Find three consecutive odd integers such that the largest decreased by three times the second is 47 less than the smallest

X = smallest y = second z = biggest

x = x

y = x + 2

z = x + 4

(x+4) - 3 (x+2) = x - 47

x+4 - 3x - 6 = x - 47

-2x - 2 = x - 47

-3x - 2 = - 47

-3x = - 45

x = 15

x = 15

y = 15 + 2

z = 15 + 4

x = 15

y = 17

z = 19