Grade 9 Math • Algebra & Functions

Systems of Inequalities — I can graph and interpret systems of linear inequalities to identify feasible regions.

Learn how overlapping shaded regions on a graph reveal every possible solution to real-world constraints.

Grade 9 Math emblem with a calculator, geometric shapes, and line graph
Inside this lesson 10 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

Mathematics has always been driven by practical problems: how to divide land, plan trade routes, or allocate scarce resources. The idea of expressing constraints as inequalities—statements that use symbols like ≤, ≥, <, and >—dates back centuries, but it was only in the twentieth century that mathematicians developed a formal, visual way to handle systems of inequalities. These systems allow us to combine multiple constraints and find the set of all solutions that satisfy every condition at once.

From military logistics in World War II to modern-day app development, the ability to graph inequalities and identify feasible regions has shaped how we solve optimization problems. Understanding this topic gives you the tools to tackle questions like: "Given a budget and a time limit, what combinations of activities can I afford?"

1826 • Fourier's Inequality Ideas

French mathematician Joseph Fourier explored systems of linear inequalities while studying heat conduction, laying early groundwork for constraint-based reasoning.

1936 • Motzkin's Formal Framework

Theodore Motzkin formalized the theory of linear inequalities in his doctoral thesis, establishing rules for when systems have solutions and describing the geometry of solution sets.

1939 • Kantorovich & Resource Allocation

Soviet mathematician Leonid Kantorovich used linear inequalities to optimize plywood production, one of the first real-world applications of what became linear programming.

1947 • Dantzig's Simplex Method

George Dantzig developed the simplex algorithm for the U.S. Air Force, enabling computers to find optimal solutions within feasible regions defined by hundreds of inequalities.

Today • Modern Optimization Everywhere

Systems of inequalities now power airline scheduling, supply chain management, video game physics engines, and even the algorithms that recommend your favorite playlists.

The central question this lesson addresses is: when you have two or more linear inequalities at the same time, how do you find and visualize every point that satisfies all of them? That overlapping region on a graph—the feasible region—is the key to answering this question.

SECTION 2

Core Principles & Definitions

Before we start graphing, let's nail down the key vocabulary and ideas. A linear inequality looks like a linear equation (such as y = 2x + 1) but replaces the equals sign with an inequality symbol: <, >, ≤, or ≥. Instead of producing a single line on the coordinate plane, a linear inequality defines an entire half-plane—all the points on one side of the boundary line. A system of inequalities is simply two or more inequalities considered together. The solution is the set of all (x, y) pairs that make every inequality true at the same time.

1

Boundary Line

The line you get when you replace the inequality symbol with an equals sign. Draw it solid for ≤ or ≥ (points on the line are included) and dashed for < or > (points on the line are excluded).
2

Shaded Region (Half-Plane)

After drawing the boundary line, shade the side that contains solutions. Use a test point (often the origin) to determine which side to shade.
3

Feasible Region

The area where all shaded regions overlap. Every point inside this region satisfies every inequality in the system simultaneously.
4

Test Point Method

Pick any point not on the boundary line (the origin (0, 0) is easiest). Substitute it into the inequality. If the statement is true, shade the side containing that point; if false, shade the opposite side.
5

Corner Points (Vertices)

The points where boundary lines intersect at the edges of the feasible region. These vertices are especially important in optimization problems.
✦

KEY TAKEAWAY

Think of each inequality as a velvet rope at a concert venue. One rope blocks you from going too far left, another from going too far forward. The feasible region is the area of the floor where you're allowed to stand—the space that satisfies every rope's restriction at once. The more ropes (inequalities) you add, the smaller and more specific that allowed area becomes.

SECTION 3

Visual Explanation — Graphing a System

The diagram below shows the system y ≤ 2x + 4 and y ≥ −x + 1 graphed on a coordinate plane. Each inequality produces a shaded half-plane, and where those half-planes overlap is the feasible region.

Intersection (−1, 2)xy24−2−424−2−4y ≤ 2x + 4y ≥ −x + 1Feasible RegionGreen overlap = all solutionsSystem: y ≤ 2x + 4 and y ≥ −x + 1
The violet region represents y ≤ 2x + 4 (below the violet line). The cyan region represents y ≥ −x + 1 (above the cyan line). The green overlap is the feasible region where both conditions are satisfied simultaneously. The boundary lines intersect at the vertex (−1, 2).

