Solve 5/x + 2/y = 3 For Integers: Maximize X + Y

Ever stared at a math problem and thought, "Dude, where do I even begin?" Well, you're not alone! Today, we're diving deep into a super cool challenge: finding integer solutions for the equation 5/x + 2/y = 3 and then figuring out the absolute biggest value x + y can hit. This isn't just about crunching numbers; it's about understanding the art of algebraic manipulation and integer constraints. So, buckle up, guys, because we're about to turn a seemingly complex fraction problem into a fun factoring adventure! We'll break down every step, making sure you not only get the answer but also why we take each turn.

Unpacking the Mystery: What Exactly Are We Solving?

Alright, let's kick things off by really understanding what we're up against. We're tasked with finding integer solutions for 5/x + 2/y = 3. This type of problem, where we're looking exclusively for whole number answers (positive or negative, but no fractions or decimals!), falls into a super interesting category of mathematics called Diophantine equations. These equations are named after Diophantus of Alexandria, an ancient Greek mathematician who loved solving problems involving only integers. What makes them particularly tricky, and often quite rewarding to solve, is that the restriction to integers significantly limits the number of possible solutions, sometimes even making solutions non-existent. But when they do exist, finding them feels like discovering a hidden treasure!

Our specific equation, 5/x + 2/y = 3, looks innocent enough, but the presence of x and y in the denominators immediately tells us a few critical things. First, neither x nor y can be zero. That's a fundamental rule of fractions: you can't divide by zero, or the universe might just glitch out! So, any potential integer solutions we find must not include x=0 or y=0. Keep that in mind as we proceed. The goal here isn't just to find any integer pairs (x, y) that satisfy the equation; the ultimate prize is to identify the pair that gives us the maximum possible sum of x + y. This means we'll need to be super thorough in our search for all valid pairs. Many folks might rush to find just one or two solutions, but to truly maximize x + y, we've got to explore every single pathway. This journey will require a keen eye for detail, a solid understanding of basic algebra, and a willingness to explore all the nooks and crannies of number theory. We'll transform this fractional beast into something much more manageable, revealing the elegant structure hidden within. It's a fantastic example of how a bit of algebraic wizardry can simplify complex problems, allowing us to pinpoint those elusive integer solutions. So, let's roll up our sleeves and get started on this exciting mathematical quest, where every step brings us closer to unraveling the full mystery and claiming that maximum sum!

Transforming the Equation: Kicking Fractions to the Curb

Okay, so we've got 5/x + 2/y = 3. The first thing any seasoned math whiz will tell you is that fractions in an equation can be a real pain when you're hunting for integer solutions. They just complicate things, making it hard to see the underlying integer relationships. So, our primary goal in this step is to eliminate those pesky denominators. How do we do that? By finding a common denominator and clearing it out! In this case, the common denominator for x and y is simply their product, xy. Let's multiply every single term in our equation by xy. Don't be shy; hit 'em all!

When we multiply 5/x by xy, the x in the denominator cancels out with the x from xy, leaving us with 5y. Sweet! Next, when we multiply 2/y by xy, the y in the denominator vanishes, leaving us with 2x. Awesome! And finally, we can't forget the right side of the equation: 3 multiplied by xy simply becomes 3xy. So, our original equation, 5/x + 2/y = 3, beautifully transforms into a much cleaner, fraction-free form: 5y + 2x = 3xy. See? Already looking way less intimidating! This new form is a nonlinear Diophantine equation, which is a fancy way of saying it involves products of our variables (like xy), but we're still only looking for integer answers. This transformation is absolutely crucial because working with integers is much more straightforward when you don't have to worry about fractional components popping up at every turn. It allows us to focus purely on the integer relationships between x and y, which is exactly what we need for solving Diophantine problems. Think of it as taking off a heavy, complicated backpack and replacing it with a lighter, more agile one. This single step fundamentally changes the landscape of the problem, setting us up for the next big move: making it factorable. This is where the real fun begins, as we start to manipulate terms to reveal a hidden structure that will lead us directly to our integer pairs. Understanding why we clear denominators is just as important as knowing how to do it; it simplifies the problem dramatically, making it accessible to integer factorization techniques. So, with 5y + 2x = 3xy in hand, we're perfectly poised for the next exciting phase of our solution quest!

The Factoring Phenomenon: Unlocking Integer Pairs with a Clever Trick

Alright, guys, we've successfully morphed our fractional equation into 5y + 2x = 3xy. Now, this is where things get super interesting and a bit clever. We need a way to find integer solutions for this, and a common strategy for equations involving an xy term alongside x and y terms is to rearrange it into a factorable form. This often involves a technique sometimes called Simon's Favorite Factoring Trick, or more generally, completing a product. The idea is to get something that looks like (Ax + B)(Cy + D) = E, where A, B, C, D, E are integers.

