SOLUTION: If there are 80 cows and hens and 208 legs, how many cows and hens are there?

Two ways.  Without algebra and with algebra:

Here's the way without algebra:

Cows and hens both have 2 front legs each,  but cows also
have two back legs each.

So since there are 80 animals there are 2•80 or 160 front legs.
The remaining 208-160=48 legs have to be back legs on the cows.
There are two back legs on each cow, so we divide 48 by 2 to get
the number of cows:


So there are 48÷2 = 24 cows and the remaining 80-24=56 animals are hens.

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Here's the way with algebra:

Let C = the number of cows
Let H = the number of hens.

Then

C + H = 80

Then the number of cows' feet = 4C
And the number of hens' feet = 2H

So

4C+2H = 208

Solve this system by substitution or elimination (addition):



and you'll get the same answer as with no algebra.

Edwin