Call the three numbers [tex] s,\ m,\ l [/tex] (small, medium and large).
Their sum is 16, so we have
[tex] s+m+l=16 [/tex]
The largest is the sum of the other two, so we have
[tex] s+m=l \iff s+m-l=0 [/tex]
Finally, we know that three times the smaller (3s) is one less than the largest (l+1), so we have
[tex] 3s = l+1 \iff 3s-l = 1 [/tex]
So, we have the following system:
[tex] \begin{cases} s+m+l=16\\ s+m-l=0 \\ 3s-l = 1 [/tex]
Subtract the second equation from the first:
[tex] (s+m+l) - (s+m-l) = 16-0 \iff 2l = 16 \iff l = 8 [/tex]
Use this value for [tex] l [/tex] in the third equation to find [tex] s [/tex]:
[tex] 3s-l = 1 \iff 3s - 8 = 1 \iff 3s = 9 \iff s = 3 [/tex]
We know that the largest is the sum of the other two, so we have
[tex] s+m=l \iff 3+m=8 \iff m=5 [/tex]
So, the three numbers are 3, 5 and 8. You can check that they have all the required features:
- Their sum is 16: [tex] 3+5+8=16 [/tex]
- The largest is the sum of the other two: [tex] 8 = 3+5 [/tex]
- Three times the smaller is one more than the largest: [tex] 3\times 3 = 8+1 [/tex]
