Question

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

Answer 1

Here's your answer !!

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It's given that,

Sum of two number is 16.
$= > x + y = 16......(1)$

Now,

It is also given that,Sum of their reciprocal is 1/3.

$= > \frac{1}{x} + \frac{1}{y} = \frac{1}{3} \\ = > \frac{y + x}{xy} = \frac{1}{3} .......(2)$

By putting value of (x+y) in equation (2) ,we get :-

$= > \frac{16}{xy} = \frac{1}{3} \\ = > xy = 48$

We can write ,

$= > x = \frac{48}{y}$

Now,

By putting value of x in equation (1) ,we get :-

$= > \frac{48}{y} + y = 16 \\ = > \frac{48 + {y}^{2} }{y} = 16$

$= > 48 + {y}^{2} = 16y$

$= > {y}^{2} - 16y + 48 = 0 \\ = > {y}^{2} - (12y + 4y) + 48 = 0$

$= > {y}^{2} - 12y - 4y + 48 \\ = > y(y - 12) - 4(y - 12) \\ = > (y - 12)(y - 4)$

Therefore,

Y can be 12 or 4.

◆If y is 12
So,
x= 16-12=4

◆If y is 4.
So,
x=16-4=12

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Hope it helps you!! :)

Answer 2

Refers to attachment ⚘⚘