If $$2y + z > 0$$, $$2z > y$$, and $$z < 3$$, find the range of possible values of $$(y + z)$$.

$$-\dfrac{\ z}{2}<\ y<\ 2z$$Given, 
$$z < 3$$
$$2z < 6$$
Also, $$ 2z > y$$
Combining both these equations, we get
$$y < 6$$
Adding $$z $$ on both sides
$$y+z < 6+z$$
Since $$ z < 3 $$
$$y+z < 6+3$$
$$y+z < 9$$

It is given that $$2y+z >0$$
$$y>-\ \dfrac{\ z}{2}$$
Given, $$ 2z > y$$
Combining both these, we get $$z>0$$
$$\left(y+z\right)_{\min}>\ -\ \dfrac{\ z}{2}+z$$
$$\left(y+z\right)_{\min}>\ \dfrac{\ z}{2}$$
$$\left(y+z\right)_{\min}>\ \ 0$$

Hence, the range of possible values of $$(y + z)$$ is $$\left(0,9\right)$$