Question

The sum of two numbers is 9. The sum of their reciprocals is $\frac{1}{2}$. Find the numbers.

Answer 1

SOLUTION :

Given : Sum of two numbers is 9 & Sum of their reciprocals is 1/2.

Let the one number be x and the other number be (9 - x).

Their reciprocals be = 1/x & 1/(9 - x)

A.T.Q

1/x + 1/(9 - x) = 1/2

(9 - x + x)/(9 - x)x = 1/2

[By taking L. C. M]

9/(9 - x)x = 1/2

9/ (9x - x²) = 1/2

9 × 2 = 9x - x²

18 = 9x - x²

x² - 9x + 18 = 0

x² - 6x - 3x - 18 = 0

[By middle term splitting]

x(x - 6) - 3(x - 6) = 0

(x - 6) (x - 3) = 0

(x - 6)  = 0  or (x - 3) = 0

x = 6 or x = 3

Case 1 :  

When x = 6 , then other number be (9 - x) = 9 - 6 = 3  

Case 2 :

When x = 3 , then other number be (9 - x) = 9 - 3 = 6

Hence, the two numbers are (6, 3) & (3,6) .

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Hey mate here's your answer

Let the numbers be = x and y

then as given

x + y = 9

and

$\frac{1}{x} + \frac{1}{y} = \frac{1}{2}$


Then from above's information


$\frac{1}{x} + \frac{1}{y} = \frac{1}{2} \\ \\ = > \frac{y + x}{xy} = \frac{1}{2} \\ \\ then \: from \: above \: by \: placing \: value \: \\ of \: x + y = 9 \\ \\ = > \frac{9}{xy} = \frac{1}{2} \\ \\ = > 18 = xy \\ \\ \: now \: as \:x + y = 9 \: \\ \: and \: also \: xy =18 \\ \\ then \: possible \: values \: of \: x \: and \: y \: by \: hit \: \\ and \: trial \: = 6 \: and \: 3$
So if

x = 3 then , y = 6

and if

x = 6 then y = 3


Hope it helps