If $m+n=5$ and $mn=3$, find $\sqrt{\frac{n+1}{m+1}} + \sqrt{\frac{m+1}{n+1}}$? It is known that $m+n=5$ and $mn=3$. So what is the value of:

$$
\sqrt{\dfrac{n+1}{m+1}} + \sqrt{\dfrac{m+1}{n+1}}
$$

I think we're suppose to solve for the system of equations first, but I'm not getting any results that's useful.

Danny's solution is short and nice. Here is an alternate, longer and less elegant solution:
Let $\sqrt{\frac{n+1}{m+1}}+\sqrt{\frac{m+1}{n+1}}=t$. Squaring both sides, we have:
$$\frac{n+1}{m+1}+\frac{m+1}{n+1}+2=t^2$$
Simplifying, we have:
$$\frac{(n+1)^2+(m+1)^2}{(m+1)(n+1)}+2=t^2$$
i.e.,
$$\frac{n^2+2n+1+m^2+2m+1}{mn+m+n+1}+2=t^2$$
i.e., 
$$\frac{m^2+n^2+2(m+n)+2}{mn+m+n+1}+2=t^2$$
But, $m^2+n^2=(m+n)^2-2mn$. Therefore, we have:
$$\frac{(m+n)^2-2mn+2(m+n)+2}{mn+m+n+1}+2=t^2$$
Substituting for $m+n=5$ and $mn=3$, we have:
$$\frac{(5)^2-2(3)+2(5)+2}{3+5+1}+2=t^2$$
i.e., 
$$\frac{31}{9}+2=t^2$$
i.e.,
$$\frac{49}{9}=t^2$$
Thus, we get $t=\frac{7}{3}$. Note that we eliminate the spurious negative solution of $-\frac{7}{3}$ as $t$ must be greater than $0$.
