Questions: Look at the system of inequalities. 2x + 3y ≥ -18 x - y ≤ 1 x - y ≥ -4 x ≤ -1 The solution set is the quadrilateral region where all the inequalities are true. What are the vertices of that quadrilateral region?

Solution Steps

To find the vertices of the quadrilateral region defined by the system of inequalities, we need to determine the points of intersection of the boundary lines of these inequalities. This involves solving pairs of equations derived from the inequalities. Once we have the intersection points, we can check which ones satisfy all the inequalities to identify the vertices of the feasible region.

We start by considering the system of inequalities: \[ \begin{align*}

1. & \quad 2x + 3y \geq -18 \\
2. & \quad x - y \leq 1 \\
3. & \quad x - y \geq -4 \\
4. & \quad x \leq -1 \end{align*} \] To find the vertices of the quadrilateral region, we convert these inequalities into equations and solve for their intersections.

We solve the following pairs of equations to find the intersection points:

• \(2x + 3y = -18\) and \(x - y = 1\) gives \((x, y) = (-3, -4)\).
• \(2x + 3y = -18\) and \(x - y = -4\) gives \((x, y) = (-6, -2)\).
• \(2x + 3y = -18\) and \(x = -1\) gives \((x, y) = \left(-1, -\frac{16}{3}\right)\).
• \(x - y = 1\) and \(x = -1\) gives \((x, y) = (-1, -2)\).
• \(x - y = -4\) and \(x = -1\) gives \((x, y) = (-1, 3)\).

We verify which of these intersection points satisfy all the original inequalities:

• \((-3, -4)\) satisfies all inequalities.
• \((-6, -2)\) satisfies all inequalities.
• \((-1, -2)\) satisfies all inequalities.
• \((-1, 3)\) satisfies all inequalities.
• \(\left(-1, -\frac{16}{3}\right)\) does not satisfy the inequality \(x - y \leq 1\).

Final Answer