SOLUTION: There are 11 animals in a barnyard. Some are chickens and some are cows. There are 38 legs in all.
Let x be the number of chickens
Let y be the number of cows
How many of ea
Question 425350: There are 11 animals in a barnyard. Some are chickens and some are cows. There are 38 legs in all.
Let x be the number of chickens
Let y be the number of cows
How many of each animal are in the barnyard? Answer by Gersid(33) (Show Source):
You can put this solution on YOUR website! x = number of chickens, each with two legs.
y = number of cows each with four legs.
1) 2x+4y = 38 "There are 38 legs in all."
2) x+y = 11 "There are 11 animals in a barnyard."
Rewrite equation 2) as:
2a) x = 11-y and substitute into equation 1) then solve for y.
1a) 2(11-y)+4y = 38 Simplify.
1b) 22-2y+4y = 38 Add the y-terms and subtract 22 from both sides.
1c) 2y = 16
1d) y = 8
2b) x = 11-y
2c) x = 11-8
2d) x = 3
There are 3 chickens and 8 cows.
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