Finding Possible Values Of 'a' In 3a + B = 21

Oct 16, 2025 by Dimemap Team 46 views

Hey guys! Let's dive into this math problem where we need to figure out the possible values for the number 'a' in the equation 3a + b = 21. We know that both 'a' and 'b' are natural numbers, which means they are positive whole numbers, and they are both less than 10. This gives us a specific range to work with, making the problem solvable. We’re essentially looking for integer solutions that fit within the given constraints. Sounds like fun, right? Let’s break it down step by step so we can understand how to approach these types of problems.

Understanding the Problem

First off, let's really understand what the problem is asking. We've got the equation 3a + b = 21, and two super important clues: 'a' and 'b' are natural numbers (1, 2, 3, and so on), and they're both less than 10. This is crucial because it limits the possible values we need to check. If we didn't have these limits, there could be tons of solutions, but thankfully, we can narrow it down. The heart of the question is: what values can we plug in for 'a' that will give us a natural number for 'b', all while keeping both numbers under 10? To figure this out, we're going to need to play around with the equation a bit and see what fits.

Setting Up Our Approach

So, how do we actually find these values? The best way to approach this is to rearrange the equation to solve for 'b'. This way, we can plug in different values for 'a' and see if we get a natural number for 'b'. Let's rewrite the equation:

b = 21 - 3a

Now we have 'b' all by itself on one side, which is perfect. This equation tells us that 'b' equals 21 minus 3 times 'a'. We can now try different values for 'a' and see what 'b' comes out to be. Remember, we're only interested in values of 'a' that are natural numbers less than 10, and that give us a 'b' that is also a natural number less than 10. This is where the fun begins – it's like a little puzzle!

Finding Possible Values for 'a'

Let's start plugging in values for 'a' and see what happens. We'll go through the natural numbers one by one, keeping in mind that 'a' must be less than 10. This systematic approach ensures we don’t miss any potential solutions. Remember, we’re looking for values of ‘a’ that result in ‘b’ also being a natural number less than 10.

Testing a = 1

First up, let's try a = 1. Plug it into our equation:

b = 21 - 3(1) = 21 - 3 = 18

Oops! b = 18, which is way more than 10. So, a = 1 doesn't work for us. We need 'b' to be less than 10, so let’s move on to the next possible value for ‘a’.

Testing a = 2

Next, let's try a = 2:

b = 21 - 3(2) = 21 - 6 = 15

Again, b = 15, which is still too big. It seems like we need to try a larger value for 'a' to bring 'b' down below 10. So, let’s keep going.

Testing a = 3

Now, let's try a = 3:

b = 21 - 3(3) = 21 - 9 = 12

Still no luck! b = 12, which is greater than 10. We're getting closer, but we haven’t found a solution yet. Let's keep increasing 'a' to see if we can get a suitable 'b'.

Testing a = 4

Let's jump to a = 4:

b = 21 - 3(4) = 21 - 12 = 9

Bingo! b = 9, which is a natural number and less than 10. So, a = 4 is a possible solution. We have our first valid value for ‘a’! But don’t stop here – there might be more. We need to continue testing to find all possible values.

Testing a = 5

Let's try a = 5:

b = 21 - 3(5) = 21 - 15 = 6

Great! b = 6, which is also a natural number less than 10. So, a = 5 is another possible solution. We’re on a roll now!

Testing a = 6

Let's check a = 6:

b = 21 - 3(6) = 21 - 18 = 3

Excellent! b = 3, which fits our criteria perfectly. So, a = 6 is another valid solution. We’re steadily adding to our list of possible values for ‘a’.

Testing a = 7

Now, let’s try a = 7:

b = 21 - 3(7) = 21 - 21 = 0

Uh oh! b = 0, which is not a natural number (natural numbers start from 1). So, a = 7 doesn’t work. We need ‘b’ to be a positive whole number, and 0 doesn’t fit that definition. Let’s see what happens with the next value.

Testing a = 8

Let's try a = 8:

b = 21 - 3(8) = 21 - 24 = -3

Nope! b = -3, which is not a natural number either. We’re getting negative values for ‘b’ now, which means we’ve gone too far. The values for ‘a’ are getting too large.

Testing a = 9

Finally, let's test a = 9:

b = 21 - 3(9) = 21 - 27 = -6

Again, b = -6, which is a negative number and not a natural number. So, a = 9 doesn't work. We've now tested all possible values for ‘a’ less than 10.

Listing the Possible Values

Okay, so we’ve tested all the natural numbers less than 10 for 'a'. Let's recap the values that worked. We found that the possible values for 'a' are 4, 5, and 6. When a = 4, b = 9; when a = 5, b = 6; and when a = 6, b = 3. All these pairs satisfy the equation 3a + b = 21 and the condition that both 'a' and 'b' are natural numbers less than 10. This systematic approach helped us narrow down the solutions and find exactly what we were looking for.

The Solutions

So, to wrap things up, the possible values for 'a' are:

These are the only natural numbers less than 10 that, when plugged into the equation 3a + b = 21, give us a natural number less than 10 for 'b'. We solved this by rearranging the equation, plugging in values, and checking if the resulting 'b' fit our criteria. See? Math can be a fun puzzle sometimes!

Conclusion

Alright guys, we've successfully found the possible values for 'a' in the equation 3a + b = 21, given that 'a' and 'b' are natural numbers less than 10. The key here was to rearrange the equation to solve for one variable (in this case, 'b') and then systematically test values for the other variable ('a'). This approach not only helps in solving this particular problem but also provides a framework for tackling similar problems in algebra. Remember, the constraints given in the problem (like 'a' and 'b' being less than 10) are super important because they help narrow down the possible solutions.

By plugging in different values for 'a' and checking if the resulting 'b' met the criteria, we were able to identify the correct solutions. It’s like detective work, but with numbers! This method of systematically testing values is a valuable tool in mathematics, especially when dealing with equations that have constraints.

So, the final answer? The possible values for 'a' are 4, 5, and 6. Keep practicing these kinds of problems, and you'll become a pro at solving algebraic equations in no time. You got this! And remember, breaking down a problem into smaller, manageable steps can make even the trickiest questions seem a whole lot easier. Keep up the great work, and happy problem-solving!