Question

divide 27 into two parts such that the sum of their reciprocal is 3 / 20​

Answer 1

Answer:

15 and 12

Step-by-step explanation:

hope this answer helps you

Answer 2

Answer:

$\huge\red{Question}$

• Divide 27 into two parts such that the sum of their reciprocal is 3 / 20

$\huge\blue{Answer}$

Let one part be x and another part b 27-x

$\mathfrak{so}$

$- > \frac{1}{x} + \frac{1}{27 - x} = \frac{3}{20}$

$- > \frac{27 - x + x}{ \times (27 - x)} = \frac{3}{20}$

$- > \frac{27}{x(27 - x)} = \frac{3}{20}$

$180 = x(27 - x)$

$- > {x}^{2} - 27x + 180 = 0$

{by middle term break theorem we can conclude}

$- > {x}^{2} - 15x - 12x + 180 = 0$

$- > ({x}^{2} - 15x)( - 12x + 180) = 0$

$- > x(x - 15) - 12(x - 15) = 0$

$- > (x - 15)(x - 12) = 0$

$\mathfrak{so}$

$\boxed{ = > x \: = \: 12 \: or \: x \: = \: 15}$

$\texttt{so, two parts are 15 and 12}$