The sum of two numbers exceeds a third number by four. If the sum of the three numbers is at least 20 and at most 28, find any three integral values satisfying the inequality.
Let 1st, 2nd, and 3rd, be F, S, and T, respectively
Then we get: F + S = T + 4 ------ eq (i)
Also,
-------- Substituting T + 4 for F + S
8 ≤ T (3rd integer) ≤ 12
From the above, 5 scenarios exist. They are:
1) With T, or 3rd being 8, the sum of F (1st), and S (2nd) is 8 + 4, or 12. Use ANY 2 integers that sum to 12 to get the 1st and 2nd integers.
2) With T, or 3rd being 9, the sum of F (1st), and S (2nd) is 9 + 4, or 13. Use ANY 2 integers that sum to 13 to get the 1st and 2nd integers.
3) With T, or 3rd being 10, the sum of F (1st), and S (2nd) is 10 + 4, or 14. Use ANY 2 integers that sum to 14 to get the 1st and 2nd integers.
4) With T, or 3rd being 11, the sum of F (1st), and S (2nd) is 11 + 4, or 15. Use ANY 2 integers that sum to 15 to get the 1st and 2nd integers.
5) With T, or 3rd being 12, the sum of F (1st), and S (2nd) is 12 + 4, or 16. Use ANY 2 integers that sum to 16 to get the 1st and 2nd integers.