Systems of Linear Inequalities — I can solve and graph systems of linear inequalities and interpret the feasible region. - Math 1

SECTION 1

Historical Context & Motivation

Long before graphing calculators existed, mathematicians and economists faced a practical problem: how do you find the best possible outcome when you have multiple constraints acting at the same time? For instance, a factory might need to maximize profit while staying within limits on labor hours, raw materials, and machine capacity. Each of those constraints can be expressed as a linear inequality, and the solution lies where all of them overlap.

The idea of solving multiple inequalities simultaneously grew out of centuries of work in algebra and geometry. As trade, industry, and warfare became more complex, the need for systematic optimization drove mathematicians to develop what we now call linear programming — a field built directly on systems of linear inequalities.

1637 • Descartes & Coordinate Geometry

René Descartes published La Géométrie, introducing the coordinate plane and making it possible to represent algebraic relationships as graphs — the foundation for graphing inequalities.

1826 • Fourier's Inequality Methods

Joseph Fourier developed early techniques for solving systems of linear inequalities, laying groundwork that would be formalized more than a century later.

1939 • Kantorovich & Linear Programming

Soviet mathematician Leonid Kantorovich used systems of linear inequalities to optimize resource allocation during World War II, pioneering linear programming.

1947 • Dantzig's Simplex Method

George Dantzig invented the simplex algorithm, which efficiently finds optimal solutions within feasible regions defined by systems of linear inequalities. This method is still used today.

Today, every time a shipping company plans delivery routes, a nutritionist designs a meal plan, or a business allocates its budget, systems of linear inequalities are working behind the scenes. The central question this lesson addresses is: How do we graph multiple linear inequalities on the same coordinate plane and identify the region where every constraint is satisfied at once?

SECTION 2

Core Principles & Definitions

Before working with a system of inequalities, you need to be comfortable with a few foundational ideas. A single linear inequality like y > 2x + 1 divides the coordinate plane into two half-planes: one where the inequality is true and one where it is false. When you combine two or more such inequalities into a system, you are looking for the set of all points (x, y) that make every inequality true simultaneously.

Linear Inequality

An inequality using ≤, ≥, <, or > with a linear expression. It represents a half-plane on one side of a boundary line.

Boundary Line

The line you get when you replace the inequality symbol with an equals sign. Draw it solid for ≤ or ≥ (included) and dashed for < or > (excluded).

Shading (Half-Plane)

After drawing the boundary line, test a point (usually the origin) to determine which side to shade. The shaded region contains all solutions to that single inequality.

Feasible Region

The overlap of all shaded half-planes in the system. Every point in this region satisfies all inequalities at once. It may be bounded (a polygon) or unbounded.

Vertices (Corner Points)

The points where boundary lines intersect on the edge of the feasible region. In optimization problems, the best solution often occurs at one of these vertices.

✦

KEY TAKEAWAY

Think of each inequality as a velvet rope at a concert venue. One rope blocks off the left side, another blocks off the back. The feasible region is like the VIP section — the area where you are inside all the ropes at the same time. If even one rope keeps you out, you are not in the feasible region.

SECTION 3

Visual Explanation — Graphing a System

The diagram below shows the system y ≤ −x + 6 and y ≥ x − 2 graphed on the same coordinate plane. The first inequality shades below and on the line y = −x + 6 (solid boundary, blue). The second inequality shades above and on the line y = x − 2 (solid boundary, pink). Where the two shaded regions overlap is the feasible region — shown in a blended purple tone.

The blue line represents y ≤ −x + 6 and the pink line represents y ≥ x − 2. The purple-shaded overlap is the feasible region. The vertex at (4, 2) is where the two boundary lines intersect.

Notice that both boundary lines are solid because the inequalities use ≤ and ≥, meaning points on the lines themselves are included in the solution set. The vertex at (4, 2) is found by solving the system of equations y = −x + 6 and y = x − 2 simultaneously. In optimization problems, this vertex — and any others forming the boundary of the feasible region — would be the first places you check for maximum or minimum values.

SECTION 4

Mathematical Framework

Graphing a system of linear inequalities follows a consistent process. Each inequality is handled one at a time, and then the individual shaded regions are combined. Let's formalize the key steps and notation.

