Linear equation problem

• May 31st 2009, 04:23 PM
Nerdfighter
Linear equation problem
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.
• May 31st 2009, 05:45 PM
pickslides
Here's a start

\$\displaystyle B+C+D = 100\$ as there are 100 animals

\$\displaystyle 11B+9C+2D = 500\$ each animals cost and the total cost
• May 31st 2009, 05:47 PM
Nerdfighter
I've successfully done that, but have been trying to figure out how to combine them or get a solution from them. I've mainly tried to combine them and solve for variables in an augmented matrix, but have had no luck so far.
• May 31st 2009, 06:15 PM
pickslides
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.

\$\displaystyle 9B+7C = 300\$
• May 31st 2009, 06:30 PM
Random Variable
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).
• May 31st 2009, 08:38 PM
Nerdfighter
Quote:

Originally Posted by Random Variable
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).

I know a variation of the Euclidean Algorithm, and I am curious as to what you did to get these 5 solutions. That does match with what my teacher told me, he said there would be five solutions.
• Jun 1st 2009, 04:54 AM
HallsofIvy
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.
• Jun 4th 2009, 06:30 PM
Nerdfighter
Thanks for the help everyone :)