Equation Solved: Find Natural Number Pairs

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Hey math whizzes, let's dive into a super interesting problem that’s all about finding pairs of natural numbers! We’re going to tackle the equation 3x + 6xy = 66 and figure out all the possible pairs of natural numbers (x, y) that make this equation true. Natural numbers, remember, are the positive whole numbers: 1, 2, 3, and so on. No negatives, no fractions, just the good old counting numbers.

This kind of problem is a classic in number theory and algebra. It tests our ability to manipulate equations and understand the properties of numbers. We’re not just looking for one solution; we’re hunting for all of them. So, let’s get our thinking caps on and break this down step-by-step. We want to make this equation friendly, so we'll do some algebraic magic to isolate variables and make it easier to spot those number pairs. Get ready, because we're about to unlock the secrets of this equation!

Unpacking the Equation: Our Starting Point

Alright guys, let's look closely at our equation: 3x + 6xy = 66. The first thing we should notice is that all the terms have a common factor. Yep, you guessed it – 3! This is a fantastic starting point because we can simplify the whole equation by dividing everything by 3. This makes the numbers smaller and, honestly, much easier to work with. So, if we divide each term by 3, we get:

(3x / 3) + (6xy / 3) = (66 / 3)

This simplifies beautifully to:

x + 2xy = 22

See? Already looking way less intimidating, right? This new form, x + 2xy = 22, is our playground now. Our mission is to find natural numbers x and y that satisfy this. Remember, natural numbers start from 1. So, x and y must be greater than or equal to 1.

Now, we need a strategy. How do we find pairs of numbers that fit? A common and super effective technique for equations like this, especially when dealing with products of variables, is factorization. We want to rearrange the equation so that we can express it as a product of two factors, something like (expression involving x and y) * (another expression involving x and y) = a constant. This often turns a tricky equation into a much simpler one where we can just test a limited number of possibilities.

Let's try to factor x + 2xy = 22. We can see that 'x' is a common factor in both terms on the left side. Let's factor that out:

x(1 + 2y) = 22

Boom! Just like that, we've transformed our equation into a product of two expressions equaling 22. This is exactly what we wanted. The two expressions are 'x' and '(1 + 2y)'. Since x and y are natural numbers, 'x' must be a natural number (i.e., x ≥ 1). Also, since y is a natural number (y ≥ 1), then 2y must be at least 2, and therefore (1 + 2y) must be at least 1 + 2 = 3. So, (1 + 2y) ≥ 3.

We are looking for two numbers, 'x' and '(1 + 2y)', that multiply together to give 22. And, critically, 'x' must be a natural number, and '(1 + 2y)' must be an integer greater than or equal to 3. This significantly narrows down our search!

Finding the Factors: Testing the Possibilities

Okay, guys, we've got x(1 + 2y) = 22. Our goal now is to find all the pairs of factors of 22 where the first factor ('x') is a natural number, and the second factor ('1 + 2y') is an integer greater than or equal to 3. Let's list out all the pairs of factors of 22. Remember, factors come in pairs that multiply to the number.

The factors of 22 are:

  • 1 and 22
  • 2 and 11
  • 11 and 2
  • 22 and 1

These are all the pairs of positive integers that multiply to 22. Now, we need to match these pairs with our expressions, 'x' and '(1 + 2y)', keeping in mind our conditions: x ≥ 1 and (1 + 2y) ≥ 3.

Let's go through each pair:

Possibility 1: x = 1 and (1 + 2y) = 22

  • Is x a natural number? Yes, x = 1 is a natural number.
  • Is (1 + 2y) ≥ 3? Yes, 22 is definitely greater than or equal to 3.
  • Now, let's solve for y: 1 + 2y = 22 => 2y = 22 - 1 => 2y = 21 => y = 21 / 2. Is y a natural number? No, 21/2 is not a whole number. So, this pair (x=1, y=21/2) is not a valid solution because y must be a natural number.

Possibility 2: x = 2 and (1 + 2y) = 11

  • Is x a natural number? Yes, x = 2 is a natural number.
  • Is (1 + 2y) ≥ 3? Yes, 11 is greater than or equal to 3.
  • Let's solve for y: 1 + 2y = 11 => 2y = 11 - 1 => 2y = 10 => y = 10 / 2. Is y a natural number? Yes! y = 5 is a natural number.
  • So, (x = 2, y = 5) is a valid solution. Let's check it in the original equation: 3(2) + 6(2)(5) = 6 + 60 = 66. Perfect!

Possibility 3: x = 11 and (1 + 2y) = 2

  • Is x a natural number? Yes, x = 11 is a natural number.
  • Is (1 + 2y) ≥ 3? No, 2 is not greater than or equal to 3. This violates our condition for (1 + 2y). Therefore, we don't even need to solve for y. This pair is not a valid solution.

Possibility 4: x = 22 and (1 + 2y) = 1

  • Is x a natural number? Yes, x = 22 is a natural number.
  • Is (1 + 2y) ≥ 3? No, 1 is not greater than or equal to 3. This violates our condition for (1 + 2y). Therefore, this pair is not a valid solution.

It looks like we've exhausted all the pairs of factors of 22 that fit our criteria. We were careful to check both conditions: 'x' being a natural number and '(1 + 2y)' being an integer greater than or equal to 3.

The Solution Revealed: Our Winning Pair!

After systematically checking all the possibilities derived from the factors of 22, we found only one pair of natural numbers (x, y) that satisfies the equation x(1 + 2y) = 22, under the conditions that x is a natural number (x ≥ 1) and (1 + 2y) is an integer greater than or equal to 3.

Let’s recap our findings:

  • We simplified the original equation 3x + 6xy = 66 to x + 2xy = 22 by dividing by 3.
  • We factored the left side to get x(1 + 2y) = 22.
  • We identified that for x and y to be natural numbers (x ≥ 1, y ≥ 1), we must have x ≥ 1 and (1 + 2y) ≥ 3.
  • We listed all the factor pairs of 22: (1, 22), (2, 11), (11, 2), (22, 1).
  • We tested each pair against our conditions:
    • For (x=1, 1+2y=22), y was not a natural number.
    • For (x=2, 1+2y=11), we found y=5, which is a natural number. This gives us the solution (x=2, y=5).
    • For (x=11, 1+2y=2), the condition (1+2y) ≥ 3 was not met.
    • For (x=22, 1+2y=1), the condition (1+2y) ≥ 3 was not met.

Therefore, the only pair of natural numbers (x, y) that satisfies the equation 3x + 6xy = 66 is (2, 5).

This is a great example of how algebraic manipulation and a solid understanding of number properties can help us solve seemingly complex problems. By simplifying, factoring, and then carefully checking the conditions, we can confidently find all possible solutions. Keep practicing these techniques, guys, and you'll become masters of number puzzles in no time!