# Pre-Algebra

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While visiting a farm, you notice that there are only chickens and rabbits in the farmyard. You can't help but wonder how many of each animal there is on the yard, but when you ask Farmer Fred he refuses to give you a direct answer. (Some say he must have been a math teacher before he became a farmer.) Fred says that there are 18 heads and 58 feet in the yard (not including themselves) Assuming there are no mutant animals how many chickens and rabbits are in the yard?

Okay, 18 heads = 18 animals my question is how do I go about finding this?

• Pre-Algebra -

Should I just do it randomly? Like first try 16 chickens and go on from there? Surely there is a simpler way to do this since a bigger problem wouldn't work this way.

• Pre-Algebra -

let the number of chickens be x
and the number of rabbits be y

so x+y = 18 and 2x + 4y = 58, divide this last equation by 2 to get x+2y=29

then
x+2y=29
x+ y=18 ..... subtract them
y=11, subbing back x = 7

7 chicken and 11 rabbits

check: 18 animals, so 18 heads
7chicken have 14 feet, 11 rabbits have 44 feet, so 14+44 = 58 feet

• Pre-Algebra -

Thank you!!

• Pre-Algebra -

There are 7 chickens- 14 Chicken feet.
There are 11 rabbits- 44 chicken feet.

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