Systems of linear inequalities
Learning Objectives
- Recognise a linear inequality in two unknowns and its solutions as ordered pairs.
- Show that the solution set of such an inequality is a half-plane.
- Graph an inequality using the boundary line and a test point.
- Find the feasible region of a system of inequalities as an intersection of half-planes.
- Model two-constraint situations and check whether a given plan is feasible.
Constraints, not conditions
An equation in two unknowns says something exact: the pair must land precisely on a line. But most real requirements are not exact — they say "at most this much money," "at least this many hours," "no more than the truck can carry." These are constraints, and the natural language for them is the inequality.
In the last unit we solved inequalities in one unknown and found intervals. With two unknowns the solutions become regions of the plane — and a system of constraints carves out the region where every requirement holds at once.
Linear inequality in two unknowns
A linear inequality in two unknowns \(x\) and \(y\) is an inequality that can be written in one of the forms
\[ax + by \leq c, \qquad ax + by < c, \qquad ax + by \geq c, \qquad ax + by > c,\]
where \(a, b, c \in \mathbb{R}\) and \(a, b\) are not both zero. A solution is an ordered pair \((x_0, y_0)\) making the statement true; the solution set is a subset of the Cartesian plane.
The solution set is a half-plane
The line \(ax + by = c\) divides the plane into two sides. The solutions of \(ax + by \leq c\) are exactly the points of the boundary line together with all the points on one of its two sides — a half-plane. For the strict inequality \(ax + by < c\), the same side without the boundary line.
Suppose first \(b > 0\). Using the transformation rules for inequalities, \(ax + by \leq c\) is equivalent to
\[y \leq \frac{c - ax}{b} = \frac{c}{b} - \frac{a}{b}\,x.\]
The right-hand side is the height of the line \(ax + by = c\) above the input \(x\). So the solutions are precisely the points that are on or below the line: for each \(x\), the heights \(y\) from the line downwards. That is one side of the line, plus the line itself.
If \(b < 0\), dividing by \(b\) reverses the inequality (the rule from the previous unit), giving \(y \geq \tfrac{c}{b} - \tfrac{a}{b}x\): the points on or above the line — again one whole side.
If \(b = 0\) (so \(a \neq 0\)), the inequality reads \(ax \leq c\), i.e. \(x \leq \tfrac{c}{a}\) or \(x \geq \tfrac{c}{a}\) depending on the sign of \(a\): everything to one side of the vertical line \(x = \tfrac{c}{a}\). In every case the solution set is a half-plane.
Graphing an inequality: the test-point method
The theorem licenses a simple recipe. To graph \(ax + by \leq c\):
Draw the boundary line \(ax + by = c\) — solid if the inequality is \(\leq\) or \(\geq\) (boundary included), dashed if it is \(<\) or \(>\) (boundary excluded).
Pick a test point not on the line — the origin \((0,0)\) is the usual choice — and substitute it. If the inequality holds, shade the side containing the test point; if not, shade the other side.
For \(x + 2y \leq 4\): the boundary passes through \((0, 2)\) and \((4, 0)\); testing the origin gives \(0 + 0 = 0 \leq 4\) — true — so we shade the origin's side.
System of inequalities and the feasible region
A system of linear inequalities is a collection of such inequalities considered together. A solution of the system is a pair satisfying all of them at once, so the solution set is the intersection of the individual half-planes.
When the inequalities express the constraints of a practical problem, this intersection is called the feasible region: the set of all plans that respect every constraint.
A triangle from three constraints
Take the system
\[\begin{cases} x \geq 0 \\ y \geq 0 \\ x + y \leq 3 \end{cases}\]
The first two constraints keep us in the upper-right quadrant; the third keeps us on the origin's side of the line \(x + y = 3\) (test: \(0 + 0 \leq 3\) ✓). The intersection is a triangle with vertices \((0, 0)\), \((3, 0)\), and \((0, 3)\) — each vertex the meeting point of two boundary lines, found exactly as in Lesson 1: by solving the system of the two boundary equations.
Two constraints in a workshop
A craftsperson makes bracelets and necklaces. A bracelet takes \(1\) hour of work and €3 of materials; a necklace takes \(2\) hours and €2 of materials. This week there are at most \(8\) hours and a materials budget of €12. With \(x\) bracelets and \(y\) necklaces, the constraints are
\[\begin{cases} x + 2y \leq 8 & \text{(hours)} \\ 3x + 2y \leq 12 & \text{(budget)} \\ x \geq 0,\; y \geq 0 \end{cases}\]
The feasible region is a quadrilateral with vertices \((0,0)\), \((4,0)\), \((2,3)\), and \((0,4)\). The interesting vertex \((2, 3)\) is where both resources are used to the full — the intersection of the lines \(x + 2y = 8\) and \(3x + 2y = 12\), found by elimination: subtracting gives \(2x = 4\), so \(x = 2\), \(y = 3\).
The region answers practical questions instantly. Is the plan "3 bracelets and 2 necklaces" feasible? Hours: \(3 + 4 = 7 \leq 8\) ✓ — but budget: \(9 + 4 = 13 > 12\) ✗. The point \((3, 2)\) lies outside the region: not feasible. The plan \((2, 2)\), by contrast, satisfies both (\(6 \leq 8\), \(10 \leq 12\)) and sits inside.
Connection to Computer Science
Checking whether a point lies in the feasible region is exactly what a scheduling or planning program does: take a proposed plan, run down the list of constraints, and accept the plan only if every single check passes. The feasible region is the set of all plans that would pass — the program never draws it, but it is implicitly exploring it, one point at a time.
What shapes can a feasible region take?
Our two examples produced a triangle and a quadrilateral — tidy, finite shapes. Is that guaranteed?
No. The region \(x \geq 0,\; y \geq 0\) (just two constraints) is an entire infinite quadrant: a feasible region can be unbounded. And the system
\[\begin{cases} x + y \leq 1 \\ x + y \geq 3 \end{cases}\]
demands a sum that is at most \(1\) and at least \(3\) at the same time — its feasible region is empty. This is the incompatible case again, in inequality form: two half-planes bounded by parallel lines, facing away from each other.
What is guaranteed is less obvious and more interesting: as an intersection of half-planes, the region can have no dents or holes — a segment between two of its points never leaves it. Regions with that property are called convex, and their corners play a starring role when one asks not just "which plans are allowed?" but "which allowed plan is best?" — a question taken up by a field called linear programming, well beyond this course but built exactly on the picture we just drew.
Exercises
Exercise 1
Decide whether the pair \((1, 2)\) satisfies each inequality.
Exercise 2
Answer each question about graphing inequalities.
Exercise 3
The workshop region from the lesson is bounded by \(x + 2y \leq 8\), \(3x + 2y \leq 12\), \(x \geq 0\), \(y \geq 0\).
Exercise 4
Decide whether each statement about feasible regions is true or false.
Exercise 5
Recall the workshop constraints: \(x + 2y \leq 8\) (hours), \(3x + 2y \leq 12\) (budget), \(x \geq 0\), \(y \geq 0\).
Summary
A linear inequality in two unknowns \(ax + by \leq c\) has a half-plane as its solution set: the boundary line \(ax + by = c\) plus one of its sides (boundary excluded for strict inequalities — drawn dashed).
To graph one: draw the boundary, test a point (usually the origin), and shade the side where the test succeeds.
A system of inequalities selects the intersection of the half-planes — the feasible region, the set of all plans satisfying every constraint. Its vertices are found by solving pairs of boundary equations, with the methods of this unit.
The region may be bounded, unbounded, or empty (incompatible constraints). A plan is feasible only if it passes all the constraints — failing one is failing.