The sum of two numbers is 18.


The sum of two numbers is 18. The sum of their reciprocals is 1/4. Find the numbers.


Let one of the number be then other number is (18 − x).

Then according to question,


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

$\Rightarrow 18 \times 4=18 x-x^{2}$

$\Rightarrow 72=18 x-x^{2}$

$\Rightarrow x^{2}-18 x+72=0$


$\Rightarrow x^{2}-12 x-6 x+72=0$

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


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


$\Rightarrow x=6$ or $x=12$

Since, being a number,


When x=12">x=12x=12 then another number will be


And when x=6">x=6x=6 then another number will be


Thus, the two numbers are 6 and 12.

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