Question

Five years ago a man was seven times as old as his son. Five years hence the father would be three times as old as his son. Find their present ages.

Answer 1

let the age of man be x
let age of his father be y

5 years ago,
his age=x-5
his son's age=y-5

ATQ
x-5=7(y-5)
x-5=7y-35
x=7y-30-(1)

now,
5 years hence,
his age=x+5
his son's age=y+5

ATQ
x+5=3(y+5)
x+5=3y+15
(7y-30)+5=3y+15(from (1))
7y-25=3y+15
7y-3y=25+15
4y=40
y=10

now put x in (1)
x=(7×10)-30
x=70-30
x=40

so,
man's age=40
his son's age=10

Answer 2

let the present age of father be x years and that of son be y years
five years ago,
x-5=7(y-5)
x-5=7y-35
x-7y= -30 ............1
five years hence,
x+5=3(y+5)
x+5=3y+15
x-3y=10............2
solving equations 1 and 2
-4y= -40
y=10
substituting y=10 in equation 2
x-3(10)=10
x-30=10
x=40
the present age of father is 40 years
and the present age of son is 10yrs