Question

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

Answer 1

Let the two numbers be x and y.
According to the question, $x+y=18\dots \left(i\right)$
$\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\dots \left(\text{ii}\right)$
Eq. (ii) can be re - written as $\frac{y+x}{xy}=\frac{1}{4}\dots \left(\text{iii}\right)$
On putting the value of $x+y=18$ in Eq. (iii), we get $\frac{18}{xy}=\frac{1}{4}$
$\Rightarrow xy=72$
$\Rightarrow x=\frac{72}{y}$
On putting the value of $x=\frac{72}{y}$ in Eq. (i), we get,
$\frac{72}{y}+y=18$
$\Rightarrow 72+y^2=18y$
$\Rightarrow y^2-18y+72=0$
$\Rightarrow y^2-12y-6y+72=o$
$\Rightarrow y(y-12)-6(y-12)=0$
$\Rightarrow (y-6)(y-12)=0$
$\Rightarrow y=6$ and $12$
If $y=6,$ then $x=\frac{72}{6}=12$
If $y=12,$ then $x=\frac{72}{12}=6$
Hence, the two numbers are $6$ and $12.$