# 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 TI-83 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 TI-83 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.