Question

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Two natural numbers differ by 3 . Find the numbers, if the sum of their reciprocals is $7/10$.

Answer 1

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Let's consider the two natural numbers to be $x$ and $x+3$. (As they differ by 3 ) Then, from the question we have
$$\frac{1}{x}+\frac{1}{x+3}=\frac{7}{10}$$
$$\begin{array}{cc}& \frac{x+3+x}{x(x+3)}=\frac{7}{10}\\ & \frac{2x+3}{x^2+3x}=\frac{7}{10}\\ & 20x+30=7x^2+21x\\ & 7x^2+x-30=0\\ & 7x^2-14x+15x-30=0\\ & 7x(x-2)+15(x-2)=0\\ & (7x+15)(x-2)=0\end{array}$$
So, $7x+15=0$ or $x-2=0$
$$x=-15/7\text{or}x=2$$
As, $x$ is a natural number. Only $x=2$ is a valid solution.
Therefore, the two natural numbers are 2 and $5.$