Algebra 2 word problem...help please?

Question

This question is fairly easy but i have a hard time figuring out how to work word problems so if some would could go step by step i would really appreicate it!

There are 40 cows and chickens in the farmyard. One quiet afternoon, Jack counted and found that there were 100 legs in all. How many cows and how many chickens are there?

a. Solve this problem by writing and solving an equation.

Answer 1

Cows have 4 legs. Chickens have 2.

Let's call cows x and chickens y

x+y = 40

4x+2y = 100

2x+2y = 80

2x = 20

x = 10

x + y = 40

y = 40-10

y = 30

10 cows, 30 chickens.

Answer 2

"40 cows and hens"

c + h = 40

"100 legs in all"

4c + 2h = 100

c = 40 - h

4(40-h) + 2h = 100

160 - 2h = 100

h = 30

c = 40-30 = 10

10 cows and 30 hens.

Answer 3

4x + 2y = 100

x + y = 40

(chickens = y; cows = x)

solve via system of equations.

(10 cows, 30 chickens)

Answer 4

do it yourself

Answer 5

Let a = number of cows, and 4a = number of cow legs

Then 40-a = number of chickens, and 2(40-a) = number of chicken legs.  The total number of legs would be 4a + 2(40-a) = 100.

Simplify to 4a + 80 - 2a = 2a + 80 = 100.  Subtract 80 from both sides.

2a = 20, so a = 10, the number of cows.  There are therefore 40-a = 40-10 = 30 chickens.

Check:  4 x 10 cows = 40 legs, 2 x 30 chickens = 60 legs.  Total = 100.