Systems of Linear Inequalities

A system of inequalities is satisfied not by a single point but by a whole half-plane region — and when you need every inequality to hold at once, the answer is wherever those regions overlap.

By the end you'll be able to graph two linear inequalities, shade each half-plane with the correct boundary style, and identify the feasible region where both are satisfied.

Predict: which region satisfies BOTH inequalities? Toggle the shading and check.

This plots Inequality A: \(y \ge 2x + b_A\), and Inequality B: \(y \le x + b_B\). Drag the intercepts to move each line, and use the comparator buttons to flip a boundary between strict (dashed) and non-strict (solid). The green overlap is the feasible region — where both hold at once.

A: \(y \ge 2x - 1.0\) (solid). B: \(y \le x + 1.0\) (solid). Corner: (2.00, 3.00).

System of two linear inequalities — boundary lines, shaded half-planes, and their overlap
A: \(y \ge 2x + b_A\) B: \(y \le x + b_B\) feasible region (overlap)

Each inequality shades a half-plane; a system of inequalities is solved by every inequality holding at once, so its solution is exactly where all the shaded regions overlap.

Visual

Graph each inequality the way you'd graph its boundary line, then shade the correct side. Use a solid line for \(\le\) or \(\ge\) (points on the line count) and a dashed line for < or > (they don't). A quick way to pick the shaded side: plug in a test point not on the line — the origin (0, 0) is easiest when the line avoids it. If it satisfies the inequality, shade its side; otherwise shade the other side. Do this for every inequality in the system, and the region where all the shadings overlap is your answer.

Formal

A system of inequalities is two or more inequalities in the same variables considered together. Its solution isn't a point — it's the set of every (x, y) that satisfies every inequality at once, which is exactly the overlap of the individual half-planes: the feasible region. If any two boundary lines aren't parallel, they intersect at a point — solve that pair as a linear system (substitution or elimination) and, if the intersection sits on the edge of the overlap, it's a corner of the feasible region.

Applied

A factory that has both a labor-hours limit and a materials-budget limit is really graphing two inequalities at once — how many of each product it can make. Every feasible production plan is a point in the overlap of those two half-planes; that's the starting picture behind linear programming, where you search the feasible region's corners for the best (most profitable) plan.

Worked example

Solve the system \(y \ge 2x - 1\) and \(y < x + 1\). Draw \(y = 2x - 1\) solid (non-strict); test (0, 0): \(0 \ge -1\) is true, so shade the side containing the origin (above the line). Draw \(y = x + 1\) dashed (strict); test (0, 0): \(0 < 1\) is true, so shade the side containing the origin (below the line). The feasible region is the wedge where both shadings overlap. Its corner is where the boundaries meet: \(2x - 1 = x + 1 \Rightarrow x = 2, y = 3\), the corner (2, 3) — found exactly the way you'd solve a linear system of equations.

Your turn

Same idea, a different system: \(y \le -x + 3\) (solid) and \(y > 2x - 2\) (dashed). Test the point (0, 1). First inequality: \(1 \le -0 + 3 = 3\) → true, so (0, 1) is on the shaded side of the solid line. Second inequality: \(1 > 2(0) - 2 = -2\) → true, so (0, 1) is also on the shaded side of the dashed line. Since both check out, (0, 1) is ____ the feasible region.

Reveal the answer

(0, 1) is inside the feasible region — it's a solution of the system, since it satisfies both inequalities simultaneously. Set \(b_A = -1\) and \(b_B = 1\) on the sliders above (the worked example's system) and check a point the same way against the diagram.

More info — why the corner matters

Every corner of a feasible region is where two of its boundary lines cross. Since two non-parallel lines form a linear system, you find that corner exactly the way you'd solve Systems of linear equations — by substitution or elimination — just applied to the boundary equations instead of the inequalities themselves. This is also the geometric idea behind linear programming: the best point in a feasible region always sits at one of its corners, so finding every corner (by solving each pair of boundary equations) is the first step. See the OpenStax link in Dive deeper for fully worked graphs.

Check your understanding

Question 1 of 4

A system requires \(y > 3x - 4\) (strict) and \(y \le -2x + 6\) (non-strict). How should the two boundary lines be drawn?

Question 2 of 4

Is (1, 1) a solution of \(y > 3x - 4\) and \(y \le -2x + 6\)?

Question 3 of 4

Two inequalities' individual shaded half-planes don't touch anywhere on the graph. What does this say about the system's solution?

Question 4 of 4

The boundary lines of a system of inequalities are \(y = 3x - 4\) and \(y = -2x + 6\). What is special about their point of intersection on the feasible-region graph?

Recap

  • A system of inequalities is solved by the overlap of the individual half-planes — the feasible region.
  • Boundary style follows the comparator: solid for \(\le\) or \(\ge\), dashed for < or >.
  • Pick the shaded side with a test point off the line (the origin, when it isn't on the line): if it satisfies the inequality, shade its side.
  • Where two non-parallel boundary lines cross is a corner of the feasible region — find it by solving those two boundary equations as a linear system.

Dive deeper

Sources

  • Systems of Linear Inequalities