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The system of inequations is given by: $$3x + y \ge 3$$ $$2x - y \ge -5$$ $$x \ge 0, y \ge 0$$ The condition $x, y \ge 0$ restricts the feasible region to the 1st quadrant.
Line 1: $3x + y = 3$. Intercepts are $(1, 0)$ and $(0, 3)$. The region $3x + y \ge 3$ is away from the origin. Line 2: $2x - y = -5$ or $y = 2x + 5$. Intercepts are $(-2.5, 0)$ and $(0, 5)$. The region $2x - y \ge -5$ (or $y \le 2x + 5$) includes the origin side.
Since the region is restricted to the 1st quadrant ($x, y \ge 0$), we look at the intersection of the half-planes. The line $3x + y = 3$ starts from $(1, 0)$ and goes to $(0, 3)$. As $x$ increases, $y$ can increase indefinitely along the line $y = 2x + 5$ while satisfying the inequalities. Because the region extends infinitely in the positive direction of the axes, it is unbounded.
Final Answer: unbounded in 1st quadrant