Question 546938
Let 
First odd integer = x+1
Second odd integer = x+3
Third odd integer = x+5
Given
Three times the smallest number is the same as nine more than twice the largest
3(x+1)=2(x+5)+9
3x+3=2x+10+9
3x+3=2x+10+9
3x-2x=19-3
x=16


First odd integer = x+1 = 16+1=17
Second odd integer = x+3=16+3=19
Third odd integer = x+5=16+5=21