Question 70032
Let the three numbers be A, B, and C with A being the smallest and C being the largest. From the problem description, you can write:
1) A+B+C = 47 The sum of three numbers is 47.
2) 3A = C+9 Three times the smallest is 9 more than the largest.
3) 2B = A+C+1 Two times the middle number is 1 more than the sum of the other two.

So, now you have three equations with three unknowns.
Rewrite equation 2) as: 
2a) C = 3A-9 and substitute into equation 3) and solve for B.
3a) 2B = A+(3A-9)+1 Simplify.
2B = 4A-8 Divide both sides by 2.
3b) B = 2A-4 Now substitute this B and the C from equation 2a) into equation 1) and solve for A.
A+(2A-4)+(3A-9) = 47 Simplify and solve for A.
6A-13 = 47 Add 13 to both sides.
6A = 60 Divide both sides by 6.
A = 10 Substitute this value of A into equation 3b) and solve for B and then into equation 2a) and solve for C.
B = 2(10)-4
B = 20-4
B = 16
C = 3(10)-9
C = 30-9
C = 21

The three numbers are:
10, 16, and 21

Check:
A+B+C = 10+16+21 = 47
3A = C+9
3(10) = 21+9
30 = 30
2B = A+C+1
2(16) = 10+21+1
32 = 32