Question

Two numbers differ by 3. the sum of their reciprocals is 7/10 . Find the no.
Thanks in Advance

Answer 1

let the numbers be x and y
given: x-y=3........(i) and 1/x + 1/y = 7/10....(ii)
from eqn..(i)
x = 3+y....(iii)
putting the value of eqn (iii) in eqn (ii)
(3+y+y)/(3+y)y= 7/10
(3+2y)/(3y+y^2) = 7/10
by cross multiplication
10(3+2y) = 7(3y + y^2)
30 + 20y = 21y + 7y^2
7y^2 + y - 30 = 0
7y^2 - 14y + 15y -30 = 0
7y( y -2 ) + 15( y -2) = 0
(7y + 15)(y - 2) = 0
equating both factors to zero we get
y = -15/7 and y = 2
x = 3-15/7=6/7 , x = 3+2=5
hence,
x = 5 and y = 2
the two numbers are 5 and 2
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Answer 2

yes lets solve this problem ...........

let one number be = x 

other numbers differs by 3 = x+3

sum of these reciprocals = 7/10

numbers in reciprocal form = 1/x , 1/x+3

according to the sum  the equation is ,

1/x + 1/x+3 = 7/10                 (take lcm of x,x+3)

3+1/ x +3 = 7/10          

4/x+3 = 7/10                            ( now cross multiply )

7x +21 = 40

7x = 40 -21

7x = 19

x= 19/7 = 2.72

this is your ans

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