1. **State the problem:** We need to determine the feasible region and vertices for the system of inequalities: $$4x - 2y < 2$$ $$3x + 5y < 21$$ $$x > 0, y > 0$$ 2. **Rewrite inequalities as equations to find boundary lines:** $$4x - 2y = 2$$ $$3x + 5y = 21$$ 3. **Find intercepts for each line:** - For $$4x - 2y = 2$$: - When $$x=0$$, $$-2y=2 \Rightarrow y=-1$$ (not in $$y>0$$ region, so ignore) - When $$y=0$$, $$4x=2 \Rightarrow x=\frac{1}{2}$$ - For $$3x + 5y = 21$$: - When $$x=0$$, $$5y=21 \Rightarrow y=\frac{21}{5}=4.2$$ - When $$y=0$$, $$3x=21 \Rightarrow x=7$$ 4. **Determine vertices by solving intersections:** - Intersection of $$4x - 2y = 2$$ and $$3x + 5y = 21$$: Multiply first equation by 5 and second by 2 to eliminate $$y$$: $$20x - 10y = 10$$ $$6x + 10y = 42$$ Add equations: $$26x = 52 \Rightarrow x = 2$$ Substitute $$x=2$$ into $$4x - 2y = 2$$: $$8 - 2y = 2 \Rightarrow -2y = -6 \Rightarrow y = 3$$ So intersection point is $$(2,3)$$. 5. **Check vertices in the first quadrant ($$x,y>0$$) and inequalities:** - Points to consider: $$(0,0)$$, $$(\frac{1}{2},0)$$, $$(0,4.2)$$, and $$(2,3)$$. 6. **Feasible region:** - Since $$4x - 2y < 2$$ and $$3x + 5y < 21$$ with $$x,y>0$$, the feasible region is bounded by these lines and the axes in the first quadrant. 7. **Vertices of the feasible region:** - $$(0,0)$$ (origin) - $$(\frac{1}{2},0)$$ (x-intercept of first inequality) - $$(2,3)$$ (intersection of two lines) - $$(0,4.2)$$ (y-intercept of second inequality) These points form the polygon of the feasible region. **Final answer:** The feasible region is the polygon with vertices at $$(0,0)$$, $$(\frac{1}{2},0)$$, $$(2,3)$$, and $$(0,4.2)$$ in the first quadrant satisfying all inequalities.