Notice that both boundary lines in the diagram are solid because the inequalities use ≤ and ≥, meaning points that lie exactly on the lines are included in the solution set. If the problem had used < or > instead, you would draw dashed lines to show that the boundary itself is not part of the solution. The green overlap zone is the feasible region: any ordered pair (x, y) you pick from that area will make both inequalities true.

SECTION 4

Mathematical Framework — Writing & Graphing

To work with systems of linear inequalities, you need to be comfortable with slope-intercept form and the process for deciding which half-plane to shade. Below are the key formulas and rules you'll rely on.

SLOPE-INTERCEPT FORM

y=mx+by = mx + b

m = slope (rise over run), b = y-intercept (where the line crosses the y-axis). Replace the = sign with <, >, ≤, or ≥ to create a linear inequality.

STANDARD FORM INEQUALITY

Ax+By≤C(or≥,<,>)Ax + By ≤ C (or ≥, <, >)

A, B, and C are constants. You can convert to slope-intercept form by solving for y. Remember: if you divide or multiply by a negative number, flip the inequality sign.

TEST POINT RULE

Substitute(0,0)intoeachinequalitySubstitute (0, 0) into each inequality

If the result is a true statement, shade the side containing (0, 0). If false, shade the opposite side. If (0, 0) lies on the boundary line, pick a different test point such as (1, 0).

⚠

Don't Forget!

When you solve an inequality for y and divide by a negative coefficient, the direction of the inequality reverses. For example, −2y ≥ 6 becomes y ≤ −3 (not y ≥ −3). This is one of the most common mistakes students make.

Here is the step-by-step algorithm you should follow every time you graph a system of linear inequalities. First, rewrite each inequality in slope-intercept form (y = mx + b) if it isn't already. Second, graph the boundary line for each inequality—solid for ≤ or ≥, dashed for < or >. Third, use the test point method to determine which side of each line to shade. Finally, identify the feasible region as the overlap of all shaded areas.

SECTION 5

Detailed Breakdown — Solid vs. Dashed & Shading Direction

One of the trickiest parts of graphing inequalities is remembering when to use a solid line versus a dashed line, and which direction to shade. The diagram below lays out all four cases side by side so you can build a clear mental picture.

Four Types of Linear Inequalitiesy < mx + bDashed lineShade belowLine NOT includedy ≤ mx + bSolid lineShade belowLine IS includedy > mx + bDashed lineShade aboveLine NOT includedy ≥ mx + bSolid lineShade aboveLine IS includedQuick-Reference Decision TreeStep 1: Rewrite in slope-intercept form (solve for y).Step 2: Draw the boundary line.≤ or ≥ → Solid< or > → DashedStep 3: Pick test point (0, 0). True → shade that side. False → shade opposite.Step 4: Overlap of all shaded regions = Feasible Region ✓
The four cases of linear inequalities. Notice that ≤ and ≥ use solid lines (boundary included) while < and > use dashed lines (boundary excluded). The shading direction (above or below) depends on whether y is greater than or less than the expression.
Inequality SymbolBoundary LineShading DirectionBoundary Included?
y < mx + bDashedBelow the lineNo
y ≤ mx + bSolidBelow the lineYes
y > mx + bDashedAbove the lineNo
y ≥ mx + bSolidAbove the lineYes

Summary of boundary line types and shading rules

SECTION 6

Worked Example — A Complete System

Let's graph the following system of inequalities and find the feasible region:

SYSTEM

y≤−x+6andy>(1/2)x−1andx≥0y ≤ −x + 6 and y > (1/2)x − 1 and x ≥ 0

This system has three inequalities, which means the feasible region is bounded by three boundary lines. The constraint x ≥ 0 limits solutions to the right side of the y-axis.

Graphing a System of Three Inequalities

  1. 1

    Step 1 — Graph the first boundary line: y = −x + 6

    This line has slope m = −1 and y-intercept b = 6. Plot the point (0, 6), then go down 1 and right 1 to reach (1, 5). Since the inequality is y ≤ −x + 6, draw a solid line and shade below it.
    Solid line through (0, 6) and (6, 0), shaded below.
  2. 2

    Step 2 — Graph the second boundary line: y = (1/2)x − 1

    This line has slope m = 1/2 and y-intercept b = −1. Plot (0, −1), then go up 1 and right 2 to reach (2, 0). Since the inequality is y > (1/2)x − 1, draw a dashed line and shade above it. The dashed line means points exactly on this line are not included.
    Dashed line through (0, −1) and (2, 0), shaded above.
  3. 3

