Required Mathematical Intelligence

Chapter 122 The Christmas Turkey Problem
(USA) Westerners regard Christmas as their most important holiday.John, Peter and Rob went to market early in the morning before Christmas to sell their turkeys.These turkeys are about the same weight, so they are sold only.

Among them, John had 10, Peter had 16, and Rob had 26.In the morning, the three sold at the same price.After lunch, since the three of them hadn't sold out and had to go home before dark, they had to sell at a reduced price, but the selling price of the three was still the same.At dusk they were all sold out of turkey.

When the money was counted, they were surprised to find that everyone got £56.Think about it, why?How much are their prices in the morning and afternoon?How many turkeys did each person sell in the morning and in the afternoon?
[Answer: If John, Peter, and Rob sell x, y, z turkeys in the morning, then they each sell 10—x, 16—y, 26—z turkeys in the afternoon.And if it is set that the price in the morning is £a each, and the price in the afternoon is £b each.From the meaning of the question, the following equations can be obtained:
ax b (10——x)=56①

ay b (16——y)=56②

az b(26——z)=56③

This is an indeterminate system of equations with 5 unknowns but only 3 equations.

①——③ get (xz) (ab)=16b, ④
②——③Get(yz)(ab)=10b, ⑤
④÷⑤得（x-z）/（y-z）=8÷5，即5x 3z=8y。⑥
According to the conditions of the question, 0<x<10, 0<y<16, 0<z<26, after substituting into ⑥ test, it can be found that only x=9, y=6, z=1 is the only set of solutions, Substitute the values ​​of x, y, and z into ① and ② to calculate a=6, b=2. Therefore, the price is 6 pounds each in the morning and 2 pounds each in the afternoon.John, Peter, and Rob each sold 9, 6, and 1 turkey in the morning, and 1, 10, and 25 turkeys in the afternoon. ]
(End of this chapter)