Let's move all terms to one side to get 3xy - 2x - 5y = 0. Now, how do we get (something x - something)(something y - something)? We can try to force it. Imagine we want (3x - A)(y - B). Expanding this gives 3xy - 3Bx - Ay + AB. Comparing this to 3xy - 2x - 5y = 0, we see 3B should be 2 and A should be 5. But wait, this would mean B = 2/3, and we need integers! This approach needs a slight tweak.

The real trick here, when you have ax + by = cxy, is to multiply the entire equation by c (or a multiple of c) to make cxy become c^2xy. Then, you can factor it. In our case, c=3. So, let's multiply 3xy - 2x - 5y = 0 by 3:

9xy - 6x - 15y = 0

Now, we can try to factor this. We're aiming for something like (3x - A)(3y - B) = Constant. Let's think about (3x - _)(3y - _). If we expand (3x - A)(3y - B), we get 9xy - 3Bx - 3Ay + AB. Comparing this to 9xy - 6x - 15y = 0, we can see:

• -3B should correspond to -6x, which means 3B = 6, so B = 2.
• -3A should correspond to -15y, which means 3A = 15, so A = 5.

So, it looks like we want factors (3x - 5) and (3y - 2). Let's check what (3x - 5)(3y - 2) expands to: 9xy - 6x - 15y + 10

We have 9xy - 6x - 15y = 0. But our factored form gives us an extra + 10. So, to make them equal, we must have: 9xy - 6x - 15y + 10 = 10

Boom! This is the golden ticket! We've successfully transformed our original equation into (3x - 5)(3y - 2) = 10. This is incredibly powerful because now we're looking for two integer factors whose product is 10. Finding integer factors is a piece of cake compared to solving the original non-linear equation. This clever maneuver essentially converts a seemingly complex problem into a straightforward factorization task, which is exactly what we need when dealing with integer solutions. The key here is recognizing the pattern and knowing that multiplying by the coefficient of the xy term (in our case, 3) often sets up the equation perfectly for this type of factorization. Without this step, solving for integer pairs would be significantly more challenging, if not nearly impossible, by inspection alone. This factoring trick truly unlocks the problem, paving the way for us to systematically list all possible integer solutions in the next stage of our adventure. It’s a testament to the elegance of algebra, transforming a mess into a clean, solvable form. So, let’s get ready to list those factors and find our x and y values!

Listing the Integer Pairs: Let the Factor Hunt Begin!

Now that we've got the awesome equation (3x - 5)(3y - 2) = 10, we're in prime position to find all the integer pairs for x and y. This is where the magic of factors comes into play! We need to list all the possible pairs of integers that multiply to 10. Remember, we're talking about integers, so positive and negative numbers are fair game. Let's list 'em out:

• Pair 1: (1, 10)
• Pair 2: (10, 1)
• Pair 3: (-1, -10)
• Pair 4: (-10, -1)
• Pair 5: (2, 5)
• Pair 6: (5, 2)
• Pair 7: (-2, -5)
• Pair 8: (-5, -2)

For each of these pairs, we'll set (3x - 5) equal to the first number in the pair and (3y - 2) equal to the second number. Then, we solve for x and y. Remember, both x and y must be integers for the solution to be valid. Also, a quick reminder: x and y cannot be zero. Let's go through each case systematically:

Case 1: (3x - 5) = 1 and (3y - 2) = 10

• From 3x - 5 = 1, we get 3x = 6, so x = 2. (Integer! Valid! And not zero.)
• From 3y - 2 = 10, we get 3y = 12, so y = 4. (Integer! Valid! And not zero.)
• Solution 1: (x, y) = (2, 4)

Case 2: (3x - 5) = 10 and (3y - 2) = 1

• From 3x - 5 = 10, we get 3x = 15, so x = 5. (Integer! Valid!)
• From 3y - 2 = 1, we get 3y = 3, so y = 1. (Integer! Valid!)
• Solution 2: (x, y) = (5, 1)

Case 3: (3x - 5) = -1 and (3y - 2) = -10

• From 3x - 5 = -1, we get 3x = 4. Uh oh! x = 4/3, which is not an integer. So, this pair is invalid. We discard this one, no biggie!

Case 4: (3x - 5) = -10 and (3y - 2) = -1

• From 3x - 5 = -10, we get 3x = -5. Again, x = -5/3, not an integer. Invalid pair. Next!