GENERAL LINEAR INEQUALITY

y ≤ mx + b or y ≥ mx + b (also < and >)

m = slope of the boundary line, b = y-intercept. The inequality symbol determines which side of the boundary line to shade.

BOUNDARY LINE TYPE

≤ or ≥ → solid line (points on line included) < or > → dashed line (points on line excluded)

A solid boundary means that points exactly on the line satisfy the inequality. A dashed boundary means they do not.

TEST POINT METHOD

Substitute (0, 0) into the inequality. If true → shade the side containing the origin. If false → shade the opposite side.

Use the origin (0, 0) as a test point unless the boundary line passes through it. In that case, choose another convenient point such as (1, 0).

FINDING A VERTEX

Set the two boundary equations equal: m₁x + b₁ = m₂x + b₂ → solve for x, then substitute back to find y.

Vertices are the corner points of the feasible region. They are found by solving pairs of boundary lines as a system of equations.

💡

Quick Check

After shading, pick any point inside your feasible region and substitute it into every inequality. If all inequalities come out true, you graphed correctly. If even one fails, re-examine your shading.

SECTION 5

Bounded vs. Unbounded Feasible Regions

Not all feasible regions look the same. Depending on the inequalities in your system, the overlap can form a closed polygon (a bounded region) or it can extend infinitely in one or more directions (an unbounded region). There is also the possibility that the shaded regions do not overlap at all, meaning no solution exists.

Three possible outcomes when graphing a system of linear inequalities: a bounded feasible region (left), an unbounded feasible region (center), and no solution (right).

Type	Description	Example Scenario
Bounded	Feasible region is a closed polygon with finite area. All vertices can be identified.	A manufacturer with limits on both inputs and outputs — every resource is capped.
Unbounded	Feasible region extends infinitely in at least one direction. Some vertices exist, but the region has no finite area.	A budget constraint with no upper limit on one variable — you could theoretically buy infinite amounts if cost allows.
No Solution	The half-planes do not overlap at all. No point satisfies every inequality simultaneously.	Contradictory constraints, like requiring y > 5 and y < 2 at the same time.

Types of feasible regions

SECTION 6

Worked Example

Let's work through a complete example. Graph the following system and identify the feasible region and its vertices:

SYSTEM OF INEQUALITIES

y ≤ −2x + 8 y ≥ (1/2)x − 1 x ≥ 0

Three inequalities with three boundary lines. The third inequality restricts us to the right side of the y-axis.

Graphing a System of Three Inequalities

1
Step 1 — Graph the first boundary line

Rewrite y ≤ −2x + 8 as the equation y = −2x + 8. The slope is −2 and the y-intercept is 8. Plot the y-intercept at (0, 8) and use the slope to find another point: from (0, 8), move right 1 and down 2 to reach (1, 6). Draw a solid line because the symbol is ≤.

Solid line through (0, 8) and (1, 6); shade below the line.
2
Step 2 — Graph the second boundary line

Rewrite y ≥ (1/2)x − 1 as y = (1/2)x − 1. The slope is 1/2 and the y-intercept is −1. Plot (0, −1) and use the slope to find (2, 0). Draw a solid line because the symbol is ≥.

Solid line through (0, −1) and (2, 0); shade above the line.
3
Step 3 — Graph the third boundary line

The inequality x ≥ 0 means the boundary is the y-axis itself. Draw a solid vertical line along x = 0 and shade to the right.

Solid vertical line at x = 0; shade to the right.
4
Step 4 — Identify the feasible region

The feasible region is where all three shaded areas overlap. It is a triangular region bounded on the left by the y-axis, above by y = −2x + 8, and below by y = (1/2)x − 1.

The feasible region is a triangle.
5
Step 5 — Find the vertices

Vertex 1 (intersection of x = 0 and y = −2x + 8): substitute x = 0 → y = 8, giving (0, 8). Vertex 2 (intersection of x = 0 and y = (1/2)x − 1): substitute x = 0 → y = −1, giving (0, −1). Vertex 3 (intersection of y = −2x + 8 and y = (1/2)x − 1): set −2x + 8 = (1/2)x − 1 → 9 = (5/2)x → x = 18/5 = 3.6, then y = −2(3.6) + 8 = 0.8, giving (3.6, 0.8).

