Dogs and Owners

Let the number of dogs be $x$ and the number of humans be $y$ .
A dog has 1 head and a human has 1 head: $x + y = 20$ .
A dog has 4 legs and a human has 2 legs: $4x+2y = 64$ .

We solve this system by eliminating $x$ .
Multiplying the first equation by 4, we get $4x + 4y = 80$ .
Substracting the second equation, we get $2y = 16$ .
Hence, $y = 8, x = 20 - y = 20 - 8 = 12$ .

Let $D$ be the number of dogs and $H$ be the number of dog-owners. Then,

$D+H=20$ $\implies$ $H=20-D$ $\color{#D61F06}(1)$

$4D+2H=64$ $\color{#D61F06}(2)$

Substitute $\color{#D61F06}(1)$ in $\color{#D61F06}(2)$ . We have

$4D+2(20-D)=64$ $\implies$ $4D+40-2D=64$ $\implies$ $2D=64-40=24$ $\implies$ $\boxed{D=12}$

We can order every dog to stand on their back legs. In that case, everybody should have 2 legs on the ground, giving us $20 \times 2 = 40$ legs. There are $64 - 40 = 24$ legs left, meaning that 24 legs are off the ground. With each dog having 2 off the ground, that gives the answer of 12 dogs. (This is merely a fun way to interpret the process of setting up variable and solving the equations.)

just think from the answers.

if there is 13 dogs ,there will be =(13 x 4 +7 x 2) legs=66legs............................(7 people with 14 legs)

if there is 12 dogs, there will be=(12 x 4+ 8 x 2)legs=64 legs............................(8 people with 16 legs)

so, 12 will be the number of dogs