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.
<pre>Let 1st, 2nd, and 3rd, be F, S, and T, respectively
Then we get: F + S = T + 4 ------ eq (i)
Also, {{{20 <= F + S + T <= 20}}}
      {{{20 <= T + 4 + T <= 28}}} -------- Substituting T + 4 for F + S
    {{{matrix(4,1, 20 <= 2T + 4 <= 28, 20 - 4 <= 2T <= 28 - 4, 16 <= 2T <= 24, 16/2 <= T <= 24/2)}}}
   <font size = 4><font color = red>8 ≤ T (3rd integer) ≤ 12</font></font>
From the above, 5 scenarios exist. They are:
<b>1)</b> 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.
<b>2)</b> 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.
<b>3)</b> 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.
<b>4)</b> 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.
<b>5)</b> 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.