Here's a start
as there are 100 animals
each animals cost and the total cost
Five hundred dollars were used to buy one hundred animals of three types. The burros cost 11 dollars each, the camels cost 9 dollars each, and the dogs cost 2 dollars each. How many animals of each kind were there?
This is a problem from a linear equation section of a number theory class I am in, and it have had no success in solving it.
noramlly you would need additional information to form a 3rd equation and solve them simultanuously.
Given the numbers for B,D & C are discrete whole numbers I would probably use trial to get a result.
You can explore this equation in 2 variables to find discrete solutions.
11b + 9c +2d = 500
b +c + d = 100
where a, b, c and all positive integers
the two equations above imply 9b + 7c = 300
Using Euclid's Extended Algorithm (I think that's what it's called) I found 5 solutions for (b,c): (31,3),(24,12),(17,21),(10,30), and (3,39).
So there are 5 solutions for (b,c,d): (31,3,66), (24,12,64),(21,17,62), (10,30,60), and (3,39,58).
To solve 9b+ 7c= 300:
7 divides into 9 once with remainder 2: 2= 9- 7
2 divides in 7 3 times with remainder 1: 1= 7- 3(2)
So 1= 7- 3(9- 7)= 4(7)- 3(9) (Of course: 28- 27= 1)
So, multiplying by 300, 1200(7)- 900(9)= 300. That means that b= -900, c= 1200 is a solution. Unfortunately, not a valid solution because a number of animals cannot be negative!
But b= -900+ 7k, c= 1200- 9k is also a solution for any integer k: 9b+ 7c= 9(-900+ 7k)+ 7(1200- 9k)= 9(-900)+ 7(1200)- 63k+ 63k= 300 as the "k" terms cancel.
In order that b= -900+ 7k be positive, k must be larger than 900/7: 129 is the smallest such number. In order that c= 1200- 9k be positive, k must be less than 1200/9: 133 is the largest such number. Take k= 129, 130, 131, 132, 133 to get the 5 valid answers.