    Step 3 — Graph the third boundary: x = 0

    The line x = 0 is simply the y-axis. Since x ≥ 0, draw a solid vertical line along the y-axis and shade to the right.
    Solid y-axis, shaded to the right.
  4. 4

    Step 4 — Verify with a test point

    Choose the point (2, 2), which appears to be inside the overlap. Check each inequality: (1) 2 ≤ −2 + 6 → 2 ≤ 4 ✓. (2) 2 > (1/2)(2) − 1 → 2 > 0 ✓. (3) 2 ≥ 0 ✓. Since all three are true, (2, 2) is in the feasible region.
    Confirmed: (2, 2) is in the feasible region.
  5. 5

    Step 5 — Identify the feasible region and vertices

    The feasible region is the triangular area where all three shaded zones overlap. The vertices (corner points) are found by solving pairs of boundary equations. The y-axis and y = −x + 6 intersect at (0, 6). The y-axis and y = (1/2)x − 1 intersect at (0, −1). The two slanted lines intersect where −x + 6 = (1/2)x − 1, giving x = 14/3 ≈ 4.67 and y = 4/3 ≈ 1.33, so the intersection point is (14/3, 4/3). However, because the second inequality is strict (y > (1/2)x − 1), the boundary line y = (1/2)x − 1 is dashed and points on it are excluded from the feasible region. This means the intersection point (14/3, 4/3) lies on the dashed boundary and is not included in the solution set. Similarly, the vertex (0, −1) lies on the dashed line and is also excluded. Only (0, 6) lies on solid boundaries and is fully included.
    Corner points: (0, 6) [included], (0, −1) [excluded — on dashed boundary], and (14/3, 4/3) [excluded — on dashed boundary]. The feasible region is an open triangular area; its interior satisfies all three inequalities, but the dashed edges and their endpoints are not part of the solution set.

SECTION 7

Common Strengths & Pitfalls

Students who master systems of inequalities tend to follow consistent habits, while those who struggle often fall into predictable traps. The table below compares best practices with common mistakes so you know exactly what to watch for.

Best Practice ✓Common Mistake ✗
Always solve for y first so you can clearly see the slope and y-intercept.Graphing directly from standard form and guessing the shading direction.
Check solid vs. dashed by looking at the inequality symbol before drawing.Drawing all boundary lines as solid, forgetting that < and > require dashed lines.
Use a test point (like (0, 0)) after drawing each line to confirm the correct shade side.Assuming 'greater than' always means shade above—this only works when the inequality is solved for y.
Flip the inequality sign when multiplying or dividing by a negative number.Forgetting to flip, which reverses the entire solution region.
Clearly label and shade each inequality differently (colors, patterns) before finding the overlap.Shading everything the same way and losing track of which region belongs to which inequality.

Best practices versus common mistakes when graphing systems of inequalities

✦

KEY TAKEAWAY

Think of graphing a system of inequalities like using multiple filters on a shopping website. One filter sets a price range, another selects a brand, a third picks a size. Each filter eliminates options, and the products that survive all filters are your feasible results. If your shading is wrong for even one inequality, your 'search results' will be completely off.

SECTION 8

Connection to Linear Programming & Advanced Topics

The skills you're building right now with systems of inequalities are the foundation for a powerful technique called linear programming. In linear programming, you don't just find the feasible region—you also have an objective function (like maximizing profit or minimizing cost) that you evaluate at each vertex of the feasible region. The optimal solution always occurs at a vertex, which is why finding corner points matters so much.

This Lesson: Systems of InequalitiesNext Level: Linear Programming
Graph inequalities and find the feasible region.Graph constraints AND optimize an objective function over the feasible region.
Any point in the feasible region is a valid solution.The best solution is at a vertex—test all vertices to find max or min.
Typically 2–3 inequalities with two variables.Real-world problems can have hundreds of variables and constraints, solved by computers.
Focus on graphing accuracy and identifying overlap.Focus on strategic formulation: translating word problems into systems.

