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Math Help - Linear equation problem

  1. #1
    Junior Member Nerdfighter's Avatar
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    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.
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  2. #2
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    Here's a start

    B+C+D = 100 as there are 100 animals

    11B+9C+2D = 500 each animals cost and the total cost
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  3. #3
    Junior Member Nerdfighter's Avatar
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    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.
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  4. #4
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    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.

    9B+7C = 300
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  5. #5
    Super Member Random Variable's Avatar
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    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).
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  6. #6
    Junior Member Nerdfighter's Avatar
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    Quote Originally Posted by Random Variable View Post
    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.
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  7. #7
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    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.
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  8. #8
    Junior Member Nerdfighter's Avatar
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    Thanks for the help everyone
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