SOLUTION: Hi. I don't know how to set this up and solve it. Thank you. There are 38 animals in a farmyard. Some are cows and some are chickens. In total there are 104 legs. How many chick

Algebra ->  Linear-equations -> SOLUTION: Hi. I don't know how to set this up and solve it. Thank you. There are 38 animals in a farmyard. Some are cows and some are chickens. In total there are 104 legs. How many chick      Log On


   

Question 703427: Hi. I don't know how to set this up and solve it. Thank you.
There are 38 animals in a farmyard. Some are cows and some are chickens. In total there are 104 legs. How many chickens are there?
Found 2 solutions by mouk, MathLover1:
Answer by mouk(232) About Me  (Show Source):
You can put this solution on YOUR website!
Let H=number of chickens
and C=number of cows

"There are 38 animals in a farmyard":
so +C+%2B+H=38+

"In total there are 104 legs":
Chickens have two legs, cows have 4 legs
so, 2H%2B4C=104
or, H%2B2C=52

But from 1st equation C=38-H so 2C=76-2H
substituting for 2C in 2nd equation gives:
H%2B%2876-2H%29=52
76-H=52
H=76-52=24

So there are 24 chicken (and 14 cows)

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

There are 38 animals in a farmyard. Some are cows=x and some are chickens=y. In total there are 104 legs.
cows have four legs, so it will be 4x cows' legs
chickens have two legs, so it will be 2y chickens' legs
since there are 104 legs, we have
4x%2B2y=104......eq. 1
since given that there are 38 animals in a farmyard, we have
x%2By=38.........eq. 2
now solve this system:
4x%2B2y=104......eq. 1
x%2By=38.........eq. 2
______________________________
start with x%2By=38.........eq. 2 and solve for x
x=38-y.....substitute it in eq. 1
4%2838-y%29%2B2y=104......eq. 1...solve for y
152-4y%2B2y=104
152-2y=104
152-104=2y
48=2y
48%2F2=y
highlight%2824=y%29....so, there are 24 chickens
go back to x=38-y and find x
x=38-24
highlight%28x=14%29