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.
