The sum of two natural numbers is 15 and the sum of their reciprocals is 3/10. Find the numbers.

Let the required natural numbers x and (15 – x)

According to given condition,

taking LCM

cross multiplying

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = – 15 c = 50

= 1.50 = 50

And either of their sum or difference = b

= – 15

Thus the two terms are – 10 and – 5

Sum = – 10 – 5 = – 15

Product = – 10. – 5 = 50

x(x – 10) – 5(x – 10) = 0

(x – 5) (x – 10) = 0

(x – 5) = 0 or (x – 10) = 0

x = 5 or x = 10

Case I: when x = 5

15 – x = 15 – 5 = 10

Case II: when x = 10

15 – x = 15 – 10 = 5

Hence required numbers are 5 and 10.

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