1. **State the problem:** Find the solution region for the system of inequalities: $$3x+y\ge15, \quad x+2y\ge10, \quad x\ge0, \quad y\ge0.$$ 2. **Graph the boundary lines:** - For $3x+y=15$: intercepts are $(0,15)$ and $(5,0)$. - For $x+2y=10$: intercepts are $(0,5)$ and $(10,0)$. 3. **Determine the shaded region:** Since the inequalities are $\ge$, shade above each line. The inequalities $x\ge0$ and $y\ge0$ restrict to the first quadrant. 4. **Find the corner points by intersections:** - Intersection of $x=0$ and $3x+y=15$ gives $(0,15)$. - Intersection of $y=0$ and $x+2y=10$ gives $(10,0)$. - Solve $3x+y=15$ and $x+2y=10$ simultaneously: $$y=15-3x$$ $$x + 2(15 - 3x) = 10$$ $$x + 30 - 6x = 10$$ $$-5x = -20$$ $$x=4$$ Substitute back: $$3(4) + y = 15 \Rightarrow y=3$$ Thus, intersection point is $(4,3)$. 5. **Characterize the solution region:** The region where all inequalities hold is bounded below by the lines but extends infinitely upwards and rightwards, so the solution is an **unbounded region**. **Answer:** The solution region is unbounded with corner points at $(0,15)$, $(4,3)$, and $(10,0)$.