Question

the difference of two positive numbers is 3 and the difference of their reciprocals in 1/6 find the number​

Answer 1

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you are welcome in my ans

let two number be x and y

x - y = 3

1/x - 1/y = 1/16

xy = - 48

(x + y)^2 = (x - y)^2 + 4xy

(x + y)^2 = 9 - 192

(x + y)^2 = 183

(x + y ) = sqrt183

2x = sqrt 183 +3

x = (sqrt183 + 3)/2

y = (sqrt183 -3)/2

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Hope it may helps you

Answer 2

Given:

The difference between the two positive numbers is 3 and the difference between their reciprocals is 1/6.

To find:

The numbers.

Solution:

To determine the numbers we have to follow the below steps as follows-

Let the numbers be x and y.

The difference between the two positive numbers is 3.

So, we can write-

$x - y = 3$

The difference in their reciprocals is 1/6.

$\frac{1}{y} - \frac{1}{x} = \frac{1}{6} \\ \frac{x-y}{xy} = \frac{1}{6} \\ 6(x-y) = xy$

The value of x=3+y.

$6(3+y - y) = y \times (3 + y) \\ {y}^{2} + 3y = 18 \\ {y}^{2} + 3y - 18 = 0 \\ {y}^{2} + 6y - 3y -18 = 0\\y(y+6)-3(y+6)=0\\(y+6)(y-3)=0\\y=-6,3$

Hence the value of y is 3.

Thus, the value of x is x-3=3

x=6.

The numbers are 3 and 6.