Jordan count 28 legs on 8 animals if the animals are either zebras or pecocks how many zabras are there

This is obviously a problem for system of two equations.

Let

Z=number of zebras (4 legged)

P=number peacocks (2 legged).

We know that

P=8-Z (total of 8 animals)

and

4Z+2P=4Z+2 (8-Z) = 28

expand and solve for Z

4Z+16-2Z=28

2Z=12

Z=6 There are 6 zebras.

Another way to solve, without pen and paper:

if all are zebras, there are 4*8=32 legs, or 4 legs too many.

Each exchange with a peacock reduces 2 legs, so we exchange 2 zebras with 2 peacocks to get 6 zebras.