Graph the solution set of a system of two or more linear inequalities as the overlap of the half-planes (LA A1: A-REI.D.12).

What this topic is asking

Standard A1: A-REI.D.12 also covers a system of linear inequalities: graph each one and find the overlap of the half-planes. On LEAP 2025 these are Type I Major Content items, graphing items, multiple choice about the solution region, or test-point questions. The solution is the region where all the shadings meet, often called the feasible region in modeling.

Graph each, then find the overlap

Treat each inequality with the half-plane method, then look for the common region.

A point like $(2, 1)$ lies in the overlap (it satisfies both), so it is a solution.

Testing a point against all constraints

A point solves the system only if it satisfies every inequality. Substitute it into each; if even one is false, the point is not a solution, even if the others hold.

The shape of the solution region

The overlap can be:

• An unbounded region (open on one or more sides), as in the wedge above.
• A bounded polygon when enough inequalities fence in a closed area (common in modeling with constraints like $x \ge 0$ and $y \ge 0$).
• Empty if the half-planes do not share any points (no solution).

How LEAP examines this topic

• Graphing item. Graph two or more inequalities and shade or identify the overlap.
• Multiple choice. Describe the solution set, or pick the graph whose overlap matches the system.
• Test-point item. Decide whether a point satisfies all inequalities.

A clarifying idea: shading each half-plane lightly and looking for the darkest region (where shadings stack) is a quick way to spot the overlap on paper or on screen.

Why the solution is the overlap of half-planes

The solution set of a system of inequalities is the intersection of the individual solution sets, which is the same logic that makes a system of equations resolve to the intersection of lines. Each inequality, on its own, is satisfied by an entire half-plane. The system demands that all the inequalities hold simultaneously, so a point qualifies only if it belongs to every half-plane at once, and the set of points lying in all of them is precisely their overlap. This is why a single point can fail the system while passing one inequality: membership in one half-plane is not membership in their intersection. The structure also explains the possible shapes. Two half-planes overlap in a wedge or strip; adding more constraints clips the region further, and constraints like $x \ge 0$, $y \ge 0$ confine it to one quadrant, often producing a closed polygon, the feasible region of an optimization problem. If the constraints contradict, the half-planes share no point and the system has no solution. Reading the overlap as "satisfy all constraints together" is exactly the reasoning Type III modeling items reward when the inequalities describe limits on resources, time, or quantities.

Try this

Q1. Is $(0, 0)$ a solution to $y < x + 1$ and $y \ge -2$? [2 points]

• Cue. $0 < 1$ true and $0 \ge -2$ true, so yes.

Q2. What does the solution set of a two-inequality system look like? [1 point]

• Cue. The overlapping (shared) region of the two shaded half-planes.