Graphical representation of inequalities

One variable, on the number line. Once an inequality is solved algebraically down to a form like $x > a$, $x \le a$, or a combined statement like $2 < x \le 7$, the solution is drawn directly on the number line.

• A strict inequality ($<$ or $>$) uses an open circle at the boundary value, because that exact value does not satisfy the inequality.
• A non-strict inequality ($\le$ or $\ge$) uses a filled circle, because the boundary value is itself part of the solution.
• Shading (or an arrow) then extends in the direction the inequality points — right for "greater than," left for "less than," and between two circles for a bounded interval such as $2 < x \le 7$.

Two variables, on the coordinate plane (formative). A linear equation $ax + by = c$ is a straight line, and it splits the plane into exactly two half-planes. An inequality such as $ax + by < c$ describes one of them:

where: $a, b, c$ are the fixed coefficients of the boundary line, $x$ and $y$ are the coordinates of a point being tested, and the boundary is solid because $\le$ includes equality (a strict version, $<$, would use a dashed line instead). Deciding which side to shade uses a single convenient test point that is not on the line — the origin $(0,0)$ whenever the line misses it. Substituting the test point's coordinates into the inequality gives a true or false statement: true means shade the side containing that point; false means shade the other side. For a system of two inequalities, both half-planes are drawn on the same axes, and the solution of the system is wherever the two shaded regions overlap.

The test-point method for shading a half-plane relies on one simple fact: the line $ax + by = c$ is the only place where the expression $ax + by - c$ changes sign, so a single check away from the line settles the whole side.

Step 1. Consider the expression $E(x,y) = ax + by - c$ for a fixed line $ax+by=c$. On the line itself, $E = 0$ by definition. Everywhere else in the plane, $E$ is either strictly positive or strictly negative, and because $E$ is a plain linear (continuous) function of $x$ and $y$, it can only pass from positive to negative by crossing through zero — that is, by crossing the boundary line.

Step 2. This means the plane splits into exactly two connected regions: one where $E > 0$ everywhere, and one where $E < 0$ everywhere. Within either region, moving around without crossing the line can never flip the sign of $E$, since flipping the sign would force $E$ through zero somewhere off the line, which is impossible for this linear expression.

Step 3. So checking the sign of $E$ at any single point inside a region tells you the sign of $E$ for the entire region. Pick a test point not on the line, substitute it into the original inequality (equivalently, check the sign of $E$ there), and note whether the inequality is true or false.

Step 4. Because every other point in that same region shares the same sign, the truth value found at the test point holds for the whole region. If the test point makes the inequality true, its entire region is the solution; if false, the other region — the one on the opposite side of the line — is the solution instead. This is exactly why testing one convenient point (usually the origin) is enough to shade correctly, with no need to check infinitely many points.

Graph the system $2x + y < 6$ and $x + y > 2$ on the same coordinate plane, and identify the overlapping solution region.

Step 1. Graph the first inequality, reusing the earlier result: the boundary line $2x+y=6$ passes through $(0,6)$ and $(3,0)$, drawn dashed since the inequality is strict. Testing the origin gives $2(0)+0 = 0 < 6$, which is true, so the half-plane containing the origin is shaded for this inequality.

Step 2. Graph the second inequality. The boundary line $x+y=2$ passes through $(0,2)$ and $(2,0)$, also drawn dashed since $x+y>2$ is strict. Testing the origin gives $0+0 = 0 > 2$, which is false, so the half-plane not containing the origin — the side farther from the origin — is shaded for this inequality.

Step 3. The solution of the system is the region where both shaded half-planes overlap: below-left of the line $2x+y=6$ and above-right of the line $x+y=2$, forming a band between the two dashed lines.

Step 4. Check a point inside that band, say $(1,2)$: $2(1)+2 = 4 < 6$ is true, and $1+2 = 3 > 2$ is true, so $(1,2)$ satisfies both inequalities and correctly lies inside the overlap. The graphical solution of the system is the open band-shaped region between the dashed lines $2x+y=6$ and $x+y=2$, on the side of each line found by the origin test — the region containing points such as $(1,2)$.

• Solid line for a strict inequality — $<$ and $>$ always get a dashed boundary (or an open circle on the number line); only $\le$ and $\ge$ get a solid boundary (closed circle).
• Shading by eye instead of testing — always substitute an actual test point rather than guessing from the direction of the inequality symbol.
• Testing a point that lies on the boundary line — this gives $E = 0$, no information at all; choose a point clearly off the line.
• Treating this as Linear Programming — there is no objective function or "feasible region" to optimise here, only the plain solution set of the inequality or system.

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