Solving Direct Variation Problems Finding Y When X Is 2

Jul 14, 2025 by ADMIN 56 views

Understanding Direct Variation and Solving for y

This problem delves into the concept of direct variation, a fundamental relationship in mathematics. Direct variation, at its core, describes a scenario where two variables are proportionally related – as one variable increases, the other increases at a constant rate, and vice versa. This consistent relationship is characterized by a constant of proportionality, which plays a crucial role in solving problems involving direct variation. To effectively tackle this problem, it's essential to understand the definition of direct variation, how to express it mathematically, and the steps involved in determining the constant of proportionality. This constant then allows us to find the value of y for any given value of x, making it a powerful tool for solving real-world problems and understanding proportional relationships.

The problem states that y varies directly as x. Mathematically, this relationship is expressed as:

$y = kx$

where k represents the constant of variation. This constant is the key to unlocking the relationship between x and y. It signifies the factor by which x must be multiplied to obtain y. In essence, k defines the slope of the linear relationship between the two variables. Understanding this equation is crucial as it forms the foundation for solving any direct variation problem.

The given information provides us with a specific instance of this relationship: when x is 6, y is 48. We can use these values to determine the constant of variation, k. Substituting these values into the equation, we get:

$48 = k(6)$

To solve for k, we divide both sides of the equation by 6:

$k = \frac{48}{6} = 8$

Now that we have found the constant of variation, k = 8, we can rewrite the direct variation equation as:

$y = 8x$

This equation now fully defines the relationship between x and y for this specific scenario. It tells us that y is always 8 times the value of x. This understanding is crucial for solving the next part of the problem, which requires us to find the value of y when x is 2.

The final step involves using the equation we derived to find the value of y when x is 2. We substitute x = 2 into the equation:

$y = 8(2) = 16$

Therefore, when x is 2, y is 16. To connect this solution to the given options, we need to identify the expression that correctly calculates this value. The expression that represents this calculation is:

$y = \frac{48}{6}(2)$

This expression first calculates the constant of variation (48/6) and then multiplies it by the new value of x (2), which aligns perfectly with our solution process.

Therefore, option A is the correct answer. This comprehensive step-by-step explanation breaks down the problem, making it easier to understand the underlying concepts and the solution process. By emphasizing the importance of the constant of variation and the direct variation equation, this explanation provides a clear and concise guide to solving similar problems.

Detailed Solution and Explanation

To reiterate, the core of this problem lies in understanding direct variation. When we say that y varies directly as x, we mean that there is a constant ratio between them. This relationship can be expressed using the equation:

$y = kx$

where k is the constant of proportionality or the constant of variation. Our first task is to find the value of k using the given information. We know that y = 48 when x = 6. Substituting these values into the equation, we get:

$48 = k * 6$

To isolate k, we divide both sides of the equation by 6:

$k = \frac{48}{6} = 8$

Now we know that the constant of variation, k, is 8. This means that the relationship between x and y is specifically defined by the equation:

$y = 8x$

This equation tells us that y is always 8 times the value of x. Now, we are asked to find the value of y when x is 2. We simply substitute x = 2 into our equation:

$y = 8 * 2 = 16$

So, when x is 2, y is 16. The question asks us to identify the expression that calculates this value. Let's examine the given options:

A. $y=\frac{48}{6}(2)$ : This expression first divides 48 by 6, which gives us the constant of variation, k = 8. Then, it multiplies this constant by 2, which is the new value of x. This is exactly how we calculated y, so this option is correct.
B. $y=\frac{6}{48}(2)$ : This expression divides 6 by 48, which is the inverse of the constant of variation. This is not the correct way to calculate y.
C. $y=\frac{(48)(8)}{2}$ : This expression multiplies 48 by 8 and then divides by 2. This does not represent the direct variation relationship we have established.
D. $y=\frac{2}{(48)(6)}$ : This expression divides 2 by the product of 48 and 6. This is also not related to the direct variation equation.

Therefore, the only expression that correctly calculates the value of y when x is 2 is option A:

$y = \frac{48}{6}(2)$

This detailed breakdown emphasizes the importance of understanding the underlying principles of direct variation and how to apply them to solve problems. By breaking down the steps and explaining the reasoning behind each calculation, this explanation provides a clear and thorough understanding of the solution.

Key Concepts and Practical Applications of Direct Variation

Direct variation is a fundamental concept in mathematics with numerous practical applications in various fields. Understanding direct variation not only helps in solving algebraic problems but also provides a framework for modeling real-world phenomena where quantities change proportionally. In this context, we will further explore the key concepts of direct variation, its mathematical representation, and its applications in diverse scenarios. This exploration will solidify your understanding of direct variation and equip you with the ability to identify and analyze directly proportional relationships in the world around you. By delving deeper into the intricacies of direct variation, you will gain a valuable tool for problem-solving and a broader appreciation for the interconnectedness of mathematical concepts.