Vertices: (0, 8), (0, −1), and (3.6, 0.8)
6
Step 6 — Verify with a test point

Choose a point inside the triangle, such as (1, 2). Check: 2 ≤ −2(1) + 8 = 6 ✓. Check: 2 ≥ (1/2)(1) − 1 = −0.5 ✓. Check: 1 ≥ 0 ✓. All three inequalities are satisfied, confirming the feasible region is correct.

Test point (1, 2) satisfies all three inequalities. ✓

SECTION 7

Common Mistakes & Tips

Even after mastering the process, students often lose points on avoidable errors. The table below compares common mistakes with the correct approach so you can spot and fix issues quickly.

Common Mistake	Why It's Wrong	Correct Approach
Using a dashed line for ≤ or ≥	≤ and ≥ include the boundary. A dashed line excludes it.	Solid line for ≤ and ≥; dashed for < and >.
Shading the wrong side	Guessing which side to shade without testing leads to an inverted region.	Always substitute a test point (like the origin) into the inequality.
Forgetting to flip the inequality when multiplying or dividing by a negative	If you rearrange an inequality and divide both sides by a negative number, the direction reverses.	Flip the inequality symbol any time you multiply or divide by a negative.
Identifying the feasible region as any shaded area instead of the overlap	The solution must satisfy all inequalities, not just one.	Look for the region where every shading overlaps.
Not labeling vertices	Vertices are often required in optimization problems; missing them loses credit.	Solve pairs of boundary-line equations to find every intersection point on the feasible region.

Common mistakes when graphing systems of linear inequalities

💡

PRO TIP

When you have more than two inequalities, color each half-plane with a different translucent shade on your graph. The feasible region is wherever all the colors stack on top of each other. This visual layering technique makes the overlap much easier to spot, especially on paper.

SECTION 8

Connection to Linear Programming

Systems of linear inequalities are the backbone of linear programming, which you may encounter in Algebra 2, Pre-Calculus, or college-level courses. In linear programming, you graph a system of inequalities to find the feasible region and then optimize an objective function — a formula you want to maximize or minimize. The key theorem states that the optimal value of the objective function always occurs at one of the vertices of the feasible region.

This Lesson	Linear Programming (Next Level)
Graph a system of inequalities and shade the feasible region.	Graph the system, shade the feasible region, then evaluate an objective function at each vertex.
Identify vertices by solving pairs of boundary equations.	Same technique — but now you plug each vertex into the objective function to compare values.
Determine whether the feasible region is bounded or unbounded.	If unbounded, the maximum (or minimum) of the objective function may not exist — critical for real-world feasibility.
Use two variables (x and y) on a 2D graph.	Advanced methods handle 3 or more variables using the simplex algorithm (beyond graphing).

How this lesson connects to linear programming

Mastering the skills in this lesson — accurate graphing, correct shading, and vertex identification — gives you the tools you need to step directly into linear programming. The only new piece is the objective function, and the feasible region work stays exactly the same.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why a system of linear inequalities can have infinitely many solutions, while a system of linear equations (two distinct, non-parallel lines) typically has exactly one solution.

Show answer

A system of two linear equations represents two lines, and two distinct non-parallel lines cross at exactly one point, which is the single solution. A system of linear inequalities, on the other hand, represents half-planes — entire regions of the coordinate plane. When those regions overlap, the overlap itself is a region (not just a point), and a region contains infinitely many points. Each of those points is a valid solution because it satisfies every inequality in the system. That is why there are infinitely many solutions rather than just one.

Question 2: BASIC CALCULATION

Graph the system y > x + 1 and y ≤ −x + 5. State whether each boundary line is solid or dashed, and find the coordinates of the vertex where the boundary lines intersect.

Show answer

The boundary for y > x + 1 is dashed because the inequality is strict (>). The boundary for y ≤ −x + 5 is solid because it uses ≤. To find the vertex, set x + 1 = −x + 5. Adding x to both sides gives 2x + 1 = 5, so 2x = 4 and x = 2. Substituting back: y = 2 + 1 = 3. The vertex is at (2, 3). Note that this point is not included in the feasible region because the first inequality is strict.

