If $$2x+y>0$$, $$2y>x$$, and $$y<3$$, what is the range of all possible values of $$x+y$$?

From the given inequalities:

1. $$2x + y > 0$$
   
2. $$2y > x \implies x < 2y$$
   
3. $$y < 3$$
   

Substitute $$x < 2y$$ into the first inequality:

$$2(2y) + y > 0 \implies 5y > 0 \implies y > 0$$

Combining this with $$y < 3$$, we get the range for $$y$$:

$$0 < y < 3$$

From the inequalities for $$x$$, we have:

$$-\frac{y}{2} < x < 2y$$

Adding $$y$$ across the inequality to form $$x + y$$:

$$-\frac{y}{2} + y < x + y < 2y + y$$

$$\frac{y}{2} < x + y < 3y$$

Using the bounds of $$y$$ ($$0 < y < 3$$):

• As $$y$$ approaches $$0$$, the lower bound $$\frac{y}{2}$$ approaches $$0$$.
  
• As $$y$$ approaches $$3$$, the upper bound $$3y$$ approaches $$9$$.
  

Thus, the range of all possible values for $$x + y$$ is from $$0$$ to $$9$$ exclusive.

The correct option is B.