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

Looking at the three given inequalities:

• $2y + z > 0$ ... (1)
• $2z > y$ ... (2)
• $z < 3$ ... (3)

First, let's find what values $z$ can take.

From inequality (1): $y > -\frac{z}{2}$

From inequality (2): $y < 2z$

For both to be true simultaneously, we need: $-\frac{z}{2} < 2z$

Solving this: $-\frac{z}{2} < 2z$ → $-z < 4z$ → $0 < 5z$ → $z > 0$

So $z$ must be positive: $0 < z < 3$

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Now let's find the range of $(y + z)$.

From the inequalities on $y$:

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

Adding $z$ to all parts:

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

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

Finding the lower bound:

Since $z > 0$, the smallest value of $\frac{z}{2}$ approaches $0$ (but never reaches it).

Therefore: $y + z > 0$

Finding the upper bound:

Since $z < 3$, we have $3z < 3(3) = 9$

Therefore: $y + z < 9$

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Combining both bounds: $0 < y + z < 9$

Option 3 is correct.