Question 3: INTERMEDIATE

Graph the system: 2x + y ≤ 10, x + 3y ≤ 15, x ≥ 0, y ≥ 0. Identify all vertices of the feasible region.

Show answer

Start by rewriting the first two inequalities in slope-intercept form: y ≤ −2x + 10 and y ≤ −(1/3)x + 5. With x ≥ 0 and y ≥ 0, the feasible region is confined to the first quadrant. Vertex at the origin: (0, 0). Intersection of x = 0 and y = −2x + 10 gives (0, 10), but check against x + 3(10) = 30 > 15, which fails, so test x = 0 with y = 5 from the second inequality: (0, 5). Intersection of y = 0 with 2x + 0 = 10 gives (5, 0); check 5 + 0 = 5 ≤ 15 ✓. Intersection of 2x + y = 10 and x + 3y = 15: from the first equation y = 10 − 2x; substitute into the second: x + 3(10 − 2x) = 15 → x + 30 − 6x = 15 → −5x = −15 → x = 3, y = 4. The vertices are (0, 0), (5, 0), (3, 4), and (0, 5).

Question 4: APPLIED

A bakery makes cupcakes (x) and cookies (y) each day. Each cupcake requires 3 minutes of oven time and each cookie requires 1 minute. The oven is available for at most 90 minutes per day (3x + y ≤ 90). The bakery also needs to produce at least 10 cookies (y ≥ 10) and at least 5 cupcakes (x ≥ 5). Write and graph the system of inequalities, then identify the feasible region's vertices.

Show answer

The system is: 3x + y ≤ 90, y ≥ 10, x ≥ 5. First, find the vertices. Intersection of x = 5 and y = 10: (5, 10). Check: 3(5) + 10 = 25 ≤ 90 ✓. Intersection of x = 5 and 3x + y = 90: 15 + y = 90, so y = 75 → vertex (5, 75). Intersection of y = 10 and 3x + y = 90: 3x + 10 = 90, so 3x = 80, x ≈ 26.67 → vertex (26.67, 10). All three boundary lines are solid. The feasible region is a triangle with vertices at (5, 10), (5, 75), and (26.67, 10). Any point inside this triangle represents a valid daily production plan that respects all constraints.

Question 5: CRITICAL THINKING

Consider the system: y ≤ 2x + 4, y ≥ 2x − 3, and x ≤ 6. Describe the shape of the feasible region without graphing, and explain why the two boundary lines y = 2x + 4 and y = 2x − 3 never produce a vertex between themselves. Could this system ever have no solution? Justify your reasoning.

Show answer

The lines y = 2x + 4 and y = 2x − 3 are parallel because they both have slope 2. Parallel lines never intersect, so they cannot form a vertex between themselves. The feasible region between them is a horizontal strip of width 7 (vertical distance between the lines) that extends infinitely to the left and is cut off on the right by x = 6. The vertices are found where x = 6 crosses each line: (6, 16) and (6, 9). The feasible region is an unbounded strip. This particular system always has a solution because the strip between the parallel lines is non-empty — the line y = 2x + 4 is always above y = 2x − 3. If the inequalities were reversed (y ≥ 2x + 4 and y ≤ 2x − 3), the system would have no solution because no point can be above the higher line and below the lower line simultaneously.

SUMMARY

Lesson Summary

Summary

A system of linear inequalities consists of two or more linear inequalities considered together. To solve the system, graph each boundary line — solid for ≤ or ≥, dashed for < or > — and shade the correct half-plane using the test-point method. The feasible region is the overlap of all shaded areas — every point in it satisfies every inequality at the same time.

The feasible region can be bounded (a closed polygon), unbounded (extending to infinity), or nonexistent if the inequalities are contradictory. The vertices of the feasible region are found by solving pairs of boundary-line equations, and they play a central role in linear programming, where the goal is to optimize an objective function. Mastering graphing and interpretation of the feasible region prepares you for these advanced applications.