Question

The sum of two number is 12 if the sum of their reciprocal is3_5 FIND NUMBER

Answer 1

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.

Answer 2

Let that  two numbers  be  x and  y

According  to question,

Sum of  numbers  = 12

x + y = 12

y = 12 - x  .......................i)

And,

Sum  of  their  reciprocal = 3

$\frac{1}{x}+\frac{1}{y}=5$

Putting  y = 12 -x,

$\frac{1}{x}+\frac{1}{12-x}=\frac{3}{5}\\\;\\\frac{12-x+x}{x(12-x)}=\frac{3}{5}\\\;\\\frac{12}{12x-x^{2}}=\frac{3}{5}\\\;\\\frac{4}{12x=x^{2}}=\frac{1}{5}\\\;\\12x-x^{2}=20\\\;\\12x-x^{2}-20=0\\\;\\x^{2}-12x+20\\\;\\x^{2}-10x-2x+20\\\;\\x(x-10)-2(x-10)\\\;\\(x-2)(x-10)\\\;\\x=2\;\;or\;\;x=10$

Hence,

Numbers are  10 and  2