A am four - digit number. My ones is twice my tens. My hundreds is five less than my ones. My thousands place is the sum of my tens and hundreds. What number I am?

Suppose we call the digits ABCD. We have
  D = 2*C
  B = D - 5
  A = C + B
Expressing these in terms of C, we have
  B = 2*C - 5
  A = C + 2*C - 5 = 3*C - 5
We know every digit must be in the range 0 to 9, so we have
  9 ≥ 2C - 5 ≥ 0
  14 ≥ 2C ≥ 5    (add 5)
  7 ≥ C ≥ 2.5    (divide by 2)

  9 ≥ 3C - 5 ≥ 0
  14 ≥ 3C ≥ 5    (add 5)
  4 2/3 ≥ C ≥ 1 2/3    (divide by 3)

In order to satisfy both these conditions, we must have C = {3, 4}. This gives rise to two possible solutions. For C=3, we have D=6, B=1, A=4. For C=4, we have D=8, B=3, A=7. The possible solutions are
  4136
  7348