Solving Equations: Comparing Quantities P And Q

Hey guys! Let's dive into a fun math problem. We're given an equation: $3x + 4y = 29$, where both $x$ and $y$ are positive integers. The real question is, how do we compare two quantities, $P$ and $Q$, based on this information? Let's break it down, step by step, and figure out the best approach to tackle this type of problem. This is a common type of question you might find in a math competition or a standardized test. The key is understanding how to manipulate the given equation and identify the possible values of $x$ and $y$. Don't worry, we'll go through it nice and easy, so you'll be a pro in no time.

Understanding the Problem: The Core Equation and Constraints

Alright, let's get down to business. The heart of our problem is the equation $3x + 4y = 29$. This equation is our main tool, but we've got some important restrictions. We know that both $x$ and $y$ are positive integers. This is crucial because it limits the possible values that $x$ and $y$ can take. Remember, positive integers are whole numbers greater than zero. So, no fractions, no decimals, and no negative numbers allowed here! This constraint is key to finding the relationship between the quantities. We need to find the possible values for $x$ and $y$ that satisfy the equation. This is a classic example of a Diophantine equation, where we are looking for integer solutions. The constraints help narrow down the possibilities, making our task manageable. Think of it like a puzzle – the more pieces you have, the easier it is to find the solution. Now, let’s get into the specifics of finding those solutions.

Now, let's dissect the equation. The coefficients 3 and 4 in front of $x$ and $y$ play a role, but the main goal is to isolate variables. One trick is to focus on one variable and see how it affects the other, especially with the integer constraint. Because we have positive integers, we know that $x$ and $y$ cannot be arbitrarily large. The equation sets a limit. We'll explore the implications of this as we work through potential values of $x$ and $y$. This initial analysis of the equation is super important. We set the stage for how to solve the problem systematically, making it easier to compare the quantities later on. The whole process is about identifying the correct setup so we can apply the right steps.

Remember, in math, understanding the question and the limitations are essential for problem-solving. This initial phase sets the groundwork for a successful approach. So, take your time, understand the conditions and the properties of positive integers. We're making sure we lay a strong foundation for the work ahead, setting us up to conquer the problem like champs.

Finding Possible Values of x and y

Okay, guys, let's get into the fun part: finding those magic numbers! Our goal is to find all the pairs of positive integers $(x, y)$ that satisfy $3x + 4y = 29$. The easiest way to do this is usually to start by considering different values for one of the variables, and then solve for the other. Since the problem is concerned with comparing x + y, let's try to identify possible combinations. We can start by isolating either x or y in the equation. Let's isolate x:

$3x = 29 - 4y$

$x = (29 - 4y) / 3$

Now, we need to find values of y that will yield a positive integer for x. Remember, y must also be a positive integer. Let's test some values of y:

• If y = 1, then $x = (29 - 4(1)) / 3 = 25 / 3$ (Not an integer, so not a solution)
• If y = 2, then $x = (29 - 4(2)) / 3 = 21 / 3 = 7$ (Integer! So, $(7, 2)$ is a possible solution!)
• If y = 3, then $x = (29 - 4(3)) / 3 = 17 / 3$ (Not an integer)
• If y = 4, then $x = (29 - 4(4)) / 3 = 13 / 3$ (Not an integer)
• If y = 5, then $x = (29 - 4(5)) / 3 = 9 / 3 = 3$ (Integer! So, $(3, 5)$ is a possible solution!)
• If y = 6, then $x = (29 - 4(6)) / 3 = 5 / 3$ (Not an integer)
• If y = 7, then $x = (29 - 4(7)) / 3 = 1 / 3$ (Not an integer)

As we increase y, x becomes smaller. Since we're limited to positive integers, we don't need to consider values of y larger than 7 because x will be negative. We found two solutions: $(7, 2)$ and $(3, 5)$.

So, what does this tell us? These are the only pairs of positive integers that meet our criteria. This systematic approach is the most reliable way to find all possible solutions. We are using simple trial and error within the confines of our knowledge of positive integers, and this works very well to give us our final answer! The point here is to be methodical, and not to miss a possible solution. Now let's calculate the values for $x + y$.

Calculating x + y and Comparing with Q

Alright, we've done the heavy lifting, guys! Now it's time to bring it all together. From the previous step, we've identified the possible pairs $(x, y)$ that satisfy our equation: $(7, 2)$ and $(3, 5)$.

Now, we need to calculate $x + y$ for each of these pairs.

• For $(7, 2)$: $x + y = 7 + 2 = 9$
• For $(3, 5)$: $x + y = 3 + 5 = 8$

So, the possible values of $x + y$ are 9 and 8. The question asks us to compare this to the quantity Q, which is 10.

Let's write it out clearly. The possible values of P (which is $x + y$) are 9 and 8. The value of Q is 10.

Now, we have to determine the relationship between $P$ (which can be either 8 or 9) and $Q$ (which is 10). Here's where we get to the core of the question.

• In both cases, $x + y$ is less than 10. That means $P < Q$ for all the possible values of $P$.

This comparison is the whole point of the problem! We had to find the values of x and y, calculate x + y, and then compare them to the given value of Q. The final step of comparing is the key. Are we sure we have found all the solutions? Have we made any mistakes along the way? Always double-check your calculations. It is a good practice to ensure the accuracy of the answer, and avoid any mistakes. In this case, we have been very methodical, and have found all the solutions. Thus, now we can proceed to the final answer. We've got our solution, and now we are ready to declare our conclusion.

Final Answer and Conclusion

Alright, guys, let's wrap this up! We've determined the possible values of $x + y$ and compared them with Q. We found that $x + y$ can be either 8 or 9, while Q is 10. So, we can confidently say that $x + y$ is always less than 10. Based on this, we can conclude that P < Q. Therefore, the correct answer is (A) P < Q. High five! We did it! This is a typical example of how to solve a comparison problem in math, and we hope this helps. Now you can use this knowledge to tackle similar problems with confidence. Keep practicing and you'll become a pro in no time.

Remember to review each step to reinforce your understanding. Always double-check your work, and don't be afraid to ask for help if you're stuck. Math can be tricky sometimes, but with the right approach and practice, you can definitely master it. Great job, everyone! Keep up the excellent work!