Question 374097
Find two integers whose product is 104 such that one of the integers is three less than twice the other integer.
Sol'n:
  Let x = the 1st integer
      y = the 2nd integer
  xy = 104 ----(1)
  x = 2y - 3 ----(2)
substitute (2)into (1)
  y(2y-3)=104
  2y^2 - 3y - 104 = 0, factor
  (y-8)(2y+13)= 0
   y = 8 & -6.5
   from (1)
  x = 104/y, subs. the value of y to find the value of x.
  let y = 8, 
      x = 104/8 = 13
  let y = -6.5
      x = 104/(-6.5) = -16
and finally:
    when x = 13, y = 8      -----answer
and, when x = -16, y = -6.5 -----answer