Question 436109
If the three numbers are consecutive and odd, and we let the first number be represented by x, then the second number would be 2 more than the x and the third number would be 4 more than x.  The integers we are looking for are:
x, (x+2), (x+4)

We want the sum of these to equal to "five less than twice the smallest integer" (we represented the smallest integer by x).  This gives us the equation
x + (x+2) + (x+4) = 2x - 5

Solve the equation for x:
x + (x+2) + (x+4) = 2x - 5
3x + 6 = 2x - 5
3x - 2x = -5 - 6
1x = -11

The integers we are looking for are -11, -9, -7, found by substituting -11 for x:
x = -11
x + 2 = -11 + 2 = -9
x + 4 = -11 + 4 = -7

Check your answer:  
The sum of the integers we found is:  -11 + (-9) + (-7) = -27
We wanted it equal to five less than twice the smallest integer.  The smallest integer is -11, twice the smallest integer is -22, less 5 is -27.