Solving Inverse Variation Problems Finding Y When X Equals 2

In the realm of mathematical relationships, inverse variation plays a crucial role in describing how two variables interact. Inverse variation occurs when one variable decreases as the other increases, and vice versa, maintaining a constant product. This concept is fundamental in various scientific and engineering applications, allowing us to model real-world phenomena accurately. To delve deeper into inverse variation, let's consider a specific scenario: If $y$ varies inversely with the square of $x$, and $y=26$ when $x=4$, find $y$ when $x=2$. This problem encapsulates the essence of inverse variation and provides a practical context for understanding its applications. The problem presents an interesting challenge that requires a clear understanding of inverse variation and the ability to translate it into a mathematical equation. By carefully analyzing the given information and applying the principles of inverse variation, we can successfully determine the value of $y$ when $x=2$. This problem serves as a valuable exercise in problem-solving and mathematical reasoning, strengthening our grasp of inverse variation and its applications in diverse contexts. Inverse variation can be expressed mathematically as $y = k/x^2$, where $y$ is inversely proportional to the square of $x$, and $k$ is the constant of variation. This constant represents the strength of the inverse relationship between $x$ and $y$. In this particular problem, we are given that $y$ varies inversely with the square of $x$, which means that as $x$ increases, $y$ decreases proportionally to the square of $x$, and vice versa. This relationship can be visualized as a curve where the product of $y$ and the square of $x$ remains constant. Understanding this relationship is key to solving the problem and determining the value of $y$ when $x=2$.

Setting up the Inverse Variation Equation

To solve this problem effectively, the initial step involves translating the given information into a mathematical equation. The statement "$y$ varies inversely with the square of $x$" can be expressed as $y = k/x^2$, where $k$ represents the constant of variation. This equation forms the foundation for our solution, allowing us to relate $x$ and $y$ mathematically. The constant of variation, k, is a crucial element in inverse variation problems. It represents the proportionality factor that links the two variables. In this case, $k$ determines the strength of the inverse relationship between $y$ and the square of $x$. To determine the value of $k$, we need to use the given information: $y=26$ when $x=4$. Substituting these values into our equation, we get $26 = k/4^2$. Solving for $k$, we multiply both sides by $4^2$, which is 16, resulting in $k = 26 * 16 = 416$. This value of $k$ is the key to solving the problem, as it allows us to determine the value of $y$ for any given value of $x$. Now that we have the constant of variation, $k = 416$, we can rewrite the inverse variation equation as $y = 416/x^2$. This equation represents the specific relationship between $y$ and the square of $x$ in this problem. It allows us to directly calculate the value of $y$ for any given value of $x$. The importance of establishing this equation cannot be overstated, as it forms the basis for the subsequent steps in solving the problem. Without a clear understanding of this equation, it would be impossible to determine the value of $y$ when $x=2$.

Determining the Constant of Variation (k)

With the inverse variation equation established, the next step is to determine the constant of variation, often denoted as $k$. This constant is essential for quantifying the relationship between $y$ and $x^2$. We are given that $y=26$ when $x=4$. By substituting these values into the equation $y = k/x^2$, we can solve for $k$. The process of solving for k involves substituting the given values of $x$ and $y$ into the inverse variation equation and then isolating $k$ on one side of the equation. In this case, we substitute $y=26$ and $x=4$ into the equation $y = k/x^2$, resulting in $26 = k/4^2$. To isolate $k$, we multiply both sides of the equation by $4^2$, which is 16. This gives us $26 * 16 = k$. Performing the multiplication, we find that $k = 416$. This value of $k$ represents the constant of variation in this specific inverse variation relationship. The constant of variation, $k=416$, represents the strength of the inverse relationship between $y$ and the square of $x$. It indicates that the product of $y$ and the square of $x$ is always equal to 416. This constant plays a crucial role in determining the value of $y$ for any given value of $x$, and vice versa. The importance of accurately determining k cannot be overstated. A slight error in calculating $k$ can lead to significant errors in the final answer. Therefore, it is essential to carefully substitute the given values and perform the calculations with precision. Once we have the correct value of $k$, we can confidently proceed to the next step in solving the problem.

Solving for y when x = 2

Now that we have determined the constant of variation, $k = 416$, we can proceed to find the value of $y$ when $x = 2$. This involves substituting $x = 2$ into the inverse variation equation $y = 416/x^2$. The substitution of x = 2 into the equation is a straightforward process. We replace the variable $x$ with the value 2, resulting in $y = 416/2^2$. Next, we need to simplify the equation to solve for $y$. First, we calculate $2^2$, which is 4. This gives us $y = 416/4$. Then, we perform the division, 416 divided by 4, which results in $y = 104$. Therefore, when $x = 2$, the value of $y$ is 104. This answer is consistent with the inverse variation relationship, as when $x$ decreases from 4 to 2, the value of $y$ increases from 26 to 104. The final answer, y = 104, represents the solution to the problem. It indicates the value of $y$ when $x$ is equal to 2, given the inverse variation relationship and the initial conditions. This answer aligns with the principles of inverse variation, where a decrease in $x$ corresponds to an increase in $y$, and vice versa. The process of solving for $y$ when $x = 2$ demonstrates the application of the inverse variation equation and the importance of accurately determining the constant of variation. By carefully substituting the given values and performing the calculations, we can confidently arrive at the correct solution.

Conclusion: Understanding Inverse Variation

In conclusion, we have successfully determined the value of $y$ when $x = 2$, given that $y$ varies inversely with the square of $x$ and $y = 26$ when $x = 4$. The solution, $y = 104$, was obtained by first establishing the inverse variation equation, $y = k/x^2$, then determining the constant of variation, $k = 416$, and finally substituting $x = 2$ into the equation. This problem serves as a valuable illustration of the principles of inverse variation and the importance of accurately translating mathematical relationships into equations. Inverse variation is a fundamental concept in mathematics and science, with applications in various fields, including physics, engineering, and economics. It describes relationships where one variable decreases as the other increases, and vice versa, while maintaining a constant product. Understanding inverse variation is crucial for modeling and analyzing real-world phenomena accurately. The ability to solve inverse variation problems, such as the one presented in this article, demonstrates a solid understanding of the concept and its applications. By carefully setting up the equation, determining the constant of variation, and substituting the given values, we can confidently arrive at the correct solution. The importance of understanding inverse variation extends beyond the realm of mathematics. It provides a framework for analyzing relationships in various fields and making predictions based on observed data. For example, in physics, inverse variation can be used to describe the relationship between pressure and volume of a gas, or the relationship between the intensity of light and the distance from the source. In economics, it can be used to model the relationship between price and demand. By mastering the principles of inverse variation, we can gain a deeper understanding of the world around us and develop valuable problem-solving skills. The solution to this problem, $y = 104$, reinforces the concept of inverse variation and provides a clear example of how to apply the principles to solve practical problems. This problem serves as a valuable learning experience for anyone seeking to enhance their understanding of inverse variation and its applications.

Therefore, the correct answer is A) 104.