Homework Help

Post by **Kills On Site** » Fri Jan 13, 2006 4:31 am

Well just slid back into algebra, as Algebra II now. Doing pretty well, reviewing now, but this one homework problem is stumping me and it really shouldn't be. Hope the answer comes to someone. The problem is, you havea farm with 40 animals total. On this farm you have chickens and cows, the total leg count is 100. What equation could be set up to solve for this?

Guest · Post by **Guest** » Fri Jan 13, 2006 4:35 am

100 = x + y
40 = x/4 + y/2

x: # of cow legs
y: # of chicken legs

The answers are:

40 cow legs = 10 cows
60 chicken legs = 30 chickens

Post by **werldhed** » Fri Jan 13, 2006 4:38 am

c = cows
h = chickens (hens, i guess)

c + h = 40
2(2c + h) = 100

Now substitute...
h = 40 - c, therefore...
2(2c + 40 - c) = 100

c + 40 = 50

c = 10
h = 30

Guest · Post by **Guest** » Fri Jan 13, 2006 4:39 am

By the way, I solved it with a matrix, sorry

Post by **werldhed** » Fri Jan 13, 2006 4:41 am

I just wanted to make myself feel better about math, since I was having such a time with that other one.

Guest · Post by **Guest** » Fri Jan 13, 2006 4:49 am

Yeah, nothing like crushing easy problems

Post by **Kills On Site** » Fri Jan 13, 2006 4:56 am

Thanks guys, I had a slight fuckup and thought I could solve it with a single equation, forgot about substitution. I remember doing this back in Algebra/Trig freshman year, and prolly Algebra I in 8th grade. Just needed the refresher, thanks again.

Guest · Post by **Guest** » Fri Jan 13, 2006 5:03 am

No problem, calculus is more fun though

Post by **menkent** » Fri Jan 13, 2006 6:54 am

ToxicBug wrote:No problem, calculus is more fun though

yea, have fun with that linear algebra II. i'm SO glad i've wiped all that crap out of my brain. :icon29: