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Question 6915: Hi,
I really need some help with my daughters homework. I have been away from alegbra for quite some time. Here is her problem.
A person has $100.00 to purchase 100 animals. Those animals are Horses for $10.00 each, Goats for $1.00 each, and Ducks 8 for $1.00. What formula would I use to solve this problem? I really need your help.
Thank you so much.
Answer by glabow(165)   (Show Source):
You can put this solution on YOUR website! I am surprised at your having a problem like this.
There is not enough information to solve in closed form.
However, you have the following information in algebraic form:
Let x be the number of horses bought
Let y be the number of goats bought
Let z be the number of ducks bought
You spend 10x + y + 1/8 z = 100 dollars
You buy x+y+z = 100 animals
So z = 100-x-y
Replacing z with 100-x-y in the first equation gives
10x + y + 100/8 - x/8 - y/8 = 100
Simplifying this gives
79x+7y=700
Since you can only buy whole animals (I assume) you are looking for integer x and y that satisfy this equation. A short table shows that
8 horses cost 632 leaving 68 -- not divisible by 7
7 horses cost 533 leaving 147 which equals 7 x 21
6 horses cost 474 leaving 226 -- not divisible by 7
5 horses cost 396 leaving 305 -- not divisible by 7
4 horses cost 316 leaving 384 -- not divisible by 7
3 horses cost 237 leaving 463 -- not divisible by 7
2 horses cost 158 leaving 542 -- not divisible by 7
1 horse costs 79 leaving 621 -- not divisible by 7
So, x=7 and y=21 are correct.
This means you buy 72 ducks (z=100-7-21).
Checking: 7x10 + 21 + 72/8 = 100 dollars
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