Systems of Linear Inequalities

Understanding the Basics

A system of linear inequalities consists of two or more linear inequalities involving the same variables. The solution to the system is the set of all coordinate points $(x, y)$ that make every inequality in the system true at the same time. When you graph the system on a coordinate plane, this solution appears as an overlapping shaded area known as the feasible region.

How to Graph a Linear Inequality

To find the feasible region of a system, you must first know how to graph each individual inequality:

1. Graph the boundary line: Treat the inequality symbol as an equal sign to find the line.
   • Use a solid line for $\leq$ (less than or equal to) or $\geq$ (greater than or equal to). This means points on the line are included in the solution.
   • Use a dashed line for $<$ (strictly less than) or $>$ (strictly greater than). This means points on the line are excluded.
2. Shade the half-plane: Pick a test point (like $(0,0)$ if it's not on the line). If the point makes the inequality true, shade the side of the line containing that point. If false, shade the opposite side.
   • Shortcut: If the equation is in $y = mx + b$ form, simply shade above the line for $>$ or $\geq$, and below the line for $<$ or $\leq$.

Example 1: Graphing Two Inequalities

Let's graph the following system: $y > 2x - 1$ $y \leq -x + 4$

Step 1: Graph $y > 2x - 1$.

• The boundary line is $y = 2x - 1$. Since the symbol is $>$, draw a dashed line.
• Because it is $y > \dots$, shade the half-plane above the dashed line.

Step 2: Graph $y \leq -x + 4$.

• The boundary line is $y = -x + 4$. Since the symbol is $\leq$, draw a solid line.
• Because it is $y \leq \dots$, shade the half-plane below the solid line.

Step 3: Find the overlap. The solution to the system is the exact region where the two shaded areas overlap. Any coordinate point picked from this overlapping section will satisfy both inequalities.

Example 2: Finding a Feasible Region with Constraints

Find the feasible region for the system: $x + y \leq 6$ $x \geq 0$ $y \geq 0$

Step 1: Graph $x + y \leq 6$.

• The boundary is the solid line $x + y = 6$ (the x-intercept is $6$, and the y-intercept is $6$).
• Using the test point $(0,0)$: $0 + 0 \leq 6$ is true, so shade towards the origin (below the line).

Step 2: Graph $x \geq 0$ and $y \geq 0$.

• $x \geq 0$ represents all points to the right of the y-axis (inclusive).
• $y \geq 0$ represents all points above the x-axis (inclusive).
• Together, these two inequalities restrict our solution entirely to the first quadrant.

Step 3: Identify the feasible region. The overlapping region is a right triangle bounded by the vertices $(0,0)$, $(6,0)$, and $(0,6)$. Every point inside and on the solid borders of this triangle is a valid solution to the system.