Find three consecutive odd integers such that six times the sum of the first and second is 5 less than 11 times the third

Question

Find three consecutive odd integers such that six times the sum of the first and second is 5 less than 11 times the third

Answer 1

Let x = 1st odd integer

x+2 = 2nd

x+4 = 3rd

6(x +x+2) = 11(x+4) - 5 (six times sum of 1st and 2nd is (=) 11 times the 3rd takeaway 5)

6(2x+2) = 11x+44-5

12x+12 = 11x+39

x = 27 (if this was not an odd integer, then the problem has no solution)

x+2 = 29

x+4 = 31

6(56) = 336

11(31)-5 = 341-5=336