An algebra teacher drove by a farmyard full of chickens and pigs. the teacher happened to notice that there were a totals of 100 heads and 270 legs. how many chickens were there? how many pigs were there ?

I'll be honest, this is a pretty interesting question. I've never seen anything like it before!

We can start off by making equations with the given. If there are 100 heads, that means that the total number of chickens and pigs combined is 100. Thus, we can form our first equation:

p + c = 100
('p' represents number of pigs, and 'c' represents number of chickens)

We also know that there are 270 legs total. Since chickens have 2 and pigs have 4, we can make the following equation:
2c + 4p = 270

I suppose the quickest way to solve for this equation is to use the subtraction method. Put the equations on top of each other:

2c + 4p = 270
c + p = 100

Multiply the bottom equation by -2;

2c + 4p = 270
-2c -2p = 100

"Add" these two equations together:

2c + 4p = 270
-2c -2p = -200

2p = 70

Divide both sides by 2:
p = 35

Ah, finally. Now that we know that there are 35 pigs, input this variable into the first equation to find the number of chickens:

c + 35 = 100
Subtract:
c = 65

There are 35 pigs and 65 chickens
-T.B.

The question is kind of confusing. if it's asking how many variations of answers can you give. then do 4^5 for your answer.this would be something like the first question could be a and the 2nd c or first b next c etc. if you can, ask the teacher for a better confirmation of what it's asking. if i'm assuming correctly what it's asking, it's 625 different variations.