Linear Inequalities in Two Variables

A linear inequality in two variables has the form $ax + by < c$ (or with $\le$, $>$, $\ge$), where $a$ and $b$ are not both zero. Whereas an inequality in one variable carves out a stretch of the number line, an inequality in two variables carves out a region of the plane — a half-plane. Its solution is a set of ordered pairs $(x, y)$, not single numbers.

The starting point is the boundary line $ax + by = c$. This line splits the whole $xy$-plane into two halves. Every point in one half makes $ax + by < c$ true; every point in the other half makes $ax + by > c$ true. The line itself is where $ax + by = c$ exactly.

Drawing the boundary line. The quickest way to plot $ax + by = c$ is by its intercepts: put $y = 0$ to get the $x$-intercept $\left(\tfrac{c}{a}, 0\right)$, and put $x = 0$ to get the $y$-intercept $\left(0, \tfrac{c}{b}\right)$, then join the two points. Special boundaries are even easier — $x = k$ is a vertical line and $y = k$ is a horizontal line.

Two decisions turn the line into a complete picture of the solution set. The first is whether the boundary is part of the solution. For a weak inequality ($\le$ or $\ge$) the points on the line satisfy it, so we draw a solid line. For a strict inequality ($<$ or $>$) the line is excluded, so we draw a dashed (broken) line.

The second decision is which half to shade. Here the test-point method is fast and foolproof: pick any point not on the line, substitute it into the inequality, and check whether the statement is true. If it is true, shade the side containing that point; if it is false, shade the other side. The origin $(0,0)$ is the easiest test point whenever the line does not pass through it.

For instance, to graph $2x + 3y \le 6$: draw the solid line $2x + 3y = 6$ through $(3, 0)$ and $(0, 2)$, then test $(0,0)$ — $2(0) + 3(0) = 0 \le 6$ is true, so shade the side that contains the origin (the side toward the lower-left). When the boundary passes through the origin (for example $y = x$), the origin gives $0 = 0$ and cannot decide; pick a clear off-line point such as $(1, 0)$ instead.

Axis inequalities appear so often they are worth memorising on sight: $x \ge 0$ is the half-plane on and to the right of the $y$-axis, $x \le 0$ is to its left, $y \ge 0$ is on and above the $x$-axis, and $y \le 0$ is below it. These four are the building blocks of the non-negativity constraints that confine many problems to the first quadrant.

Deeper Insight — why one test point settles everything: The reason a single test point decides the shading for an entire half-plane is that the quantity $ax + by$ changes monotonically as you move directly across the line. On the boundary it equals $c$; step to one side and it grows larger, step to the other and it shrinks — it can never "come back" to satisfy the opposite inequality without re-crossing the line. So every point on the same side of the line gives the same true-or-false verdict, which is precisely why checking one representative point is enough to colour all of them. Choosing the origin is simply a convenience because the arithmetic collapses to comparing $0$ with $c$; the only time you must pick a different point is when the line itself passes through $(0,0)$, in which case any off-line point such as $(1,0)$ does the job. This half-plane idea is the foundation of linear programming, where overlapping half-planes fence off the feasible region in which an optimal solution is hunted.