prove the combinatorial identity: $ {2n \choose n} + {2n\choose n-1} = (1/2){2n + 2 \choose n + 1}$ So the first thing I did was to multiply and then isolate the left-most term. 
That leaves me with  $$ 2{2n \choose n} ={2n + 2\choose n + 1} - 2{2n \choose n -1} $$
No specific reasoning except that it looks cleaner, at least to me. Then the way I thought about it was that we are trying to make two $n$ member committees out of $n$ koalas and $n$ pandas. I was told when seeing subtraction I should think of complement but I don't really know how to incorporate the complement into this. Any way, am I on the right path or completely off. Thank you in advance for your time. 

Whenever you see these types of identities, at least I don't think like that. I think algebraically. Maybe long, but always correct. We want to prove
$$2{2n \choose n} ={2n + 2\choose n + 1} - 2{2n \choose n -1}$$
We know that $${2n\choose n}+{2n\choose n-1}={2n+1\choose n}$$
Multiplying by $2$ $$2{2n\choose n}+2{2n\choose n-1}=2{2n+1\choose n}$$
Now we have to prove that $$2{2n+1\choose n}={2n+2\choose n+1}$$
$$2.\frac{(2n+1)!}{(n+1)!n!}=\frac{(2n+2)!}{(n+1)!(n+1)!}$$
Let us assume the above, if it is true, we will get RHS=LHS, else, we will get a contradiction. So , by cancellation
$$2=\frac{2n+2}{n+1}$$
Which is true. Hence, $$2{2n\choose n}+2{2n\choose n-1}={2n+2\choose n+1}$$
So this expression is equivalent to your question. 
As long as the proof of ${2n\choose n}+{2n\choose n-1}={2n+1\choose n}$ is concerned, I will prove the more general case- $${n\choose k}+{n\choose k-1}={n+1\choose k}$$ 
Proof: The given equation is same as $$\frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}=\frac{(n+1)!}{k!(n-k+1)!}$$
Let us assume that with the same condition as before, 
$$\frac{n!}{(k-1)!(n-k)!}(\frac1k+\frac1{n-k+1})=\frac{(n+1)!}{k!(n-k+1)!}$$
By cancelling and adding on the LHS, $$\frac{n+1}{k(n-k+1)}=\frac{n+1}{k(n-k+1)}$$
So LHS=RHS. Hence proved.
