Answer:
(2k-6)/3
Explanation:
Let the 3 consecutive odd integers be n, n+2 and n+4
Sum: n + n+2 + n+4 = k
3n+6 = k
3n = k - 6
n = (k - 6)/3
The 2 smaller of these integers are n and n+2
n = (k - 6)/3
n+2 = (k -6)/3 + 2/1 = (k-6+6)/3 = k/3
Sum of the 2 smaller integers
(k - 6)/3 + k/3 = (k-6+k)/3 = (2k-6)/3