The essence of direct variation lies in the proportional relationship between two variables. Mathematically, we express this relationship as:

$y = kx$

where:

y is the dependent variable
x is the independent variable
k is the constant of variation or the constant of proportionality

The constant of variation, k, is the crucial element that defines the specific direct variation relationship. It represents the factor by which the independent variable (x) is multiplied to obtain the dependent variable (y). A larger value of k indicates a steeper increase in y for a given increase in x, while a smaller value of k indicates a gentler increase. Understanding the significance of k is key to interpreting and applying direct variation relationships.

For instance, if y represents the distance traveled by a car at a constant speed and x represents the time traveled, then k represents the speed of the car. A higher k (speed) means the car covers more distance in the same amount of time. This simple example illustrates how direct variation can model real-world scenarios involving constant rates.

To identify if a relationship is a direct variation, we look for the following characteristics:

Linear Relationship: The graph of a direct variation equation is a straight line that passes through the origin (0, 0). This is because when x is 0, y is also 0 ( $y = k * 0 = 0$ ).
Constant Ratio: The ratio of y to x is constant. This constant ratio is the constant of variation, k. In other words, for any two points (x₁, y₁) and (x₂, y₂) on the line, the ratio y₁/x₁ will be equal to y₂/x₂.

Now, let's explore some practical applications of direct variation:

Physics: Ohm's Law states that the voltage (V) across a conductor is directly proportional to the current (I) flowing through it, with the constant of proportionality being the resistance (R). Mathematically, this is expressed as V = RI. This is a classic example of direct variation in physics.
Chemistry: The amount of product formed in a chemical reaction is often directly proportional to the amount of reactants used, assuming other factors are constant. This principle is crucial in stoichiometry and chemical calculations.
Economics: The earnings of an hourly employee are directly proportional to the number of hours worked, with the constant of variation being the hourly wage. This is a fundamental concept in labor economics.
Engineering: The elongation of a spring is directly proportional to the force applied to it, according to Hooke's Law. This principle is used in designing various mechanical systems and structures.

These examples demonstrate the wide range of applications of direct variation across different disciplines. By understanding the principles of direct variation, we can model and analyze various real-world phenomena, make predictions, and solve problems effectively.

In summary, direct variation is a powerful mathematical tool that describes proportional relationships between variables. Its applications extend beyond textbook problems, providing a framework for understanding and modeling real-world phenomena. By grasping the key concepts and characteristics of direct variation, you can enhance your problem-solving skills and gain a deeper appreciation for the role of mathematics in various fields.

Conclusion: Mastering Direct Variation

In conclusion, this problem effectively illustrates the concept of direct variation and how to solve problems involving this relationship. The key to solving these problems lies in understanding the definition of direct variation, which states that two variables are directly proportional if their ratio is constant. This constant ratio is known as the constant of variation, and it plays a pivotal role in establishing the relationship between the variables. By identifying the constant of variation, we can construct an equation that accurately models the direct proportionality, allowing us to predict the value of one variable given the value of the other. This understanding is crucial for solving not only mathematical problems but also for applying the concept of direct variation to real-world scenarios.

The step-by-step approach used to solve the problem highlights the importance of breaking down complex problems into manageable steps. First, we translated the problem statement into a mathematical equation, recognizing that "y varies directly as x" implies a linear relationship of the form y = kx. Next, we used the given information (y = 48 when x = 6) to determine the constant of variation, k. This step involved substituting the known values into the equation and solving for k. Once we found k, we could rewrite the direct variation equation with the specific constant for this relationship. Finally, we used this equation to find the value of y when x = 2, answering the original question. This methodical approach can be applied to a wide range of mathematical problems, fostering a systematic and confident problem-solving style.

Furthermore, this problem underscores the significance of connecting mathematical concepts to real-world applications. Direct variation is not merely an abstract mathematical idea; it is a fundamental principle that governs many phenomena in the world around us. From the relationship between distance, speed, and time to the relationship between the amount of ingredients and the yield of a recipe, direct variation provides a framework for understanding proportional relationships. By recognizing these connections, we can appreciate the practical relevance of mathematics and develop a deeper understanding of the world.

Therefore, mastering direct variation is an essential skill for students and professionals alike. It not only enhances our ability to solve mathematical problems but also equips us with a valuable tool for analyzing and interpreting the world around us. By understanding the principles of direct variation and practicing problem-solving techniques, we can confidently tackle challenges and gain a deeper appreciation for the power and versatility of mathematics.