Solving Systems of Linear Inequalities

Introduction to Systems of Inequalities

A system of linear inequalities in two variables consists of two or more linear inequalities that must be satisfied simultaneously. The solution to the system is the set of all ordered pairs (x, y) that satisfy every inequality in the system.

The Solution Region (Feasible Region)

Graphically, the solution to a system of inequalities is the intersection of the solution regions of each individual inequality. This common, overlapping area is known as the solution region or feasible region.

Method for Solving a System Graphically

1. Graph Each Inequality: For every inequality in the system, follow the procedure from the previous section:

   • Draw the boundary line (solid for $\le, \ge$; dashed for $<, >$).
   • Choose a test point (like (0,0)).
   • Shade the appropriate half-plane.

2. Identify the Common Region: The solution to the system is the region on the graph where all the shaded half-planes overlap. If there is no region where all shadings overlap, the system has no solution.

3. Find Corner Points: The vertices or corner points of the feasible region are particularly important in optimization problems (Linear Programming). A corner point is found at the intersection of two boundary lines.

Example 1: A Simple System

Question: Solve the following system of inequalities graphically: $x+y \ge 5$ and $x-y \le 3$.

Solution:

1. Graph $x+y \ge 5$: Draw the solid line $x+y=5$. Test (0,0): $0+0 \ge 5$ is false. Shade the half-plane that does not contain the origin.
2. Graph $x-y \le 3$: Draw the solid line $x-y=3$. Test (0,0): $0-0 \le 3$ is true. Shade the half-plane that contains the origin.
3. Find the Solution Region: The solution is the region where the two shaded areas overlap.

Example 2: Bounded Feasible Region

Question: Find the solution region for the system: $x+2y \le 8$, $2x+y \le 8$, $x \ge 0$, $y \ge 0$.

Solution:

1. Graph all four inequalities:
   • $x+2y \le 8$: Solid line through (0,4) and (8,0). Shaded towards the origin.
   • $2x+y \le 8$: Solid line through (0,8) and (4,0). Shaded towards the origin.
   • $x \ge 0$: The y-axis and the region to its right.
   • $y \ge 0$: The x-axis and the region above it.
2. Find the Solution Region: The intersection of all four regions is a bounded quadrilateral (the feasible region) in the first quadrant.
3. Find Corner Points: The vertices of this region are:
   • (0,0): Intersection of $x=0$ and $y=0$.
   • (4,0): Intersection of $2x+y=8$ and $y=0$.
   • (0,4): Intersection of $x+2y=8$ and $x=0$.
   • To find the fourth point, solve $x+2y=8$ and $2x+y=8$ simultaneously. This gives the point $(8/3, 8/3)$.

Example 3: Unbounded Feasible Region

Question: Solve the system: $x+y > 4$, $2x-y > 0$.

Solution:

1. Graph $x+y > 4$: Draw a dashed line through (0,4) and (4,0). Shade the side that does not contain the origin.
2. Graph $2x-y > 0$ (or $y < 2x$): Draw a dashed line through (0,0) and (1,2). Test a point not on the line, e.g., (1,0). $2(1)-0 > 0$ is true. Shade the side containing (1,0).
3. Find the Solution Region: The overlapping shaded area is an unbounded region extending infinitely in one direction.