Question 212940
Let x = first odd integer
x+2 = second 
x+4 = third


The sum of these integers (add them up!) is 3x+6.


You must find x such that 
135<3x+6<147


Bubtract 6 from the center, left, and right parts of this inequality:
135-6<3x+6-6<147-6
129<3x<141


Divide by 3:
129/3 <x<141/3
43< x< 47


Now remember, x has to be an ODD NUMBER, BETWEEN (not including!) 43 and 47!!


So the first number must be x=45 
The second number must then be x+2=  47
The third number is x+4 = 49


The solution:   45, 47, 49.
The sum is :     141


R^2


Dr. Robert J. Rapalje, Retired
Seminole Community College
Altamonte Springs Campus
Florida