Graph the feasible region for the following system of inequalities: 2x + 3y <= 6, x - 2y >= -3, x >= 0, y >= 0.

For inequality 1: $2x + 3y \le 6$
The boundary line is $2x + 3y = 6$.
To find two points on this line:

• If $x=0$, then $3y=6 \implies y=2$. Point: $(0,2)$.
• If $y=0$, then $2x=6 \implies x=3$. Point: $(3,0)$. Plot these points and draw a solid line connecting them. To determine the shaded region, test a point not on the line, for example, $(0,0)$: $2(0) + 3(0) = 0 \le 6$. This is true, so the region satisfying $2x+3y \le 6$ is below or on the line $2x+3y=6$.

For inequality 2: $x - 2y \ge -3$
The boundary line is $x - 2y = -3$.
To find two points on this line:

• If $x=0$, then $-2y=-3 \implies y=\frac{3}{2}$. Point: $(0, 1.5)$.
• If $y=0$, then $x=-3$. Point: $(-3,0)$. Plot these points and draw a solid line connecting them. To determine the shaded region, test a point not on the line, for example, $(0,0)$: $0 - 2(0) = 0 \ge -3$. This is true, so the region satisfying $x-2y \ge -3$ is above or on the line $x-2y=-3$.

For inequality 3: $x \ge 0$
This represents all points to the right of or on the y-axis.

For inequality 4: $y \ge 0$
This represents all points above or on the x-axis.

The feasible region is the area where all four shaded regions overlap. Since $x \ge 0$ and $y \ge 0$, the feasible region will be restricted to the first quadrant.

The vertices are the intersection points of the boundary lines that define the feasible region.

1. Intersection of $x=0$ and $y=0$: $$(0,0)$$
2. Intersection of $x=0$ and $x-2y=-3$: Substitute $x=0$ into $x-2y=-3$: $0 - 2y = -3 \implies y = \frac{3}{2}$ $$(0, \frac{3}{2})$$
3. Intersection of $y=0$ and $2x+3y=6$: Substitute $y=0$ into $2x+3y=6$: $2x + 3(0) = 6 \implies 2x = 6 \implies x = 3$ $$(3,0)$$
4. Intersection of $2x+3y=6$ and $x-2y=-3$: From the second equation, $x = 2y-3$. Substitute this into the first equation: $2(2y-3) + 3y = 6$ $4y - 6 + 3y = 6$ $7y - 6 = 6$ $7y = 12$ $y = \frac{12}{7}$ Now substitute $y=\frac{12}{7}$ back into $x=2y-3$: $x = 2\left(\frac{12}{7}\right) - 3 = \frac{24}{7} - \frac{21}{7} = \frac{3}{7}$ $$\left(\frac{3}{7}, \frac{12}{7}\right)$$

The solution to the system of inequalities is the feasible region, which is a polygon defined by its vertices.

The vertices of the feasible region are:

• $(0,0)$
• $(0, \frac{3}{2})$
• $(\frac{3}{7}, \frac{12}{7})$
• $(3,0)$

The feasible region is the quadrilateral bounded by these four points.