Question 1183093: 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.
Found 2 solutions by Boreal, MathTherapy:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! numbers are x, y, z
x+y=z+4
x+y+z=20; x+y+z=28
substitute
2z+4=20
z=8
z+4=12
so 2 integers whose sum is 12
5 and 7, with z=8
4 and 8, with z=8
----
2z+4=28
z=12
x+y=16
0 and 16 works with z=12
1 and 15 works with z=12
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can set up for other integers between 20 and 28
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website! 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.
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