Question

The sum of two numbers is 16 and the sum of their reciprocals is 1/3 .Find the numbers. 

Answer 1

Two numbers are : x and y
Sum is 16
⇒x+y = 16
⇒x = 16-y
Sum of reciprocals = 1/3

⇒$\frac{1}{x} + \frac{1}{y} = \frac{1}{3}$

⇒$\frac{1}{16-y} + \frac{1}{y} = \frac{1}{3}$

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

⇒$\frac{16}{y(16-y)} = \frac{1}{3}$

⇒16×3 = 16y - y² 

⇒ y² - 16y + 48 = 0

⇒ y² - 12y - 4y + 48 = 0

⇒ (y-12)(y-4) = 0 

⇒ y=4 or 12 

⇒x=12 or 4 

So numbers are 4,12

Answer 2

Let the two numbers be x and y
ATQ;
x+y=16 .................................................(i)
x=16-y...................................................(ii)
and
$\frac{1}{x}$+$\frac{1}{y}$=1/3
(taking LCM)
->($\frac{x+y}{xy}$)=1/3
    from (i)
->$\frac{16}{xy}$=1/3
->xy=48
from (ii)
->(16-y)y=48
->16y-$y^{2}$=48
->$y^{2}$-16y+48=0
->$y^{2}$-12y-4y+48=0
->y(y-12)-4(y-12)=0
->y=12,4
Hence the value of x and y is 12 and 4