Question 94817
Four brothers are born in successive intervals of three years. The eldest is twice the age of the youngest. What is the age of their father, who is eighteen years older than the eldest son?
:
Let x = age of the youngest
Then the remaining brothers' ages would be: x+3, x+6, x+9
: 
It says,"The eldest is twice the age of the youngest."
 (x+9) = 2x; you can do this in your head, but continuing in algebra
  9 = 2x - x
  x = 9 age of the youngest
:
Oldest, obviously is x + 9 = 18, + 18 = 36 is father