Direct Variation: Finding Y When X = 3/10

Hey guys! Let's dive into a classic direct variation problem. These problems might seem tricky at first, but once you understand the core concept, they become super manageable. In this article, we're going to break down a specific problem step-by-step: "If y varies directly with x, and y = 5/6 when x = 8, what is the value of y when x = 3/10?" We'll go through the theory behind direct variation, the setup, and the solution, making sure you're confident in tackling similar problems on your own. Let’s get started!

Understanding Direct Variation

Okay, before we jump into solving the problem, let's make sure we're all on the same page about what direct variation actually means. Direct variation, at its core, describes a relationship between two variables where one variable is a constant multiple of the other. Imagine you're buying candy – the more candy you buy, the higher your total cost. That’s a direct variation! In mathematical terms, we say that y varies directly with x if there's a constant k such that:

y = kx

Here:

• y is the dependent variable (its value depends on x).
• x is the independent variable (you can choose its value).
• k is the constant of variation (this value stays the same throughout the problem).

The constant of variation, k, is the key to unlocking these problems. It tells us the ratio between y and x. If k is a larger number, y changes more dramatically with changes in x. If k is a smaller number, y changes less dramatically. Think of it like the steepness of a hill – a steeper hill means a bigger change in height for every step you take horizontally.

So, how do we find this magical k? Simple! If we know one pair of x and y values that fit the relationship, we can plug them into the equation y = kx and solve for k. This is exactly what the first part of our problem gives us: "y = 5/6 when x = 8." We’ll use this information to find k and then use k to find the value of y for a different value of x.

Think of direct variation in real-world terms. More hours worked often mean more money earned (directly proportional!), more ingredients in a recipe usually mean a larger dish, and faster you drive, the more distance you'll cover in the same amount of time (assuming the speed is constant). Recognizing these direct relationships in everyday situations can really help solidify your understanding of the concept. Remember, the core idea is that the ratio between the two variables remains constant. As one increases, the other increases proportionally, and as one decreases, the other decreases proportionally.

Setting Up the Problem

Alright, now that we've got a good grasp of direct variation, let's set up our specific problem. The question states: "If y varies directly with x, and y = 5/6 when x = 8, what is the value of y when x = 3/10?" We've already identified the core concept – y and x have a direct relationship, meaning they're linked by that y = kx equation. Our mission is to find the value of y under a new set of conditions.

Here's how we can break down the given information into actionable steps:

1. Identify the knowns: We know that y varies directly with x. We also have a specific pair of values: y = 5/6 when x = 8. Finally, we have a new value for x, which is x = 3/10, and we need to find the corresponding y.
2. Establish the equation: The fundamental equation for direct variation is y = kx. This is our starting point. We need to find k.
3. Find the constant of variation (k): We'll use the given values of y and x (y = 5/6 and x = 8) and plug them into the equation y = kx. This will allow us to solve for k. Think of k as the bridge connecting x and y – once we know k, we can easily hop between x and y values.
4. Use k to find the new y: Once we've calculated k, we'll use it along with the new value of x (x = 3/10) in the equation y = kx to find the corresponding value of y. This is the final step – the answer we're looking for!

This structured approach helps us avoid getting lost in the numbers. We're not just blindly plugging things in; we're following a logical process. By clearly identifying the knowns, the equation, and the steps involved, we're setting ourselves up for success. Remember, in math, a clear plan is half the battle! Now, let’s move on to the actual calculations and find that elusive k value.

Solving for the Constant of Variation (k)

Okay, let's roll up our sleeves and get our hands dirty with some math! As we established in the previous section, the key to solving this direct variation problem is finding the constant of variation, k. We know that y = kx, and we're given that y = 5/6 when x = 8. This is our golden ticket to finding k. All we need to do is substitute these values into the equation and solve.

So, here's the equation:

• y = kx

Now, let's substitute y = 5/6 and x = 8:

• 5/6 = k * 8

Our goal is to isolate k, meaning we want to get k by itself on one side of the equation. To do this, we need to get rid of the 8 that's being multiplied by k. The way we do that is by dividing both sides of the equation by 8. Remember, whatever we do to one side of the equation, we must do to the other to keep it balanced. Think of it like a seesaw – if you add weight to one side, you need to add the same weight to the other side to keep it level.

So, let's divide both sides by 8:

• (5/6) / 8 = (k * 8) / 8

This simplifies to:

• 5/6 ÷ 8 = k

Now we have a fraction divided by a whole number. To divide a fraction by a whole number, we can rewrite the whole number as a fraction (8 = 8/1) and then multiply by the reciprocal. The reciprocal of 8/1 is 1/8.

So, we have:

• 5/6 * 1/8 = k

Multiplying the fractions, we get:

• 5/48 = k

Boom! We've found our k. The constant of variation is 5/48. This means that for this particular relationship between x and y, y is always 5/48 times x. This is a crucial piece of information, as we can now use it to find y for any given value of x. Let’s move on to the final step: finding the value of y when x = 3/10.

Finding y When x = 3/10

Awesome! We've successfully navigated the trickiest part – finding the constant of variation, k. We know that k = 5/48, and we're now ready to use this information to find the value of y when x = 3/10. Remember our fundamental equation for direct variation: y = kx. We have k, we have the new x, and we're hunting for y. This is a straightforward substitution problem now.

Let's write down the equation again:

• y = kx

Now, substitute k = 5/48 and x = 3/10:

• y = (5/48) * (3/10)

We're now faced with multiplying two fractions. Before we jump into multiplying the numerators and denominators, let's see if we can simplify anything. Simplifying fractions before multiplying makes the arithmetic much easier. Notice that both 5 and 10 share a common factor of 5, and both 3 and 48 share a common factor of 3. Let’s divide them out!

Divide 5 and 10 by 5:

• 5 ÷ 5 = 1
• 10 ÷ 5 = 2

Divide 3 and 48 by 3:

• 3 ÷ 3 = 1
• 48 ÷ 3 = 16

Our equation now looks like this:

• y = (1/16) * (1/2)

Much simpler, right? Now we can multiply the numerators and the denominators:

• y = (1 * 1) / (16 * 2)

• y = 1/32

And there you have it! When x = 3/10, y = 1/32. We've successfully solved the problem. We started with the concept of direct variation, found the constant of variation, and used it to find the value of y for a given x. This is a common type of problem in algebra, and mastering this process will definitely come in handy.

Conclusion

Alright guys, we did it! We successfully solved a direct variation problem, and hopefully, you now feel much more confident in tackling these types of questions. Remember, the key to direct variation problems is understanding the relationship y = kx and finding that constant of variation, k. Once you've got k, you can unlock the value of y for any given x, or vice versa.

Let's quickly recap the steps we took:

1. Understood the concept: We defined direct variation and the importance of the constant of variation.
2. Set up the problem: We identified the knowns and the unknown and established the fundamental equation y = kx.
3. Solved for k: We used the given values of y and x to find the constant of variation, k.
4. Found the new y: We used the calculated k and the new value of x to find the corresponding value of y.

These problems might seem a little daunting at first, but breaking them down into these logical steps makes them much more approachable. Practice is key! The more direct variation problems you solve, the more comfortable you'll become with the process. Look for real-world examples of direct variation around you – like the relationship between hours worked and money earned, or the number of ingredients in a recipe and the size of the dish. This will help solidify your understanding even further.

So, next time you encounter a direct variation problem, remember the y = kx equation, find that k, and you'll be golden! Keep practicing, keep exploring, and you'll become a direct variation master in no time. You got this!