The answer - I believe it's unique, but it's very late and I'm tired - is $15$ cows, $3$ lambs, and $82$ piglets.
This was my approach.
First, you have a system of two linear equations. The usual thing to do is to try to eliminate some of the variables. Multiply the first equation by $120$ to get $120c+120l+120p=12000$, and subtract the other equation from that to eliminate $c$, leaving $70l+95p=8000$. Dividing by the GCD, we get $14l+19p=1600$, and now we can set up a linear relation between $p$ and $l$:
$$p=\frac{1600-14l}{19}$$
I chose to divide by $19$ and not $14$ because $19$ happens to be prime. This is a good thing because we're now going to work modulo $19$ to determine when the right hand side is an integer. We want $1600-14l\equiv0\mod19$, and since $19$ is prime we get to do division, and find that the only possibility is $l\equiv3$. So:
$$l=19k + 3$$
And, applying the previous relation for $p$ and simplifying:
$$p=14k+82$$
Great, we've expressed $l$ and $p$ in terms of one variable $k$ - we now have a system of two equations in two variables. Plug what we've found back into the original two equations and we get:
$$\begin{align}
33k+85+c&=100\\
1300k+2200+120c&=4000
\end{align}$$
From which we deduce the solution I gave above.
