Direct Variation: Solving For Y When X=11

Understanding direct variation is a fundamental concept in mathematics, and it's all about how two variables relate to each other. When we say $y$ varies directly as $x$, it means that as $x$ increases, $y$ increases proportionally, and as $x$ decreases, $y$ decreases proportionally. This relationship can be expressed by the equation $y = kx$, where $k$ is the constant of variation. This constant $k$ is the key to unlocking the relationship between $x$ and $y$. It represents the factor by which $x$ is multiplied to get $y$. So, if you know the value of $k$ and $x$, you can easily find $y$, or if you know $y$ and $x$, you can find $k$. In this article, we'll dive deep into how to find this constant and use it to solve for unknown values, much like in the scenario where $y$ is 18 when $x$ is 5, and we need to find $y$ when $x$ is 11. This kind of problem is common in algebra and has practical applications in various fields, from physics to economics. We'll break down the steps clearly, making sure you grasp the concept of direct variation and feel confident in solving these types of problems. Remember, the core idea is proportionality; one variable is always a constant multiple of the other.

Finding the Constant of Variation ($k$)

The first crucial step in solving any direct variation problem is to determine the constant of variation, often denoted by $k$. This constant is the heart of the direct variation equation, $y = kx$. It tells us the specific ratio between $y$ and $x$ for any given pair of values. In our specific problem, we are given that $y$ is 18 when $x$ is 5. We can use this information to find $k$. By substituting these values into our direct variation equation, we get $18 = k(5)$. To isolate $k$, we need to divide both sides of the equation by 5. This gives us $k = \frac{18}{5}$. This fraction, $\frac{18}{5}$, is our constant of variation. It means that for this particular relationship, $y$ is always $\frac{18}{5}$ times $x$. It's important to note that $k$ will remain the same for all pairs of $(x, y)$ values that satisfy this direct variation. This is why it's called a 'constant.' It doesn't change, no matter what $x$ and $y$ are, as long as they maintain their direct proportional relationship. So, once we've found $k = \frac{18}{5}$, we have essentially unlocked the secret code of this variation. We can now use this value of $k$ to predict or calculate $y$ for any given $x$, or vice versa. This process of finding $k$ is foundational, and mastering it will make solving subsequent steps much more straightforward. Keep this value of $k$ handy, as it will be used in the next stage of our problem-solving journey.

Using the Constant to Find a New Value of $y$

Now that we have successfully determined the constant of variation ($k$), which is $\frac{18}{5}$, we can proceed to find the value of $y$ when $x$ is 11. Our direct variation equation is $y = kx$. We already know $k = \frac{18}{5}$ and we are given the new value of $x = 11$. So, we can substitute these values back into the equation: $y = \left(\frac{18}{5}\right)(11)$. To find the value of $y$, we simply perform the multiplication. Multiply the numerator of the fraction by 11: $18 \times 11 = 198$. The denominator remains 5. Therefore, $y = \frac{198}{5}$. This is the value of $y$ when $x$ is 11, based on the direct variation established by the initial conditions. This step highlights the predictive power of direct variation. Once the relationship (defined by $k$) is established, you can confidently calculate values for $y$ given any $x$, or even solve for $x$ if $y$ is known. The expression we used to find $y$ is indeed $y = \frac{18}{5}(11)$. Looking at the options provided in the original question, this corresponds to option B. It's crucial to recognize how the formula is applied: the constant $k$ (which is $\frac{y_{initial}}{x_{initial}}$) is multiplied by the new value of $x$. This straightforward substitution and calculation is the essence of solving direct variation problems once the constant has been identified. It’s a systematic approach that ensures accuracy.

Analyzing the Options

Let's take a moment to carefully examine the provided options and see why option B is the correct one. We established that the relationship is $y = kx$, and that $k = \frac{18}{5}$. When $x = 11$, the equation becomes $y = \frac{18}{5} \times 11$. This is precisely what option B states: $y = \frac{18}{5}(11)$. Now, let's look at why the other options are incorrect. Option A suggests $y = \frac{5}{18}(11)$. This expression uses the reciprocal of our constant of variation ($k$). If the relationship were inverse variation, this might be relevant, but for direct variation, it's incorrect. Option C presents $y = \frac{(18)(5)}{11}$. This expression seems to be trying to form a fraction with the initial values of $x$ and $y$ and then dividing by the new $x$, which doesn't align with the direct variation formula $y = kx$. Option D, $y = \frac{11}{(18)(5)}$, is also incorrect as it manipulates the numbers in a way that doesn't follow the established mathematical principle of direct variation. The beauty of understanding the formula $y=kx$ and how to find $k$ is that it eliminates guesswork. You can directly construct the correct expression. In this case, finding $k$ as $\frac{18}{5}$ and then substituting it into $y=kx$ with the new $x$ value of 11 clearly leads to $y = \frac{18}{5}(11)$. This systematic approach confirms that option B is the only mathematically sound choice based on the principles of direct variation.

The Essence of Direct Variation

At its core, direct variation is about a consistent, proportional relationship between two variables. Think of it like a recipe: if you double the amount of flour, you must also double the amount of sugar to keep the taste the same. In mathematics, this proportionality is captured by the constant of variation, $k$. Our problem started with the statement that $y$ varies directly as $x$. This immediately tells us the form of our equation: $y = kx$. The initial condition, $y=18$ when $x=5$, is our anchor. It allows us to solve for $k$. We plugged these values in: $18 = k imes 5$. To find $k$, we rearranged the equation: $k = \frac{18}{5}$. This value, $\frac{18}{5}$, is the scaling factor that connects $x$ and $y$ in this specific scenario. It means $y$ is always $\frac{18}{5}$ times the value of $x$. Once we have this constant, solving for a new value of $y$ is straightforward. Given a new $x$ value of 11, we substitute $k$ and the new $x$ into our equation: $y = \frac{18}{5} imes 11$. This leads directly to the expression $y = \frac{18}{5}(11)$, which matches option B. This entire process reinforces the idea that direct variation is predictable. If you know one pair of values and the type of variation, you can determine the exact relationship and use it to find any other corresponding value. It’s a powerful concept for modeling relationships where quantities change in lockstep. Understanding this concept is crucial for many areas of study, from understanding how distance relates to time at a constant speed to how the cost of a product relates to the quantity purchased when there are no bulk discounts. The consistent ratio ($k$) is the key.

Conclusion

In conclusion, when $y$ varies directly as $x$, and we are given that $y=18$ when $x=5$, we first establish the constant of variation, $k$. Using the formula $y=kx$, we substitute the known values to find $k$: $18 = k(5)$, which gives us $k = \frac{18}{5}$. With the constant of variation determined, we can then find the value of $y$ when $x=11$ by substituting these values back into the direct variation equation: $y = \left(\frac{18}{5}\right)(11)$. Therefore, the expression that can be used to find the value of $y$ when $x$ is 11 is $y = \frac{18}{5}(11)$. This corresponds to option B. Direct variation is a fundamental concept in algebra that describes a proportional relationship between two variables, and understanding how to find and use the constant of variation is key to solving such problems accurately. For further exploration into the fascinating world of mathematical relationships and variations, you might find it helpful to visit Khan Academy's section on ratios and proportional relationships or Math is Fun's explanation of direct proportion.