Finding The Number To Add To The Additive Inverse Of -6/5 To Get 13/7
Introduction
In the realm of mathematics, understanding concepts like additive inverses and fractions is crucial for solving various problems. This article delves into a specific problem that combines these concepts. We aim to determine what number should be added to the additive inverse of -6/5 to obtain 13/7. This exploration will not only enhance your problem-solving skills but also solidify your understanding of fundamental mathematical principles. Let's embark on this journey of discovery together, breaking down each step and ensuring clarity along the way.
Understanding Additive Inverses
To effectively tackle the problem, a solid grasp of additive inverses is essential. An additive inverse, also known as the opposite, is a number that, when added to the original number, results in a sum of zero. For example, the additive inverse of 5 is -5 because 5 + (-5) = 0. Similarly, the additive inverse of -3 is 3 because -3 + 3 = 0. This concept is fundamental in various mathematical operations and is particularly important when dealing with fractions and algebraic equations. Understanding additive inverses allows us to manipulate equations and simplify expressions, making complex problems more manageable. In this context, we'll apply this knowledge to find the additive inverse of a fraction, which is a crucial step in solving the main problem.
Finding the Additive Inverse of -6/5
Now, let's apply the concept of additive inverses to the fraction -6/5. The additive inverse of -6/5 is the number that, when added to -6/5, equals zero. To find this number, we simply change the sign of the fraction. Therefore, the additive inverse of -6/5 is 6/5. This is because (-6/5) + (6/5) = 0. This step is pivotal as it sets the foundation for the rest of the problem. By correctly identifying the additive inverse, we can proceed to determine what number needs to be added to it to reach the target value of 13/7. The ability to quickly and accurately find additive inverses is a valuable skill in mathematics, and mastering it will significantly aid in solving more complex problems.
Setting Up the Equation
With the additive inverse of -6/5 determined to be 6/5, we can now formulate an equation to solve the problem. The question asks: what number should be added to 6/5 to get 13/7? Let's represent the unknown number by the variable x. The problem can then be translated into the following equation:
6/5 + x = 13/7
This equation is the core of our problem, and solving it will reveal the answer. Setting up the equation correctly is crucial because it ensures that we are addressing the problem accurately. The equation clearly states the relationship between the additive inverse (6/5), the unknown number (x), and the target value (13/7). From this point forward, the task is to isolate x and find its value, which will tell us the number that needs to be added to 6/5 to obtain 13/7. The next step involves using algebraic techniques to solve for x, a process that requires careful manipulation of the equation.
Solving for the Unknown
To solve the equation 6/5 + x = 13/7 for x, we need to isolate x on one side of the equation. This can be achieved by subtracting 6/5 from both sides of the equation. Doing so maintains the equality and allows us to gradually reveal the value of x. The steps are as follows:
6/5 + x - 6/5 = 13/7 - 6/5
This simplifies to:
x = 13/7 - 6/5
Now, we need to subtract the fractions 6/5 from 13/7. To do this, we must find a common denominator. The least common multiple (LCM) of 7 and 5 is 35. Therefore, we will convert both fractions to have a denominator of 35. This involves multiplying the numerator and denominator of each fraction by the appropriate factor. Once the fractions have a common denominator, we can subtract the numerators and keep the common denominator. This process is a fundamental skill in fraction arithmetic and is essential for solving this type of equation.
Finding a Common Denominator
Before we can subtract the fractions, we need to find a common denominator for 13/7 and 6/5. As mentioned earlier, the least common multiple (LCM) of 7 and 5 is 35. This means we will convert both fractions to have a denominator of 35. To convert 13/7 to an equivalent fraction with a denominator of 35, we multiply both the numerator and the denominator by 5:
(13/7) * (5/5) = 65/35
Similarly, to convert 6/5 to an equivalent fraction with a denominator of 35, we multiply both the numerator and the denominator by 7:
(6/5) * (7/7) = 42/35
Now that both fractions have the same denominator, we can proceed with the subtraction. This step is crucial because subtracting fractions with different denominators can lead to incorrect results. By finding a common denominator, we ensure that we are comparing and subtracting equivalent parts, leading to an accurate solution. The next step involves performing the subtraction and simplifying the result.
Subtracting the Fractions
Now that we have the fractions with a common denominator, we can subtract them. Our equation is:
x = 65/35 - 42/35
To subtract fractions with the same denominator, we subtract the numerators and keep the denominator the same:
x = (65 - 42) / 35
This simplifies to:
x = 23/35
Therefore, the number that should be added to 6/5 to get 13/7 is 23/35. This result is the solution to our equation and answers the original question. It is crucial to double-check the solution to ensure its accuracy, especially in mathematical problems. In the next section, we will verify this solution to confirm that it is indeed correct.
Verification
To ensure the accuracy of our solution, we will now verify that adding 23/35 to 6/5 indeed results in 13/7. This step is crucial in problem-solving as it confirms that our calculations are correct and that we have arrived at the right answer. The verification process involves substituting the value we found for x back into the original equation and checking if the equation holds true.
Adding 23/35 to 6/5
We need to check if 6/5 + 23/35 equals 13/7. First, we need to find a common denominator for 6/5 and 23/35. The least common multiple (LCM) of 5 and 35 is 35. So, we convert 6/5 to an equivalent fraction with a denominator of 35:
(6/5) * (7/7) = 42/35
Now we can add the fractions:
42/35 + 23/35 = (42 + 23) / 35
This simplifies to:
65/35
Simplifying the Result
We have 65/35, and we need to check if this is equal to 13/7. To simplify 65/35, we can divide both the numerator and the denominator by their greatest common divisor (GCD), which is 5:
(65/5) / (35/5) = 13/7
Since 65/35 simplifies to 13/7, our verification is successful. This confirms that our solution, x = 23/35, is correct. The verification process not only validates our answer but also reinforces our understanding of the problem and the steps taken to solve it. It is a valuable practice in mathematics and ensures that we have confidence in our results.
Conclusion
In conclusion, we have successfully determined that the number that should be added to the additive inverse of -6/5 to get 13/7 is 23/35. This problem required a solid understanding of additive inverses, fraction arithmetic, and algebraic equation-solving techniques. By breaking down the problem into smaller, manageable steps, we were able to systematically arrive at the solution. We started by identifying the additive inverse of -6/5, which is 6/5. Then, we set up an equation to represent the problem: 6/5 + x = 13/7. We solved for x by finding a common denominator, subtracting the fractions, and simplifying the result. Finally, we verified our solution by adding 23/35 to 6/5 and confirming that it indeed equals 13/7.
This exercise highlights the importance of mastering fundamental mathematical concepts and applying them to solve problems. The ability to manipulate fractions, understand additive inverses, and solve equations are essential skills in mathematics. By practicing and reinforcing these skills, one can confidently tackle more complex problems and excel in mathematics. The process of problem-solving, from understanding the question to verifying the solution, is a valuable skill that extends beyond mathematics and into various aspects of life. Therefore, continuous learning and practice are key to success in mathematics and beyond.
