Question

Sum of two numbers is 18, if the sum their reciprocal is 1/4.Then what are the numbers?

Answer 1

two numbers are 12 and 6
12+6=18
12*6=72
a+b=18. a=18-b equation 1
reciprocal =1/a1/b=1/4
b+a/an=1/4
therefore b square -18b+72=0equation is formed
by factorisation we got values b=12 or b=6

put b=12 in equation 1
a=18-b
=18-12
a=6
therefore a=6 and b=12

Answer 2

Let the numbers be x and y.

Given that the sum of two numbers = 18.

x + y = 18 

x = 18 - y  ---- (1).


Given that the sum of their reciprocals is 1/4.

1/x + 1/y = 1/4

1/(18-y) + 1/y = 1/4

$\frac{18 - y + y}{(18-y)(y)}$ = 1/4

$\frac{18}{(18y-y^2)}$ = 1/4

18 * 4 = (18y - y^2)

72 = 18y - y^2

y^2 - 18y + 72 = 0
 
y^2 - 12y - 6y + 72 = 0

y(y - 12) - 6(y - 12) = 0

(y - 12)(y - 6) = 0

y = 12 (or) y = 6.



If y = 12:

Substitute y = 12 in (1) , we get

x = 18 - y

   = 18 - 12

   = 6.


If y = 6:

Then x = 18 - y

             = 18 - 6

            = 12.


Therefore the numbers are 6 and 12.


Verification:


$\frac{1}{x} + \frac{1}{y} = \frac{1}{4}$

[tex] \frac{1}{6} + \frac{1}{12} = \frac{1}{4} [/tex]

$\frac{2 + 1}{12} = \frac{1}{4}$

$\frac{3}{12} = \frac{1}{4}$

$\frac{1}{4} = \frac{1}{4}$


Hope this helps!