Day 2: Simultaneous Turtle Doves

When we add the first two equations together we get $(d + o)^2 = 400$ and since $d$ and $o$ are positive, $d + o = 20$ . Similarly for the other two we get $v + e = 8$ .

Therefore the answer is $20 + 8 =\boxed{28}$ .

We must verify that there are positive solutions; in order to do this simply divide the first two by $d+o$ to solve and similarly by $v+e$ for the other two. This shows that there are indeed positive solutions.

Let me answer it systematically.

$\begin {cases} d^2 + do = 250 &...(1) \\ o^2 + do = 150 &...(2) \\ v^2 + do = 30 &...(3) \\ e^2 + do = 34 &...(4) \end {cases}$

Eqn 1 + Eqn 2 and Eqn 3 + Eqn 4:

$\Rightarrow \begin {cases} d^2 + 2do + o^2 = 250 +150 & \Rightarrow (d+o)^2 = 400 & \Rightarrow d+o = 20 \\ v^2 + 2ve + e^2 = 30+34 & \Rightarrow (v+e)^2 = 64 & \Rightarrow v+e = 8 \end {cases}$

Therefore, $d+o+v+e = (d+o)+(v+e) = 20+8 = \boxed{28}$

simply first add first 2 equation - we get- (d+o)^2=400. so d+o =20 and similarly v+e=8 so d+o+v+e=28

After adding the first two equations i.e. (d^2+do=250) ,(o^2+do=150), we get an equation (d^2+o^2+2do=400) which can also be written as {(d+o)^2=(20)^2}.

Therefore, (d+o=20).

Now adding the 3rd and 4th equation i.e. (v^2+ve=30),(e^2+ve=34) , we get an equation (v^2+e^2+2ve=64) which can also be written as {(v+e)^2=(8)^2}.

Therefore, (v+e=8).

Now adding the equations (d+o=20),(v+e=8) we get (d+o+v+e=28).

d^2 + 2do + o^2 = 400 (d+o)^2 = 400 d + o = 20 v^2 + 2ve + e^2 = 64 (v+e)^2 = 64 v + e = 8 d + o + v + e = 28

d^2+2od+o^2=400 (d+o)^2=400 d+o=20 v2+2ve+e^2=64 (v+e)^2=64 v+e=8 d+o+e+v=20+8=28

d(d+o)=250........d=250/d+o, o(d+o)=150........o=150/d+o. d+o=(250/d+o)+(105/d+o) =400/d+o, (d+o)^2=400, d+o=20 ...................v+e=8, d+o+v+e=28