You can do it without algebra this way:

All the animals have 1 head and exactly 2 FRONT legs. so 70 animals accounts
for 70*2 or 140 front legs.  The remaining 196-140 or 56 hind legs are owned by
the rabbits.  Each rabbit has two hind legs, so the 56 hind legs amount to 56÷2
or 28 pairs of hind legs, so there are 28 rabbits and 70-28 or 42 hens, and so
there are 42-28 or 14 more hens that rabbits. 

By algebra:

Let h = number of hens
Let r = number of rabbits

Every animal has one head so 

number of hens + number of rabbits = number of heads

or 

h + r = 70


Hens have exactly 2 legs and rabbits have exactly 4 legs, so

2h + 4r = 196

Solve the system of equations

 h +  r =  70
2h + 4r = 196

and get h = 42 and r = 28, 42 hens and 28 rabbits.

However the question was not how many of each, but,

So 42 - 28 = 14.  So 42 is 14 more than 28, so the answer is 14 
more hens than rabbits.

Checking:

Together that's 42+28=70 animals.  

The 42 hens have 84 legs and the 28 rabbits have 112 legs.

That's a total of 84+112 = 196 legs.

Edwin