1. **State the problem:** We are given a system of inequalities: $$5x + y \leq 10$$ $$x + y \leq 6$$ $$x \geq 0$$ $$y \geq 0$$ We want to understand the feasible region defined by these inequalities. 2. **Explain the inequalities:** - The first inequality $5x + y \leq 10$ represents all points on or below the line $y = 10 - 5x$. - The second inequality $x + y \leq 6$ represents all points on or below the line $y = 6 - x$. - The third and fourth inequalities $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant. 3. **Find intercepts for each line:** - For $5x + y = 10$: - When $x=0$, $y=10$. - When $y=0$, $5x=10 \Rightarrow x=2$. - For $x + y = 6$: - When $x=0$, $y=6$. - When $y=0$, $x=6$. 4. **Find intersection of the two lines:** Solve the system: $$\begin{cases} 5x + y = 10 \\ x + y = 6 \end{cases}$$ Subtract second from first: $$5x + y - (x + y) = 10 - 6$$ $$5x + y - x - y = 4$$ $$4x = 4$$ $$x = 1$$ Substitute $x=1$ into $x + y = 6$: $$1 + y = 6 \Rightarrow y = 5$$ 5. **Summary of key points:** - Intercepts of $5x + y = 10$: $(0,10)$ and $(2,0)$ - Intercepts of $x + y = 6$: $(0,6)$ and $(6,0)$ - Intersection point: $(1,5)$ 6. **Feasible region:** The feasible region is bounded by the lines $5x + y \leq 10$, $x + y \leq 6$, and the axes $x \geq 0$, $y \geq 0$. It is a polygon with vertices at $(0,0)$, $(0,6)$, $(1,5)$, and $(2,0)$. This region can be graphed on the coordinate plane with the given axis ranges and labels.