Question

The sum of two number 12 if the sum of their reciprocal is 3 upon 5
fine the number

Answer 1

Let the numbers be x and y

x+y = 12

1/x  +  1/y = 3/5

simplifying the second by lcm method,

(x+y)/xy = 3/5

12/xy= 3/5

xy = 20

x= 20/y

substituting in first eqn.

20/y  + y = 12

(20+y²)/y = 12

20+y² = 12y

y²-12y+20 = 0

y-10y-2y+20 = 0

(y-10) (y-2) = 0

y= 10 or y= 2

substituting in first eqn

x = 2 or 10

the numbers are hence 10 and 2

[10×2 = 20, 10+2 =12]

Answer 2

Hey there !!

Let the required numbers be x and y.

Then,

→ x + y = 12............(1) .

And, the sum of its reciprocal :-

→ $\bf \frac{1}{x} + \frac{1}{y} = \frac{3}{5} .$

⇒ $\bf \frac{ x + y }{xy} = \frac{3}{5} .$

⇒ $\bf \frac{12}{xy} = \frac{3}{5} .$

⇒ 3xy = 12 × 5 .

⇒ 3xy = 60 .

⇒ xy = 60/3 .

⇒ xy = 20.

Now, using identity :-

→ ( x - y )² = ( x + y )² - 4xy .

∴ ( x - y ) = $\bf \sqrt{(x + y )^{2} - 4xy  }$ .

⇒ x - y = $\sqrt{(12)^{2} - 4 \times 20 .$

⇒ x - y = √(144 - 80 ) .

⇒ x - y = √(64) .

⇒ x - y = 8..........(2).

On substracting the equation (1) and (2), we get

x + y = 12.

x - y = 8.

-  +     -

________

⇒ 2y = 4 .

∴ y = 2 .

Putting the value of y in equation (1), we get

⇒ x + 2 = 12.

⇒ x = 12 - 2 .

∴ x = 10 .

Hence, the required number are 10 and 2 .

THANKS

#BeBrainly.