Lesson 4.7: Graphing Systems of Linear Inequalities

New Concept

A system of linear inequalities groups two or more inequalities. Its solution is the set of all points $(x, y)$ satisfying every inequality. We find this solution by graphing each one and identifying the overlapping shaded region on the coordinate plane.

What’s next

Now, you'll work through interactive examples of graphing these systems. Then, test your understanding with practice cards and application problems.

Property

Two or more linear inequalities grouped together form a system of linear inequalities. A system of two linear inequalities is shown here.

Examples

• Determine if $(2, 5)$ is a solution to the system $\begin{cases} y > 2x \\ x+y < 8 \end{cases}$. For the first inequality, $5 > 2(2)$ or $5 > 4$ is true. For the second, $2+5 < 8$ or $7 < 8$ is true. Since both are true, $(2, 5)$ is a solution.
• Determine if $(4, 1)$ is a solution to the system $\begin{cases} x - y \leq 3 \\ 2x + y > 10 \end{cases}$. For the first inequality, $4 - 1 \leq 3$ or $3 \leq 3$ is true. For the second, $2(4) + 1 > 10$ or $9 > 10$ is false. Therefore, $(4, 1)$ is not a solution.
• Determine if $(0, 0)$ is a solution to the system $\begin{cases} y \geq x + 2 \\ x \geq 1 \end{cases}$. For the first inequality, $0 \geq 0 + 2$ or $0 \geq 2$ is false. We do not need to check the second inequality. Therefore, $(0, 0)$ is not a solution.

Explanation

Think of this as a set of rules. A point is only a 'solution' if it follows every single rule (inequality) in the system. The complete solution is a whole region of points that satisfy all conditions simultaneously.

Property

Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true. The solution of a system of linear inequalities is shown as a shaded region in the $x, y$ coordinate system that includes all the points whose ordered pairs make the inequalities true. To determine if an ordered pair is a solution to a system of two inequalities, we substitute the values of the variables into each inequality.

Examples

• Is $(-1, 5)$ a solution to the system $\begin{cases} x + 2y \geq 8 \\ 3x - y < 1 \end{cases}$? Check 1: $-1 + 2(5) = 9 \geq 8$ (True). Check 2: $3(-1) - 5 = -8 < 1$ (True). Yes, it is a solution.
• Is $(6, 1)$ a solution to the system $\begin{cases} y < x - 4 \\ y > -x + 8 \end{cases}$? Check 1: $1 < 6 - 4$ is $1 < 2$ (True). Check 2: $1 > -6 + 8$ is $1 > 2$ (False). No, it is not a solution.
• Is $(10, 2)$ a solution to the system $\begin{cases} x - 4y > 0 \\ 2x + y < 21 \end{cases}$? Check 1: $10 - 4(2) = 2 > 0$ (True). Check 2: $2(10) + 2 = 22 < 21$ (False). No, it is not a solution.

Explanation

An ordered pair must be a team player! To be a solution for the system, it has to work for every single inequality involved. If it fails even one, it's out. The graph's solution is where all the 'true' zones overlap.

Property

1. Graph the first inequality. Graph the boundary line. Shade in the side of the boundary line where the inequality is true.
2. On the same grid, graph the second inequality. Graph the boundary line. Shade in the side of that boundary line where the inequality is true.
3. The solution is the region where the shading overlaps.
4. Check by choosing a test point.

Examples

• For the system $\begin{cases} y > x \\ y < 4 \end{cases}$, the solution is the region above the dashed line $y=x$ and below the dashed horizontal line $y=4$.
• For the system $\begin{cases} x+y \leq 6 \\ x \geq 2 \end{cases}$, the solution is the region to the left of the solid line $x+y=6$ and to the right of the solid vertical line $x=2$, including the boundary lines.
• For the system $\begin{cases} y \geq 2x-3 \\ y \leq -x+6 \end{cases}$, the solution is the region above the solid line $y=2x-3$ and below the solid line $y=-x+6$, including the boundary lines.

Explanation

Graph each inequality separately on the same grid. The solution is the 'double-shaded' zone where the shaded regions from all inequalities overlap. This area contains every point that solves the entire system.

Property

Systems of linear inequalities where the boundary lines are parallel might have no solution. If the shaded regions for each inequality do not overlap, there is no point that satisfies both inequalities, and the system has no solution. There is no point in both shaded regions, so the system has no solution.

Examples

• The system $\begin{cases} y > x + 3 \\ y < x - 1 \end{cases}$ has no solution because the shaded regions for the two parallel lines do not overlap.
• The system $\begin{cases} 3x + 2y \geq 12 \\ 3x + 2y \leq 4 \end{cases}$ has no solution. The regions are on opposite sides of two parallel lines and never intersect.
• The system $\begin{cases} y \geq 2 \\ y \leq -1 \end{cases}$ has no solution. A number cannot be both greater than or equal to 2 and less than or equal to -1 at the same time.

Explanation

This happens with parallel lines when the shaded areas go in opposite directions, never overlapping. It's like being told to stand north of one line and south of another parallel line—it's impossible!