## Question:

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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?

### Tag:

**Asked About:**Algebra 2, Pg 23, Ex 54

Method: system of equations

cows (x) = 4 legs, chickens (y) = 2 legs

x + y = 40

4x + 2y = 100

In other words # of cows + # of chickens = 40 animals, AND the # of cows times 4 legs + # of chickens times 2 legs = 100 legs

Subtract the equations. You'll get 3x + y = 60. Solve for y --> y = 60 - 3x.

Plug this into either of the original equations (x + y = 40 is the easier choice). You'll get x=10. Now plug this into the original equation again (x+y=40) and you'll get y=30.

Answer: 10 cows, 30 chickens

cows (x) = 4 legs, chickens (y) = 2 legs

x + y = 40

4x + 2y = 100

In other words # of cows + # of chickens = 40 animals, AND the # of cows times 4 legs + # of chickens times 2 legs = 100 legs

Subtract the equations. You'll get 3x + y = 60. Solve for y --> y = 60 - 3x.

Plug this into either of the original equations (x + y = 40 is the easier choice). You'll get x=10. Now plug this into the original equation again (x+y=40) and you'll get y=30.

Answer: 10 cows, 30 chickens