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

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by dweejareddy

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by dweejareddy

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Two numbers are : x and y

Sum is 16

⇒x+y = 16

⇒x = 16-y

Sum of reciprocals = 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**

Sum is 16

⇒x+y = 16

⇒x = 16-y

Sum of reciprocals = 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

ATQ;

x+y=16 .................................................(i)

x=16-y...................................................(ii)

and

+=1/3

(taking LCM)

->()=1/3

from (i)

->=1/3

->xy=48

from (ii)

->(16-y)y=48

->16y-=48

->-16y+48=0

->-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