Solve Systems of Inequalities with Two Variables

This step-by-step tutorial explains how to solve systems of inequalities with two variables. You will learn to graph each inequality, identify the feasible region, and calculate its vertices using worked examples.

To solve a system of inequalities, graph the solution of each individual inequality on the same coordinate plane. The area common to all individual solutions is the feasible region — the solution set of the system.

Example 1: Solve a System of Two Linear Inequalities

Solve graphically the system of inequalities:

$$ \begin{cases} \ x + 2y \ge - 2 \\ \ x + y \lt 0 \\ \end{cases} $$
View Solution to Example 1

We first use standard methods for inequalities with two variables to graph each given inequality.

Below (in red) is the solution set of the first inequality: $x + 2y \ge - 2$. Because the inequality includes the equal sign, the boundary line $x + 2y = - 2$ is solid.

Graph of the inequality x + 2y ≥ −2 with a solid boundary line, red shaded region

We then solve the second inequality $x + y \lt 0$; it is shown below in blue. Because the inequality strictly does not include the equal sign, the boundary line $x + y = 0$ is a dashed (broken) line.

Graph of the inequality x + y < 0 with a dashed boundary line, blue shaded region

We next find the region common to both solution sets. The intersection is the set of points where the red and blue hashed areas overlap.

Overlap of the two shaded regions showing the intersection of the inequalities

Finally, the solution set of the entire system of inequalities is represented by the solid blue region. This is the feasible region.

Feasible region of Example 1: the intersection of x + 2y ≥ −2 and x + y < 0

Example 2: Three Inequalities — Find the Vertices

Solve graphically the system of inequalities and find the vertices of the region representing the solution set.

$$ \begin{cases} \ x \gt y - 1 \\ \ 2x + y \le 2 \\ \ 4y \ge -3x - 12 \\ \end{cases} $$
View Solution to Example 2

We graph all three inequalities independently to find their overlapping region.

  • The solution set for $x \gt y - 1$ is the region with black hash lines.
  • The solution set for $2x + y \le 2$ is the region with red hash lines.
  • The solution set for $4y \ge -3x - 12$ is the region with blue hash lines.
Graph of the inequality x > y − 1 with a dashed boundary line Graph of the inequality 2x + y ≤ 2 with a solid boundary line Graph of the inequality 4y ≥ −3x − 12 with a solid boundary line

The solution set is the intersection of all three regions found above. It forms a triangle.

Intersection of the three inequalities forming a triangular region

This triangular solution set is the feasible region, shown below.

Feasible region for Example 2 with vertices A, B, and C

To find the coordinates of the vertices $A$, $B$, and $C$, we solve the systems of linear equations formed by the intersecting boundary lines.

Vertex A: Intersection of line AB and line AC

Line $AB$: $x = y - 1$
Line $AC$: $4y = -3x - 12$

Solve the system:
Substituting $y = x + 1$ into $4y = -3x - 12$ gives $4(x + 1) = -3x - 12$, which yields $x = -16/7$ and $y = -9/7$.
$A(-16 / 7 , -9 / 7)$

Vertex B: Intersection of line AB and line BC

Line $AB$: $x = y - 1$
Line $BC$: $2x + y = 2$

Solve the system:
Substituting $y = x + 1$ into $2x + y = 2$ gives $2x + (x + 1) = 2$, which yields $x = 1/3$ and $y = 4/3$.
$B(1 / 3 , 4 / 3)$

Vertex C: Intersection of line AC and line BC

Line $AC$: $4y = -3x - 12$
Line $BC$: $2x + y = 2$

Solve the system:
Substitute $y = 2 - 2x$ into $4y = -3x - 12$ to get $4(2 - 2x) = -3x - 12$, which yields $x = 4$ and $y = -6$.
$C(4 , -6)$

The three vertices have coordinates: $A(-16 / 7 , -9 / 7)$, $B(1 / 3 , 4 / 3)$, and $C(4 , -6)$.

Example 3: Five Inequalities — Polygon Feasible Region

Solve graphically the inequality system and find the vertices of the region representing the solution set.

$$ \begin{cases} \ x \ge 0 \\ \ y \ge 0 \\ \ y \le x + 1 \\ \ 4y + x \le 10 \\ \ y - x \ge -3 \\ \end{cases} $$
View Solution to Example 3

We first graph each inequality to determine the common area (feasible region) shown below as a polygon.

Polygonal feasible region for Example 3 with vertices A, B, C, D, and E

There are 5 vertices: $A, B, C, D,$ and $E$. Each vertex is determined by solving the system of equations corresponding to the two intersecting boundary lines.

  • Vertex A: Intersection of $x = 0$ and $y = 0$. Solution: $A(0 , 0)$.
  • Vertex B: Intersection of $x = 0$ and $y = x + 1$. Solution: $B(0 , 1)$.
  • Vertex C: Intersection of $y = x + 1$ and $4y + x = 10$. Solution: $C(6/5 , 11/5)$.
  • Vertex D: Intersection of $4y + x = 10$ and $y - x = -3$. Solution: $D(22/5 , 7/5)$.
  • Vertex E: Intersection of $y - x = -3$ and $y = 0$. Solution: $E(3 , 0)$.

Frequently Asked Questions

What is a system of inequalities with two variables?

A system of inequalities with two variables is a set of two or more inequalities that must all be satisfied at the same time. Its solution is the set of ordered pairs $(x, y)$ that make every inequality in the system true.

What is the feasible region?

The feasible region is the intersection of all the individual solution sets of a system of inequalities. It is the area on the coordinate plane where every inequality in the system is satisfied simultaneously.

How do you find the vertices of a feasible region?

Each vertex of the feasible region is the intersection point of two boundary lines. Replace the inequality signs with equal signs and solve the resulting system of two linear equations in two variables.

When is the boundary line solid or dashed?

Use a solid line when the inequality includes “equal to” ($\le$ or $\ge$). Use a dashed line when the inequality is strict ($<$ or $>$). The dashed line indicates that the boundary itself is not part of the solution.