Cats and pigeons for sale

Let $c =$ Cats , $p =$ Pigeons , $h =$ Head & $l =$ Leg

Cats have one head and four legs, pigeons however have one head and two legs, there are a total of thirty heads and eighty legs. Put this into equation form and you get

$30h + 80l = c(h + 4l) + p(h + 2l)$

Since heads and legs don't mix we can split it into two equations

$30h = c(h) + p(h)$

$80l = c(4l) + p(2l)$

These can now be used as simultaneous equations but first let's simplify and re-arrange

$c = 30 - p$

$80 = 4c + 2p$

Next we substitute to get

$80 = 4(30 - p) + 2p$

This expands and simplifies into

$80 = 120 - 2p$

Re-arranging this gives us a value for $p$

$p = 20$

So there are $20$ pigeons in the shop

Its really simple

Suppose there are "x" number of pigeons in that shop so number of cats - (30-x)

Here ,

Cats have 4 legs

and Pigeons have 2

So , the equation

2x + 4(30-x) = 80

Solve it and you'll get x=20.

Let the no. of pigeons be x and no. of cats be y. Total no.of heads = no.of pigeons + no.of cats =x + y =30. Pigeons have 2 legs where as cats have 4 legs. Total no.of legs = 80 =2x + 4y. Or, x + 2y =40 By solving the equations, we get x =20 &y =10 . Hence there are 20 pigeons