Sum of two natural numbers is $8$ and the difference of their reciprocal is $\frac{2}{15}$. Find the numbers.

Let 1st no. be $x$ and 2nd be $8-x$.

By the given condition,

$\frac{1}{x}-\frac{1}{8-x}=\frac{2}{15}$

$\Rightarrow \frac{8-x-x}{x(8-x)}=\frac{2}{15}$

$\Rightarrow \frac{8-2x}{8x-x^2}=\frac{2}{15}$

$\Rightarrow 120-30x=16x-2x^2$
$2x^2-46x+120=0$
$x^2-23x+60=0$
$x^2-20x-3x+60=0$
$x(x-20)-3(x-20)=0$
$(x-3)(x-20)=0$

$x=3$ or $x=20$ not possible

$1^{st}$ no. $=3$ and $2^{nd}$ no. $=8-3=5$.