Question

Find three consecutive odd integers such that six times the second decreased by twice the first is equal to twenty more than the sum of the second and third

Answer 1

let
x--------> the first odd integer
x+2-----> the second odd integer
x+4-----> the third odd integer

we know that
6(x+2)-2x=20+(x+2)+(x+4)
6x+12-2x=26+2x
4x-2x=26-12
2x=14
x=7

the answer is
the numbers are
7, 9 and 11