Linear Inequalities

An inequality is a statement that two quantities are not necessarily equal — one is less than, greater than, or comparable to the other. A linear inequality in one variable is any statement of the form $ax + b < 0$, $ax + b \le 0$, $ax + b > 0$ or $ax + b \ge 0$, where $a \ne 0$. The strict symbols $<$ and $>$ exclude equality; the weak symbols $\le$ and $\ge$ allow it.

Solving an inequality means finding every value of the variable that makes the statement true. Unlike an equation, which usually pins the variable to one or two numbers, an inequality almost always has a whole range of solutions — an interval on the number line.

Most of the rules you already use for equations carry over unchanged. You may add or subtract the same number from both sides, and you may multiply or divide both sides by the same positive number. The one rule that catches students out is the sign-flip:

For example, $-3x \ge 6 \Rightarrow x \le -2$ — dividing by $-3$ turns $\ge$ into $\le$. The table below summarises which operations preserve the direction of the inequality and which reverse it:

The solution is then written either in set-builder form, such as $\{x : x \le -2,\ x \in \mathbb{R}\}$, or as an interval, such as $(-\infty, -2]$. On the number line we use an open circle ($\circ$) at the endpoint for a strict inequality and a closed circle ($\bullet$) for a weak one, then shade the ray of all values that satisfy it.

When the variable is restricted to the integers ($x \in \mathbb{Z}$) or naturals ($x \in \mathbb{N}$) rather than all reals, the same algebra applies but the final answer is a discrete list, for instance $\{-2, -1, 0, 1\}$, not a continuous interval.

Deeper Insight — why the sign-flip is unavoidable: The reversal rule is not an arbitrary convention you must memorise; it is forced by what "less than" actually means on the number line. Consider the plain truth $2 < 5$. Now multiply both sides by $-1$: the numbers $-2$ and $-5$ sit on the negative side, where the order is mirrored — $-5$ lies further left than $-2$, so $-2 > -5$. Multiplying by any negative number reflects every point across zero, and reflection turns "to the left of" into "to the right of", which is exactly what flipping the symbol records. This is also why you should never multiply an inequality by an expression like $x$ whose sign you do not know — you would not know whether to flip. Keep the operations transparent: isolate the variable using additions and positive multipliers wherever possible, and reach for a negative multiplier only when you consciously remember to reverse the symbol.

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.

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.

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 $2x + 3y = 6$, 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).

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.

A system of linear inequalities is two or more linear inequalities in the same two variables that must hold simultaneously. A point is a solution of the system only if it satisfies every inequality at once. Geometrically, each inequality contributes one half-plane, and the solution of the whole system is the overlap — the region common to all of them.

This common region is called the feasible region (or solution region). It is found by graphing every inequality on the same axes and then keeping only the area where all the shadings agree.

A reliable procedure for any system:

Many systems include the non-negativity constraints $x \ge 0$ and $y \ge 0$, which confine the solution to the first quadrant. These appear constantly in word problems where the variables count physical things — units produced, hours worked, items bought — that cannot be negative.

The feasible region may be bounded (enclosed on all sides, forming a polygon) or unbounded (extending to infinity in some direction). It is also possible for a system to have no solution at all: if the half-planes never overlap, the feasible region is empty.

For example, the system $x + y \le 4,\ x \ge 0,\ y \ge 0$ has a feasible region that is the triangle with corners $(0,0)$, $(4,0)$ and $(0,4)$ — a bounded region trapped in the first quadrant beneath the line $x + y = 4$.

Deeper Insight — the feasible region is where linear programming begins: Finding the common region is not an end in itself; it is the first half of one of the most useful techniques in applied mathematics, linear programming. In a typical problem the inequalities are constraints (limited materials, budget, labour) and you want to maximise or minimise some linear quantity such as profit or cost over all points that satisfy them. A beautiful theorem guarantees that the optimum, when it exists, always occurs at a corner (vertex) of the feasible region — never in the smooth interior — so once the region is drawn, the search shrinks to checking a handful of corner points. That is why precision matters here: shade the wrong half-plane and the whole optimisation collapses. Whether the region is bounded also carries meaning: a bounded region guarantees both a maximum and a minimum exist, while an unbounded one may let a quantity grow without limit. Mastering the overlap now makes the optimisation you meet later feel like a short final step rather than a new topic.