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October 25, 2014

Posted by **lia mowin** on Wednesday, December 1, 2010 at 10:58pm.

- Math -
**Reiny**, Wednesday, December 1, 2010 at 11:56pmnumber of chicken --- x

number of cows ---- y

number of horses ---- 100-x-y

.01x + .10y + .50(100-x-y) = 5

times 100

x + 10y + 50(100-x-y) = 500

x+10y+5000-50x-50y = 500

-49x-40y = -4500

49x + 40y = 4500

We need only integer solutions here,

after a few trial and errors I got

x = 60

y = 39

so 60 chickens --- .60

39 cows and -----3.90

1 horse ---------0.50 for a total of $5.00

Is this from an 1850 textbook?

- Math -
**lia mowin**, Thursday, December 2, 2010 at 12:35amyes my teacher assigned it to my seventh grade class, thank you so much for your time!! =)

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