How this lesson connects to more advanced optimization

You might also encounter systems of inequalities in higher math courses where the boundaries are curves instead of lines—these are called nonlinear systems of inequalities. For now, mastering the linear case gives you a rock-solid foundation. Every step you've learned—drawing boundary lines, choosing solid or dashed, using test points, finding the overlap—carries directly into these more advanced topics.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why the boundary line for y < 3x + 2 is dashed, while the boundary line for y ≤ 3x + 2 is solid. What does the difference mean for the solution set?
Show answer
A dashed boundary line indicates that points lying exactly on the line are not part of the solution set. The strict inequality < means y must be strictly less than 3x + 2, so any point where y equals 3x + 2 fails the condition. A solid line means the boundary is included because ≤ allows y to equal 3x + 2. In practical terms, the only difference between the two solution sets is whether or not the infinitely many points on the line itself are counted as solutions.

Question 2: BASIC CALCULATION

Determine whether the point (3, 1) is a solution to the system: y ≤ −2x + 8 and y ≥ x − 4. Show your substitution work.
Show answer
Substitute (3, 1) into the first inequality: 1 ≤ −2(3) + 8 → 1 ≤ 2. This is true. Now substitute into the second inequality: 1 ≥ 3 − 4 → 1 ≥ −1. This is also true. Since the point satisfies both inequalities, (3, 1) is in the feasible region and is a valid solution to the system. If even one inequality had been false, the point would not be a solution.

Question 3: INTERMEDIATE

Graph the system y > −(1/3)x + 4 and y ≤ 2x − 1. Find the intersection point of the boundary lines and describe the feasible region.
Show answer
To find the intersection, set −(1/3)x + 4 = 2x − 1. Adding (1/3)x to both sides gives 4 = (7/3)x − 1, then 5 = (7/3)x, so x = 15/7 ≈ 2.14. Substituting back: y = 2(15/7) − 1 = 30/7 − 7/7 = 23/7 ≈ 3.29. The boundary lines cross at approximately (2.14, 3.29). The first inequality (y > −(1/3)x + 4) is shaded above with a dashed line. The second (y ≤ 2x − 1) is shaded below with a solid line. The feasible region is the wedge-shaped area above the first line and below the second line, extending upward and to the right from the intersection point.

Question 4: APPLIED

A student has at most 20 hours per week to split between a tutoring job (x hours) and a retail job (y hours). The tutoring job requires at least 5 hours per week, and the retail job requires at least 4 hours per week. Write a system of inequalities that models these constraints, and identify one point in the feasible region.
Show answer
The constraints translate to: x + y ≤ 20 (total hours cap), x ≥ 5 (minimum tutoring), and y ≥ 4 (minimum retail). One point in the feasible region is (8, 10): 8 + 10 = 18 ≤ 20 ✓, 8 ≥ 5 ✓, and 10 ≥ 4 ✓. The feasible region is a polygon on the first-quadrant side of the coordinate plane, bounded by the line x + y = 20 on top, the vertical line x = 5 on the left, and the horizontal line y = 4 on the bottom. Any point inside or on the boundary of this polygon represents a valid weekly schedule.

Question 5: CRITICAL THINKING

Consider the system y ≤ x + 3, y ≥ x + 5. Explain why this system has no solution. Then modify one of the inequalities so that the system does have a non-empty feasible region, and describe what changes.
Show answer
The two boundary lines are y = x + 3 and y = x + 5. Both have slope 1, so they are parallel and never intersect. The first inequality shades below y = x + 3, while the second shades above y = x + 5. Since y = x + 5 is entirely above y = x + 3, there is no region that is simultaneously below the lower line and above the higher line—the shaded regions don't overlap at all. To create a feasible region, you could change the second inequality to y ≥ x + 1 (moving it below the first line). Now the region between the lines y = x + 1 and y = x + 3 satisfies both conditions, forming an infinite horizontal band as the feasible region.

SUMMARY

Lesson Summary

Summary

A system of linear inequalities consists of two or more inequalities graphed on the same coordinate plane. Each inequality defines a half-plane bounded by a boundary line that is drawn solid for ≤ and ≥ or dashed for < and >. You determine which side to shade using a test point, most commonly (0, 0).

The feasible region is the overlap of all shaded areas—every point inside it satisfies all inequalities simultaneously. The vertices (corner points) of the feasible region are found by solving pairs of boundary equations and are critical for optimization. Remember to flip the inequality sign when multiplying or dividing by a negative number, and always verify your feasible region with a test point inside it. These skills connect directly to linear programming, where feasible regions are used to find optimal solutions to real-world problems.

Varsity Tutors • Grade 9 Math • Systems of Inequalities — I can graph and interpret systems of linear inequalities to identify feasible regions.