Answer 1

Let the smaller number be x then the other number be 3 + x.

Then according to question,

\[\frac{1}{x} - \frac{1}{3 + x} = \frac{3}{28}\]

\[ \Rightarrow \frac{3 + x - x}{x(3 + x)} = \frac{3}{28}\]

\[ \Rightarrow \frac{3}{3x + x^2} = \frac{3}{28}\]

\[ \Rightarrow 28 = 3x + x^2 \]

\[ \Rightarrow x^2 + 3x - 28 = 0\]

\[ \Rightarrow x^2 + 7x - 4x - 28 = 0\]

\[ \Rightarrow x(x + 7) - 4(x + 7) = 0\]

\[ \Rightarrow (x - 4)(x + 7) = 0\]

\[ \Rightarrow x - 4 = 0 \text { or } x + 7 = 0\]

\[ \Rightarrow x = 4 \text { or }x = - 7\]

Since, x being a natural number,

Therefore, x = 4.

Then another number will be \[3 + x = 3 + 4 = 7\]

Thus, the two natural numbers are 7 and 4.