Case 5: (3x - 5) = 2 and (3y - 2) = 5

• From 3x - 5 = 2, we get 3x = 7. Another one! x = 7/3, not an integer. Invalid pair.

Case 6: (3x - 5) = 5 and (3y - 2) = 2

• From 3x - 5 = 5, we get 3x = 10. x = 10/3, not an integer. Invalid pair.

Case 7: (3x - 5) = -2 and (3y - 2) = -5

• From 3x - 5 = -2, we get 3x = 3, so x = 1. (Integer! Valid!)
• From 3y - 2 = -5, we get 3y = -3, so y = -1. (Integer! Valid!)
• Solution 3: (x, y) = (1, -1)

Case 8: (3x - 5) = -5 and (3y - 2) = -2

• From 3x - 5 = -5, we get 3x = 0, so x = 0. Wait a minute! Remember our initial rule? x cannot be zero because it was in the denominator of the original equation (5/x). So, even though it's an integer, this solution is invalid because it would lead to division by zero in the original problem. This is a super important check that many might miss! Always go back to the original equation's domain.

After diligently checking all eight factor pairs, we are left with only three valid integer solutions for (x, y):

1. (2, 4)
2. (5, 1)
3. (1, -1)

This systematic approach ensures we haven't missed any possible valid solutions and correctly filtered out those that don't meet the integer criteria or the domain restrictions. The process of meticulously checking each pair is paramount; a single oversight could lead to missing the optimal solution. It truly highlights the necessity of precision when solving Diophantine equations. Now that we have our definitive list of solutions, the final step is to calculate x + y for each and find our maximum!

Calculating X + Y and Pinpointing the Maximum Value

Alright, team! We've done the hard work of identifying all the valid integer pairs (x, y) that satisfy our equation 5/x + 2/y = 3. This was a crucial step, and we ended up with three solid contenders. Now, for the grand finale: we need to figure out the value of x + y for each of these pairs and then, naturally, pick the biggest one. This is where all our efforts come together, and we see which pair truly reigns supreme.

Let's take each valid solution we found and calculate their respective sums:

Solution 1: (x, y) = (2, 4)

• For this pair, x + y = 2 + 4 = 6.
• This is a strong candidate, a nice positive sum. Remember, we're looking for the maximum, so positive values are generally a good sign!

Solution 2: (x, y) = (5, 1)

• Here, x + y = 5 + 1 = 6.
• Whoa, another 6! This means we have multiple pairs that give us the same sum. This is perfectly normal in these kinds of problems, and it just means we've got to keep our eyes peeled for any sum that might be even larger.

Solution 3: (x, y) = (1, -1)

• For this pair, x + y = 1 + (-1) = 0.
• A sum of zero. While a valid integer solution to the equation, it's clearly not going to be our maximum when compared to the 6s we found earlier. But it's super important we still calculated it, just in case other circumstances might have made it relevant or if other sums were negative.

After evaluating all the possible x + y sums from our valid integer pairs, we can clearly see the results: we got 6, 6, and 0. Comparing these values, the maximum value of x + y is undoubtedly 6. It's pretty cool how two different pairs (2, 4) and (5, 1) both lead to the same maximum sum. This underscores the idea that often, there isn't just one unique solution set that yields the desired outcome; sometimes, multiple paths lead to the same peak! This systematic evaluation is the final verification step, ensuring that we've not only correctly identified the integer solutions but also extracted the specific piece of information (the maximum sum) that the problem was asking for. Without this careful comparison, we might have incorrectly guessed or assumed a maximum, missing the full scope of possibilities. It’s the satisfying conclusion to our algebraic detective work, bringing clarity and a definitive answer to the original challenge.

Wrapping Up: The Power of Algebraic Insight

And there you have it, folks! We've successfully navigated the twists and turns of finding integer solutions for 5/x + 2/y = 3 and, more importantly, identified the maximum value for x + y, which turned out to be a solid 6. This journey wasn't just about getting an answer; it was about appreciating the incredible power of algebraic manipulation and logical deduction when tackling Diophantine equations. Remember, the key steps were eliminating fractions, using that super clever factoring trick to transform the equation into a more manageable form, and then systematically checking every single integer factor pair. We also learned how crucial it is to remember those initial restrictions, like x and y cannot be zero, as they can quietly invalidate otherwise seemingly correct solutions. This process of breaking down a complex problem into smaller, solvable parts is a skill that extends far beyond mathematics, into everyday problem-solving. So, next time you see a problem that looks daunting, just remember this journey. Take a deep breath, break it down, and trust in the power of careful, step-by-step thinking. Keep exploring, keep questioning, and most importantly, keep having fun with math! You've got this, guys!