The sum of two numbers is 16. The sum of their reciprocals is $\frac{1}{3}$. Find the numbers.

Given:

The sum of two numbers is 16. The sum of their reciprocals is $\frac{1}{3}$.

 To do:

We have to find the numbers.

Solution:

Let one of the numbers be $x$.

This implies,

The other number $=16-x$.

According to the question,

$\frac{1}{x}+\frac{1}{16-x}=\frac{1}{3}$

$\frac{1(16-x)+1(x)}{x(16-x)}=\frac{1}{3}$

$\frac{16-x+x}{16x-x^2}=\frac{1}{3}$

$\frac{16}{16x-x^2}=\frac{1}{3}$

$3(16)=1(16x-x^2)$

$48=16x-x^2$

$x^2-16x+48=0$

Solving for $x$ by factorization method, we get,

$x^2-12x-4x+48=0$

$x(x-12)-4(x-12)=0$

$(x-12)(x-4)=0$

$x-12=0$ or $x-4=0$

$x=12$ or $x=4$

If $x=4$, then $16-x=16-4=12$.

If $x=12$, then $16-x=16-12=4$

The required numbers are $4$ and $12$.