6-5 Systems of Inequalities

Property

A system of linear inequalities consists of two or more linear inequalities considered simultaneously.

The solution to a system is any ordered pair $(x, y)$ that makes EVERY inequality in the system true at the exact same time. On a graph, the solution set is represented by the intersection—the specific region where the shaded half-planes of all the individual inequalities overlap perfectly.

Examples

• Example 1 (Checking Solutions): Determine if $(4, 1)$ is a solution to the system:

$x + 2y \leq 8$
$3x - y > 5$
Check Inequality 1: $4 + 2(1) \leq 8 \rightarrow 6 \leq 8$ (True).
Check Inequality 2: $3(4) - 1 > 5 \rightarrow 11 > 5$ (True).
Since both are true, $(4, 1)$ is a solution to the system.

• Example 2 (Checking a False Solution): Is $(2, 3)$ a solution to the same system?

Check Inequality 1: $2 + 2(3) \leq 8 \rightarrow 8 \leq 8$ (True).
Check Inequality 2: $3(2) - 3 > 5 \rightarrow 3 > 5$ (False).
Since it fails the second rule, $(2, 3)$ is NOT a solution to the system.

Explanation

Think of a system of inequalities as an extremely strict VIP club with multiple bouncers. To get into the club (the solution region), you must follow every single rule at the same time. If you pass the dress code but forget your ID, you don't get in. Visually, the solution isn't just a single dot; it is an entire zone on the map where all the different shaded colors crash into each other and mix together!

Property

To solve a system of linear inequalities by graphing, follow a step-by-step layering process:

1. Graph the boundary line for the first inequality (solid for $\leq, \geq$; dashed for $<, >$). Shade its solution half-plane lightly.
2. On the exact same coordinate grid, graph the boundary line for the second inequality. Shade its solution half-plane lightly.
3. Identify the intersection where the shadings from both inequalities overlap. Make this overlapping region noticeably darker. This dark region is the final solution to the system.

Examples

• Example 1: Solve the system $y > x - 2$ and $y \leq -x + 4$.

Graph a dashed line for $y = x - 2$ and shade above it.
Graph a solid line for $y = -x + 4$ and shade below it.
The final solution is the dark wedge-shaped region where the two shadings overlap.

• Example 2: Solve the system $x < 3$ and $y \geq -2$.

Graph a dashed vertical line at $x = 3$ and shade everything to its left.
Graph a solid horizontal line at $y = -2$ and shade everything above it.
The solution is the top-left rectangular quadrant defined by the crossing of these two lines.

Examples

• Solve the system:
  $$\begin{cases} y > x - 2 \\ y \leq -x + 4 \end{cases}$$
  Graph a dashed line for $y = x - 2$ and shade above it. Graph a solid line for $y = -x + 4$ and shade below it. The overlapping region is the solution.
• Solve the system:
  $$\begin{cases} x + y < 5 \\ y > 1 \end{cases}$$
  Graph a dashed line for $x+y=5$ and shade below it. Graph a dashed horizontal line for $y=1$ and shade above it. The solution is the overlapping triangle-like region.
• Solve the system:
  $$\begin{cases} x < 3 \\ y \geq -2 \end{cases}$$
  Graph a dashed vertical line at $x=3$ and shade to the left. Graph a solid horizontal line at $y=-2$ and shade above. The solution is the top-left quadrant defined by these lines.