lies in its systematic approach, allowing us to unravel complex expressions into simple solutions. Keep in mind that each operation performed on one side of the equation must be mirrored on the other side to preserve the balance. This fundamental principle is the cornerstone of solving equations. Now, let's dive into the first step of simplifying this equation.

2. Distribute the Constants

The next step in solving our equation, 0.13−0.01(x+1)=−0.03(4−x)0.13 - 0.01(x + 1) = -0.03(4 - x), involves distributing the constants outside the parentheses. This means multiplying −0.01-0.01 by both terms inside the first parentheses (xx and 11) and multiplying −0.03-0.03 by both terms inside the second parentheses (44 and −x-x).

  • Distributing −0.01-0.01 on the left side gives us −0.01∗x=−0.01x-0.01 * x = -0.01x and −0.01∗1=−0.01-0.01 * 1 = -0.01. So, the left side of the equation becomes 0.13−0.01x−0.010.13 - 0.01x - 0.01.
  • Distributing −0.03-0.03 on the right side gives us −0.03∗4=−0.12-0.03 * 4 = -0.12 and −0.03∗−x=0.03x-0.03 * -x = 0.03x. Thus, the right side of the equation becomes −0.12+0.03x-0.12 + 0.03x.

After distribution, our equation now looks like this: 0.13−0.01x−0.01=−0.12+0.03x0.13 - 0.01x - 0.01 = -0.12 + 0.03x. This step is crucial because it eliminates the parentheses, making it easier to combine like terms and isolate the variable 'x'. Careful attention to signs during distribution is essential to avoid errors. With the constants distributed, we're one step closer to solving for 'x'.

3. Combine Like Terms

Following the distribution of constants in the equation 0.13−0.01x−0.01=−0.12+0.03x0.13 - 0.01x - 0.01 = -0.12 + 0.03x, our next objective is to combine like terms. Like terms are those that have the same variable raised to the same power. In this equation, we have constant terms (numbers without variables) and terms with the variable 'x'.

On the left side of the equation, we have two constant terms: 0.130.13 and −0.01-0.01. Combining these, we get 0.13−0.01=0.120.13 - 0.01 = 0.12. So, the left side simplifies to 0.12−0.01x0.12 - 0.01x.

The right side of the equation, −0.12+0.03x-0.12 + 0.03x, already has its terms simplified, with one constant term and one term with 'x'.

After combining like terms, our equation is now: 0.12−0.01x=−0.12+0.03x0.12 - 0.01x = -0.12 + 0.03x. This step streamlines the equation, making it more manageable for the subsequent steps. Combining like terms is a fundamental algebraic technique that simplifies expressions and equations, paving the way for isolating the variable. With the equation simplified, we can now proceed to gather the 'x' terms on one side and the constants on the other.

4. Isolate the Variable Terms

Now that we have simplified the equation to 0.12−0.01x=−0.12+0.03x0.12 - 0.01x = -0.12 + 0.03x, the next critical step is to isolate the variable terms. This involves moving all terms containing 'x' to one side of the equation and all constant terms to the other side. To achieve this, we'll employ the principle of inverse operations.

Let's start by adding 0.01x0.01x to both sides of the equation. This will eliminate the −0.01x-0.01x term on the left side. The equation becomes:

0.12−0.01x+0.01x=−0.12+0.03x+0.01x0.12 - 0.01x + 0.01x = -0.12 + 0.03x + 0.01x

Simplifying, we get:

0.12=−0.12+0.04x0.12 = -0.12 + 0.04x

Next, we need to isolate the term with 'x' further by adding 0.120.12 to both sides. This will eliminate the −0.12-0.12 term on the right side. The equation transforms into:

0.12+0.12=−0.12+0.12+0.04x0.12 + 0.12 = -0.12 + 0.12 + 0.04x

Simplifying, we have:

0.24=0.04x0.24 = 0.04x

At this stage, all the 'x' terms are on one side, and all the constant terms are on the other. This isolation is a pivotal step in solving for 'x'. Isolating the variable is a common strategy in algebra, allowing us to focus on the term containing the unknown and ultimately determine its value. With the variable term isolated, the final step involves solving for 'x' by dividing both sides by the coefficient of 'x'.

