Graph linear inequalities in two variables and systems of linear inequalities, identifying the solution region and testing whether a point is a solution, including in real-world constraint contexts (MA.912.AR.9.6).

What this topic is asking

MA.912.AR.9 extends inequalities to two variables and to systems. You graph each inequality as a shaded half-plane and, for a system, find the overlap where all are satisfied. You also test a point to decide if it is a solution. The B.E.S.T. Algebra 1 EOC tests boundary style, shading, the overlap region, and point-testing, often in a real constraint context.

Graphing one inequality

Treat the inequality like a line, then decide the boundary style and the shading:

• Boundary style. Solid line for $\le$ or $\ge$ (the line is included); dashed line for $<$ or $>$ (excluded).
• Shading. Test a point not on the line (the origin $(0,0)$ is easiest when it is not on the line). If the inequality is true there, shade that side; if false, shade the other side.

Testing a point

To check whether $(x, y)$ is a solution of an inequality, substitute and evaluate the truth of the statement. For a system, the point must satisfy all the inequalities.

How the B.E.S.T. EOC examines this topic

• GRID and hot-spot. Choose the boundary style and shade, or click the solution region.
• Multiple choice and multiselect. Decide whether a point is a solution, or select all points in the region.
• Context items. Model a constraint (budget and time, materials) and identify feasible combinations.

A clarifying idea: a two-variable inequality divides the plane into two half-planes, and the solutions are an entire half-plane, not a line. A system narrows the solutions to the intersection of those half-planes, the only zone where every condition holds at once.

Why the overlap is the solution

For a system, an ordered pair must make every inequality true simultaneously, just as a system of equations needed a point on every line. Each inequality's solutions form a half-plane, so a pair satisfying all of them must lie in all of those half-planes at once, which is exactly their overlap. Anywhere outside the overlap, at least one condition fails, so that point is rejected even if it satisfies the others. This is why shading every inequality and reading off the common region works: the multiply-shaded zone is the set of points that passed all the tests. In a constraint problem this overlap is the feasible region, the combinations that meet all the real-world limits.

Real-world constraints

Many EOC system items model limits: a budget (\text{cost} \le \100$), a time cap, a minimum requirement. Each limit is one inequality; the feasible region is the combinations that satisfy all of them. Because the quantities are often counts, the realistic solutions are the whole-number points inside the region.

Reading a region back into an inequality

The EOC sometimes runs the process backward: it shows a shaded half-plane and asks which inequality it represents. Work from the boundary out. First read the boundary line's equation (its slope and $y$-intercept) to get the $y = mx + b$ part. Next check the line style: a solid line means the inequality is $\le$ or $\ge$, a dashed line means $<$ or $>$. Finally use the shaded side to pick the direction: test a point in the shaded region, and whichever inequality direction makes that point true is the answer. If the region above a solid line $y = x + 1$ is shaded, testing a high point like $(0, 5)$ gives $5 \ge 1$, so the inequality is $y \ge x + 1$. Doing this for each boundary lets you reconstruct an entire system from its feasible region, which is a common matching item.

Try this

Q1. Should the boundary of $y > 3x - 2$ be solid or dashed? [1 point]

• Cue. Dashed, because $>$ excludes the line.

Q2. Is $(0, 0)$ a solution of $y \le 2x + 1$? [1 point]

• Cue. $0 \le 1$ is true, so yes.