Question

The sum of two numbers is 16. The sum of their reciprocals is $\frac{1}{3}$. Find the numbers.

Answer 1

SOLUTION :

Given : Sum of the two natural numbers is 16

Let the two natural numbers be x and (16 - x) .

Their reciprocals are 1/x  & 1/(16 - x)

A.T.Q

1/x + 1/(16 − x) = 1/3

(16− x + x) / x(16 − x) = ⅓

[By taking LCM]

16/ 16x − x² = ⅓

16x − x² = 3 × 16

16x - x² = 48

- x² + 16x -  48 = 0

x² - 16x +  48 = 0

[Multiplying by - 1]

x² - 12x  - 4x +  48 = 0

x(x - 12) - 4(x - 12) = 0

[By middle term splitting]

(x - 12)(x - 4) = 0

(x - 12) or (x - 4) = 0

x = 12  or x = 4  

Hence, the two numbers are 4 and 12.

HOPE THIS  ANSWER WILL HELP YOU..

Answer 2

Answer:12;4

Step-by-step explanation:

Le the no. Be x and y

Now...

X+y=16

X=16-y

Also

1/x + 1/y = 1/3

(X+y)/xy = 1/3

Xy/x+y = 3

(16-y)y / 16-y+y = 3

16y-y^2 = 48

Y^2 -16y + 48 = 0

Middle term splitting

Y^2 -12y-4y+48=0

Y(y-12) -4(y-12) = 0

Y= 4 , 12

Y = 4 y= 12

X= 16-y=12 x=16-y=4

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