1. **State the problem:** We need to solve the system of inequalities graphically and determine if the solution region is bounded or unbounded. The system is: $$3x + y \geq 15$$ $$x + 2y \geq 10$$ $$x \geq 0$$ $$y \geq 0$$ 2. **Rewrite the inequalities as equations to find boundary lines:** - Boundary 1: $$3x + y = 15$$ - Boundary 2: $$x + 2y = 10$$ - Boundary 3: $$x = 0$$ (y-axis) - Boundary 4: $$y = 0$$ (x-axis) 3. **Find intercepts for each boundary:** - For $$3x + y = 15$$: - When $$x=0$$, $$y=15$$ - When $$y=0$$, $$x=5$$ - For $$x + 2y = 10$$: - When $$x=0$$, $$2y=10 \Rightarrow y=5$$ - When $$y=0$$, $$x=10$$ 4. **Determine which side of each line satisfy the inequalities:** - For $$3x + y \geq 15$$, test point $(0,0)$: $$3*0 + 0 = 0 \not\geq 15$$, so the region is above the line. - For $$x + 2y \geq 10$$, test point $(0,0)$: $$0 + 0 = 0 \not\geq 10$$, so the region is above this line. - $$x \geq 0$$ is to the right of the y-axis. - $$y \geq 0$$ is above the x-axis. 5. **Find the corner points by intersection of boundary lines within feasible region:** - Intersection of $$3x + y = 15$$ and $$x + 2y = 10$$: From $$x + 2y = 10$$, solve for $$x = 10 - 2y$$. Substitute into $$3x + y = 15$$: $$3(10 - 2y) + y = 15$$ $$30 - 6y + y = 15$$ $$30 - 5y = 15$$ $$-5y = -15$$ $$y = 3$$ Then $$x = 10 - 2(3) = 10 - 6 = 4$$. So intersection point is $$ (4,3) $$ - Intersection of $$3x + y = 15$$ and $$y=0$$: $$3x + 0 = 15 \Rightarrow x=5$$ Point: $$(5,0)$$ - Intersection of $$x + 2y = 10$$ and $$y=0$$: $$x + 0 = 10 \Rightarrow x=10$$ Point: $$(10,0)$$ (but check inequalities) - Intersection with axes and constraints for $$x, y \geq 0$$: As above. 6. **Check which corner points are in solution region:** - Point $$(4,3)$$: $$3(4) + 3 = 12 + 3 = 15 \geq 15$$ true $$4 + 2(3) = 4 + 6 = 10 \geq 10$$ true $$4 \geq 0$$ and $$3 \geq 0$$ true - Point $$(5,0)$$: $$3(5) + 0 = 15 \geq 15$$ true $$5 + 0 = 5 \geq 10$$ false, so not feasible - Point $$(10,0)$$: $$3(10) + 0 = 30 \geq 15$$ true $$10 + 0 = 10 \geq 10$$ true $$x,y \geq 0$$ true 7. **Determine boundary points within solution:** - Intersection of $$x=0$$ with $$x + 2y \geq 10$$: When $$x=0$$, $$2y \geq 10 \Rightarrow y \geq 5$$. - Intersection of $$x=0$$ with $$3x + y \geq 15$$: When $$x=0$$, $$y \geq 15$$. Since $$y\geq 15$$ is more restrictive, on $$x=0$$, the solution region starts at $$(0,15)$$ upward. - Check if $$(0,15)$$ satisfies both inequalities: $$3(0)+15=15 \geq 15$$ $$0 + 2(15) = 30 \geq 10$$ So that point is feasible. 8. **Summary of corner points of solution region:** - $$(0,15)$$ - Intersection $$(4,3)$$ - $$(10,0)$$ 9. **Check if region is bounded:** The solution region satisfies $$x \geq 0$$ and $$y \geq 0$$ and lies above lines $$3x+y=15$$ and $$x+2y=10$$. Graphically, the region is the intersection of half-planes to the upper right of these lines and axes. Because the inequalities are \geq and constraints are $$x,y \geq 0$$ this region extends infinitely to larger $x,y$, so the solution region is **unbounded**. **Final answers:** - Coordinates of corner points are $$\boxed{(0,15), (4,3), (10,0)}$$. - The solution region is **unbounded**.