If 37/13 = 2+1/(x+1/(y+1/z)),where x,y,z are natural number..what are x,y,z=?

• x = 1, y=5, z=2
  
  37/13 = 2+1/(x+1/(y+1/z))
  2 + 11/13 = 2 + 1/(x+1/(y+1/z))
  ===>11/13 = 1/(x+1/(y+1/z))
  
  Taking reciprocal on both sides
  13/11 = x+1/(y+1/z)
  1 + 2/11 = x + 1/(y+1/z)
  ===> x=1 and 1/(y+1/z) = 2/11
  
  1/(y+1/z) = 2/11
  Taking reciprocal
  
  y+1/z = 11/2
  y + 1/z = 5 + 1/2
  ===> y = 5
  
  1/z = 1/2
  z = 2
  
• 10 years agoHelpfull: Yes(5) No(1)