Question

find two parts of 15 such that the sum of their reciprocal is 3/10​

Answer 1

let the first part be x. Then the other part will be 15-x.

According to the question

$\frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10} \\ \frac{15 - x + x}{x(15 - x)} = \frac{3}{10} \\ \frac{15}{15x - {x}^{2} } = \frac{3}{10} \\ 150 = 3(15x - {x}^{2} ) \\ 50 = 15x - {x}^{2} \\ {x}^{2} - 15x + 50 = 0 \\ {x}^{2} - 5x - 10x + 50 = 0 \\ x(x - 5) - 10(x - 5) = 0 \\ (x - 5)(x - 10) = 0 \\ x = 5 \: or \: 10$

therefore the two required parts are

5 and 10

here is your answer :)