Question

the sum of two unmbers is 18 ,the sum of their reciprocal is 1/4.find the number.

Answer 1

let two numbers be x and y,their reciprocals be 1/x and 1/y
x+y=18
1/x+1/y=1/4
y+x/xy=1/4
18/xy=1/4
so xy=72 is eq2
from x+y=18 we get y=18-x
sub in eq 2
x(18-x)=72
18x-x^2=72
x^2-18x+72=0
x^2-6x-12x+72=0
x(x-6)-12(x-6)=0
so x=6 or 12
if x=6,y=12
so numbers are 6,12.
hope it helps
please mark it as brainliest

Answer 2

let the two numers be x and y

x+y=18

1/x+1/y=1/4

x+y/xy=1/4

4x+4y=xy

4(x+y)=xy

we know x+y =18-----eqn 1

xy=72------eqn 2

x+y=18

x= 18 -y  substitute in eqn 2

(18-y)*y= 72

18y-y²=72

y²-18+72 

y² - 12xy- 6y - 72 = 0

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

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

y = 6 and y= 12

x+y =18

substitute y value as  6 

thn x=12

substite y value as  12

thn x=6