Please show your work!

-1. There are several chickens and rabbits in a cage (with no other types of animals). There are 72 heads and 200 feet inside the cage. How many chickens are there, and how many rabbits?

-2. You've been asked to buy 100 squeegies, using 100 dollars to do so. You may buy no more or less than 100 squeegies, and the total price must be exactly 100 dollars. There is no sales tax. Red squeegies cost $6.00. Yellow squeegies cost $3.00. Blue squeegies cost $0.10. How many of each must you buy?

-3. On a man's tombstone, it is said that one sixth of his life was spent in childhood and one twelfth as a teenager. One seventh of his life passed between the time he became an adult and the time he married; five years later, his son was born. Alas, the son died four years before he did. He lived to be twice as old as his son did. How old did the man live to be?

1. c=chickens r=rabbits
c+r=72 (1)
2c+4r=200 (2)
2c+2r=144 (3)from(1)x2
2r=56 (2)-(3)
r=28
c=44
44 chickens 28 rabbits

I'm hoppy you got that one right, manxman!

2. r+y+b=100 (1)
6r+3y+0.1b=100 (2)
60r+30y+b=1000 (3) 10x(2)
subtracting (1) from (3) gives
59r+29y=900 (r and y must be whole numbers)
solving this gives r=1 and y=29 which means b=70

I'd like a yellow one, please.

3: Seems to be a "trick" question (make it appear difficult).

x = age at death:
since 1/12 of age was spent as teenager (13 to 19 : 7 year span),
then (1/12)*x = 7 : x = 12*7 = 84

You spotted the trick, Denis!

Of course, it could also have been solved the regular way:

Let x be the age of the man at the time of his death. His age consists of the number of years he spent as a child plus the number of years he spent as a teenager plus the number of years between his coming of age and his marriage plus five plus his son's age (half his own) plus four. In mathematical terms, this is:
x = x/6 + x/12 + x/7 + 5 + x/2 + 4
x = 14x/84 + 7x/84 + 12x/84 + 42x/84 + 9
x = 75x/84 + 9
84x/84 - 75x/84 = 9
9x/84 = 9
x/84 = 1
x = 84