https://file.aiquickdraw.com/mathbot/file/172568724593379brrhxb.jpg

The problem from the image is:

"One natural number is 3 times another natural number, and their sum is 84. Find the natural numbers."

### Solution:

Let the two natural numbers be \( x \) and \( y \), where \( x \) is 3 times \( y \). Therefore, we can write the following equations:

1. \( x = 3y \)
2. \( x + y = 84 \)

Substitute the value of \( x \) from equation (1) into equation (2):

\[
3y + y = 84
\]
\[
4y = 84
\]
\[
y = \frac{84}{4} = 21
\]

Now, substitute \( y = 21 \) into equation (1) to find \( x \):

\[
x = 3 \times 21 = 63
\]

Thus, the two natural numbers are **21** and **63**.

Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:
1. How can we generalize this approach for different multipliers other than 3?
2. What if the sum of the two numbers were different, say 90?
3. Can you set up a similar problem where the first number is 4 times the second?
4. How would the solution change if the sum were given as 100 instead of 84?
5. What is the smallest value for the sum that would still have integer solutions?

**Tip:** For these types of problems, it's helpful to translate word problems into algebraic equations and solve step by step.

One natural number is 3 times another natural number, and their sum is 84. Find the natural numbers.

Solve this algebraic problem where one natural number is 3 times another, and their sum is 84. Follow a step-by-step solution using linear equations to find the two natural numbers. Suitable for students in Grades 6-8.

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