Question

the sum of two numbers is 16 and the sum of their reciprocals is 1/3 find the numbers

Answer 1

Answers:
    Let one number be x.
therefore , another number = 16 - x 
       (because sum of two numbers is 16)
also ,
       1/x + 1/16 - x  = 1/3 (Given)
   16 -x +x / x(16 -x) = 1/3
   16*3 = x(16 - x)
   48 = 16x - x²
or x² - 16x + 48 = 0
or x² - 12x - 4x + 48 =0
 x(x-12) -4(x-12) =0
x-4 = 0 or x-12 =0
x=4 or x=12
hence one Number is 4 .
and another number = 16 - x =16 - 4 = 12 .
Hope this helps you!!

Thank You!!

Answer 2

Let one number be x and the other be 16-x

their reciprocals will be 1/x and1/16-x

sum of reciprocals is given as 1/x + 1/16-x = 1/3

Explanation:

$\frac{1}{x} + \frac{1}{16-x} = \frac{1}{3}$

solving this,

$\frac{16-x+x}{(x)(16-x)} = \frac{1}{3}$

$\frac{16}{16x-x^2} = \frac{1}{3}$

by cross multiplication,

$48 = 16x - {x}^{2}$

by taking (-) sign common,

${x}^{2} - 16x + 48 = 0$

by factorisation,

${x}^{2} - 12x - 4x + 48 = 0$

$x(x - 12) - 4(x - 12) \\ (x - 4)(x - 12)$

value of x from this will be 4 and 12

other number be 16-x= 16-4=12

in the case if x=12, 16-12=4