We are given three natural numbers whose sum is 70. The first number is two-thirds of the second number, and the third number is the sum of the first and second numbers. We need to find these three numbers.

1. Problem Description

We are given three natural numbers whose sum is

7

0. The first number is two-thirds of the second number, and the third number is the sum of the first and second numbers. We need to find these three numbers.

2. Solution Steps

Let the three numbers be $a$, $b$, and $c$.

We are given that:

$a + b + c = 70$ (1)

$a = \frac{2}{3}b$ (2)

$c = a + b$ (3)

Substitute (2) and (3) into (1):

$\frac{2}{3}b + b + ( \frac{2}{3}b + b ) = 70$

$\frac{2}{3}b + b + \frac{2}{3}b + b = 70$

Combine like terms:

$(\frac{2}{3} + 1 + \frac{2}{3} + 1)b = 70$

$(\frac{4}{3} + 2)b = 70$

$(\frac{4}{3} + \frac{6}{3})b = 70$

$\frac{10}{3}b = 70$

Multiply both sides by $\frac{3}{10}$:

$b = 70 \cdot \frac{3}{10}$

$b = 7 \cdot 3$

$b = 21$

Now, find $a$ using (2):

$a = \frac{2}{3}b = \frac{2}{3}(21) = 2 \cdot 7 = 14$

$a = 14$

Now, find $c$ using (3):

$c = a + b = 14 + 21 = 35$

$c = 35$

We can check our answer:

$a + b + c = 14 + 21 + 35 = 70$

This matches the given information.

3. Final Answer

The three numbers are 14, 21, and 35.