The Proprietor of a managerie was asked how many 
parrots and how many elephants it included. He replied, 
" well, the lot have 32 heads and 100 feet." How many 
of each kind were there?

You don't need algebra to figure it out, but I will show 
you the algebra way. However first let's reason it out 
without using algebra.

Since there are 100 feet, that means there are fifty pairs 
of feet. Therefore there are 50 - 32 or 18 more pairs of 
feet than heads.  These 18 extra pairs of feet must belong
to the elephants.  So there are 18 elephants, which
accounts for 18 heads, thus the remaining 32 - 18 or 14 
heads must belong to the parrots.  

Answer: 18 elephants, 14 parrots

Check:

The 18 elephants have 18 heads and 18×4 or 72 feet
The 14 parrots have 14 heads and 14×2 or 28 feet.
That's 18+14=32 heads and 72+28=100 feet.

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Now, by algebra:

Let P = the number of parrots.
Let E = the number of elephants.

Parrots have one head, so there are P parrot heads.
Elephants have one head, so there are E elephant heads.
Since there are 32 heads in all,

P + E = 32

Parrots have 2 feet so there are 2P parrot feet.
Elephants have 4 feet so there are 4E elephant feet. 
Since there are 100 feet in all,

2P + 4E = 100

So you have the system of equations:

 P +  E =  32
2P + 4E = 100

Solve that system and get P = 14 and E = 18   

Edwin