5. Solve for x

Having successfully isolated the variable term, we arrive at the equation 0.24=0.04x0.24 = 0.04x. The final step in solving for 'x' is to divide both sides of the equation by the coefficient of 'x', which in this case is 0.040.04. This will give us the value of 'x'.

Dividing both sides by 0.040.04, we get:

0.24/0.04=(0.04x)/0.040.24 / 0.04 = (0.04x) / 0.04

Performing the division, we find:

x=6x = 6

Therefore, the solution to the equation 0.13−0.01(x+1)=−0.03(4−x)0.13 - 0.01(x + 1) = -0.03(4 - x) is x=6x = 6. This value of 'x' is the unique solution that satisfies the equation. We have successfully navigated through the steps of distributing, combining like terms, isolating the variable, and finally solving for 'x'. This process exemplifies the systematic approach to solving linear equations. Finding the value of x marks the culmination of our algebraic journey, providing us with the answer we sought. Now, let's verify our solution to ensure its accuracy.

6. Verify the Solution

To ensure the accuracy of our solution, x=6x = 6, we must verify it by substituting this value back into the original equation: 0.13−0.01(x+1)=−0.03(4−x)0.13 - 0.01(x + 1) = -0.03(4 - x). This process of verification is a crucial step in problem-solving, as it helps to identify any potential errors made during the solution process. By substituting the value back into the original equation, we are essentially checking if the left-hand side (LHS) equals the right-hand side (RHS). If they are equal, our solution is correct; otherwise, we need to revisit our steps.

Substitute x=6x = 6 into the original equation:

0.13−0.01(6+1)=−0.03(4−6)0.13 - 0.01(6 + 1) = -0.03(4 - 6)

Simplify the equation step by step:

0.13−0.01(7)=−0.03(−2)0.13 - 0.01(7) = -0.03(-2)

0.13−0.07=0.060.13 - 0.07 = 0.06

0.06=0.060.06 = 0.06

Since the LHS equals the RHS, our solution x=6x = 6 is indeed correct. Verifying the solution provides us with confidence in our answer and reinforces the understanding of the equation's properties. This final check underscores the importance of thoroughness in mathematical problem-solving. With our solution verified, we can confidently present our answer.

7. Final Answer

After meticulously solving the linear equation 0.13−0.01(x+1)=−0.03(4−x)0.13 - 0.01(x + 1) = -0.03(4 - x) and verifying our solution, we have arrived at the final answer. Through a series of algebraic manipulations, including distributing constants, combining like terms, isolating the variable, and performing the division, we determined that the value of 'x' that satisfies the equation is 6.

Therefore, the final answer is:

x = 6

This solution represents the single point where the equation holds true. Our journey through this equation highlights the power of algebraic techniques in unraveling mathematical problems. From the initial equation to the final solution, each step was carefully executed and verified, ensuring the accuracy of our result. Presenting the final answer marks the culmination of our problem-solving process, providing a clear and concise resolution to the equation.

Conclusion

In this comprehensive guide, we have successfully navigated the process of solving the linear equation 0.13−0.01(x+1)=−0.03(4−x)0.13 - 0.01(x + 1) = -0.03(4 - x). From the initial distribution of constants to the final verification of the solution, each step was meticulously explained to provide a clear understanding of the underlying algebraic principles. We found that the solution to the equation is x=6x = 6, a single, unique value that satisfies the given condition.

This exercise underscores the importance of systematic problem-solving in mathematics. By breaking down a complex equation into manageable steps, we can effectively arrive at the correct answer. The techniques employed here, such as combining like terms, isolating variables, and verifying solutions, are fundamental to algebra and applicable to a wide range of mathematical problems.

Mastering these skills not only enhances our ability to solve equations but also deepens our understanding of mathematical relationships. The journey of solving this equation serves as a valuable lesson in algebraic thinking, empowering us to tackle future challenges with confidence and precision. Remember, the key to success in mathematics lies in practice and a thorough understanding of the fundamental concepts.

Therefore, the correct choice is A. x = 6