The sum of two numbers is 18. The sum of their reciprocals is $\frac{1}{4}$. Find the numbers.

Given:

The sum of two numbers is 18. The sum of their reciprocals is $\frac{1}{4}$.

 To do:

We have to find the numbers.

Solution:

Let the two numbers be $x$ and $18-x$.

According to the question,

$\frac{1}{x}+\frac{1}{18-x}=\frac{1}{4}$

$\frac{1(18-x)+1(x)}{x(18-x)}=\frac{1}{4}$

$\frac{18-x+x}{x(18-x)}=\frac{1}{4}$

$\frac{18}{18x-x^2}=\frac{1}{4}$

$4(18)=1(18x-x^2)$

$72=18x-x^2$

$x^2-18x+72=0$

Solving for $x$ by factorization method,

$x^2-12x-6x+72=0$

$x(x-12)-6(x-12)=0$

$(x-12)(x-6)=0$

$x-12=0$ or $x-6=0$

$x=12$ or $x=6$

The required numbers are $6$ and $12$.