Explain a couple steps in proof that ${n \choose r} = {n-1 \choose r-1} + {n-1 \choose r}$ Show  ${n \choose r}$ = ${n-1 \choose r-1}$ + ${n-1 \choose r}$ 
I found a similar question on here but I am looking for a little bit more of an explanation on how they simplified 
Right Hand Side $$= \frac{(n-1)!}{(r-1)!(n-r)!} + \frac{(n-1)!}{r!(n-r-1)!}$$
                $$= \frac{r(n-1)!+(n-r)(n-1)!}{r!(n-r)!}$$
               $$= \frac{n!}{r!(n-r)!}$$
               $$= \binom{n}{r}$$,
the first equality because of definition
,the second equality because of summing fractions
the third because of $n(n-1)!=n!$,
the fourth by definition.
I would like more of an explanation on equality 2 and equality 3.  I understand that they are summing fractions but I am interested in how they came up with the common denominator and then simplified. 

For a combinatorial proof, $\binom nr$ is the number of $r$-element subsets of $\{1,2,\ldots,n\}$. Each such subset either contains $n$ or does not. $\binom {n-1}{r-1}$ counts the former, $\binom {n-1}r$ counts the latter. Therefore the quantities $\binom nr$ and $\binom {n-1}{r-1} + \binom {n-1}r$ are equal.
