Question 116439
LET X, X+6 & X+12 BE THE 3 NUMBERS.
6<(3X+18)<50 
CHECK FOR THE LOWER BOUNDRY
6<3X+18
6-18<3X
-12<3X
-12/3<X
-4<X 
LET X=-3 & SEE IF IT IS A SOLUTION
-3+(-3+6)+(-3+12)<50
-3+3+9<50
9<50 SO -3,3&9 ARE ACCEPTABLE NUMBERS.
NOW WE TRY FOR THE UPPER LIMIT
X+(X+6)+(X+12)<50
3X+18<50
3X<50-18
3X<32
X<32/3
X<10.67
LET X=10
10+16+22<50
48<50 SO 10,16&22 ARE ALSO ACCEPTABLE NUMBERS.
THEREFORE THE ACCEPTABLE NUMBERS FOR X ARE ALL NUMBERS BETWEEN -3 & 10:
-3,-2,-1,0,1,2,3,4,5,6,7,8,9&10 ARE THE ANSWERS FOR X.