Answer:
- Jim's: 7
- Duke's: 21
- Hank's: 11
Step-by-step explanation:
Let J, D, and H represent the drove sizes of Jim, Duke, and Hank.
2(H -6 + 1) = (J +6 -1) . . . . . Hank's offer
3(D -14 +1) = (H +14 -1) . . . . Duke's offer
6(J -4 +1) = (D +4 -1) . . . . . . Jim's offer
These equations can be simplified to ...
2H - J = 15
3D - H = 52
6J - D = 21
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<u>Solution</u>
Adding twice the second equation to the first gives ...
2(3D -H) +(2H -J) = 2(52) +(15)
6D -J = 119
Adding 6 times the third equation to this one gives ...
6(6J -D) +(6D -J) = 6(21) +(119)
35J = 245
J = 7
Using this value in the third equation gives ...
6(7) -D = 21
42 -21 = D = 21
Using the value of J in the first equation gives ...
2H -7 = 15
H = (15 +7)/2 = 11
Then the solution is (J, D, H) = (7, 21, 11).
Jim's drove had 7 animals; Duke's drove had 21 animals; Hank's drove had 11 animals.
