Math
posted by Miranda .
There are 18 animals in the barnyard. There are 50 legs. How many cows and how many chickens are there in the barnyard?
Technically, I wonder if a chicken is an animal. And are chickens and cows the ONLY "animals" in the barnyard.
If we assume a chicken is an animal and that we have only chickens and cows in the barnyard, then let
X = number of chickens
Y = number of cows.
You can write two equations.
X + Y = 18 (total chickens and cows)
2X + 4Y = 50 (number of legs).
Solve for X and Y. Post your work if you get stuck.
Put your equations in the TI83 plus put it in matrices. Go to Edit then type in where it says 1X1 put 2 X3 then type in the first equation like puting it straight across.Like 1 1 18 and then you do the second equations like you did the first equation. Then exit out then go back to matrices go to B mash go back the matrices then go to names then go to the thing that you type in mash that then mash enter again. The answer for x and y is at the end the answer to this problem is x=11 and y=7.
Put your equations in the TI83 plus put it in matrices. Go to Edit then type in where it says 1X1 put 2 X3 then type in the first equation like puting it straight across.Like 1 1 18 and then you do the second equations like you did the first equation. Then exit out then go back to matrices go to B mash go back the matrices then go to names then go to the thing that you type in mash that then mash enter again. The answer for x and y is at the end the answer to this problem is x=11 and y=7.
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