Question

there is a total of 50 chickens and cows. If there are 118 legs, how many chickens are there?

Answer 1

I have to use the method of substitution and linear equations.

Answer 2

I CAME UP WITH:
Let x be chickens
let y be cows

1. 50x+y=118
2. 2x+4y=118
1. y=118-50x
2. 2x+4(118-50x)=118

is that right!?!

Answer 3

there are 50 chickens and cows altogether, total 50

so say h = chicken c = cows

h + c = 50

so we know that chicken have 2 legs, and cows have 4 legs
so the amount of chicken legs are
2 times whatever "h" is 2h

the amount of cows legs are
4 times whatever "c" is 4c

we know there is a total of 118 legs

so we could say that \(\bf 2h + 4c = 118\)

notice that a cow has twice as many legs as a chicken does
so one can say that \(\bf c = 2h\)

so \(\bf 2h + 4c = 118\implies 2h+4(2h)=118\)

Answer 4

hmmm ohh shoot... forgot something ahemm

Answer 5

anyhow

Answer 6

so cows have 4 legs, and chickens have two legs. Let cows be x and Chickens be n.
2n+4x=226
we can use the substitution property by changing n to 2(x-50)
so lets enter that in the equation.
2(50-x)+4x= 118
100-x + 4x = 118
we can simplify to
2x=18
divide by 2
x=9
so if there are 9 cows, then 50 - 9 =41.

Well I got 9 cows and 41 Chickens.
(i could be totally wrong)

Answer 7

I CAME UP WITH:
Let x be chickens
let y be cows

1. \(\bf x+y=50\)
2. 2x+4y=118

Answer 8

well we need to substitute, since you can't solve it with two variables.

Answer 9

so you can change the x to 2(50-y)