HSA.REI.D.12: Graphing Linear Inequalities and Systems of Inequalities in Two Variables
In plain English: HSA.REI.D.12 is the Common Core algebra standard that asks students to graph the solutions of a linear inequality in two variables as a half-plane, excluding the boundary for a strict inequality, and to graph the solution set of a system of linear inequalities as the intersection of the half-planes. A test point shows which side to shade. It is usually taught in Algebra I.
Graph the solutions to a linear inequality in two variables as a half-plane (excluding the boundary in the case of a strict inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the corresponding half-planes.
Common Core State Standards for Mathematics · Domain: Reasoning with Equations and Inequalities (REI) · Cluster: Represent and solve equations and inequalities graphically Also written as HSA-REI.D.12 or A-REI.12 · Official standard
In this lesson, students learn that the solutions to a linear inequality in two variables fill a whole region of the coordinate plane, a half-plane, rather than a single line. They graph the boundary line, decide whether it is solid (≤ or ≥) or dashed (< or >), and use a test point to choose which side to shade. They then graph a system of linear inequalities and identify its solution set as the region where the half-planes overlap.
The central idea is that every point in the shaded region is an ordered pair that makes the inequality true, and every point outside it makes the inequality false. Students check this by substituting points, including points on the boundary and at the corner where two boundaries meet.
Learning Objectives
By the end of this lesson, students will be able to:
Graph the boundary line of a linear inequality in two variables and decide whether it is solid or dashed
Use a test point to decide which half-plane contains the solutions, including when the inequality is not in slope-intercept form
Graph the solution set of a system of two or more linear inequalities as the intersection of half-planes
Decide whether a given point, including a boundary point or a corner point, belongs to the solution set and justify the decision by substitution
Interpret points in a graphed solution region for a system given in context
Prior Knowledge Required
Students should already be comfortable with:
Graphing a line from slope-intercept form or from its intercepts 8.EE.B.6
Solving and graphing one-variable inequalities on a number line 7.EE.B.4
Solving a linear inequality for one variable, including reversing the symbol when multiplying or dividing by a negative number HSA.REI.B.3
Understanding that the graph of an equation is the set of all its solutions HSA.REI.D.10
Project a coordinate grid with the line y = x + 1 drawn on it. Give each student three sticky dots, or have them list three ordered pairs on paper.
Warm-Up Prompt
"Find three points where y = x + 1 is true, three points where y > x + 1 is true, and three points where y < x + 1 is true. Where on the grid did each group of points end up?"
Collect points from the class and plot them. Students will see that the points for y > x + 1 all sit above the line and the points for y < x + 1 all sit below it. Ask: "Is there any point above the line that makes y < x + 1 true?" This sets up the idea that the line splits the plane into two half-planes.
Direct Instruction20 minutes
Present the three-step routine for graphing one linear inequality, then extend it to systems:
Graph the boundary: replace the inequality symbol with = and graph that line. Use a solid line for ≤ or ≥ (boundary points are solutions) and a dashed line for < or > (boundary points are not solutions).
Pick a test point that is not on the line, usually (0, 0). Substitute it into the original inequality.
Shade the half-plane that contains the test point if the result is true, and the other half-plane if it is false.
For a system, repeat for each inequality on the same axes. The solution set is the region shaded by every inequality. Check a point in the overlap in every inequality.
Work these four examples on the board. Examples 1 and 2 match Diagram 1's routine, and Example 4 is shown in Diagram 2.
Strict inequality, slope-intercept form
Graph y > 2x - 3.
Equation: Dashed line with y-intercept -3 and slope 2. Test (0, 0): 0 > -3 is true, so shade the side containing (0, 0), which is above the line.
Standard form with a negative coefficient
Graph 3x - 2y ≤ 12.
Equation: Solve for y: -2y ≤ -3x + 12, so y ≥ 1.5x - 6 (the symbol reverses when dividing by -2). Solid line through (4, 0) and (0, -6). Test (0, 0): 0 ≤ 12 is true, so shade above the line.
Vertical and horizontal boundaries
Graph the system x ≥ -2 and y < 3.
Equation: Solid vertical line x = -2, shade to the right. Dashed horizontal line y = 3, shade below. The solution set is the overlap. The corner (-2, 3) is not included because it lies on the dashed line.
System of two inequalities
Graph the system y ≥ x - 2 and x + 2y < 8.
Equation: Solid line y = x - 2, shade above. Dashed line through (0, 4) and (8, 0), shade below. The boundaries meet at (4, 2), which is not a solution because 4 + 2(2) < 8 is false.
Guided Practice15 minutes
Students work in pairs on four inequalities: y ≤ -x + 4, 2x + y > 6, y < 3x (the boundary passes through the origin, so they must choose a different test point such as (1, 0)), and the system y > -1 and y ≤ 2x + 1. For each graph, one partner states the boundary type and the other chooses and checks the test point. Circulate and listen for three common errors: shading from the original form without reversing the symbol after dividing by a negative, using (0, 0) as a test point when it lies on the boundary, and drawing a solid line for a strict inequality.
Independent Practice15 minutes
Students graph three systems on their own: one with two slanted boundaries, one with a vertical or horizontal boundary, and one given in context with x ≥ 0 and y ≥ 0 (for example, 4x + 6y ≤ 48 for the cost of x smoothies at $4 and y sandwiches at $6 on a $48 budget). For each system, students label the solution region, mark every corner point as included (closed dot) or not included (open dot), and verify one point inside the region in every inequality.
Closure5-10 minutes
Exit ticket: Show a graph of the system y < x + 2 and y ≥ -x. Ask students to (a) decide whether (-1, 1) is a solution and explain using the graph and substitution, and (b) explain in one sentence why the line y = x + 2 is dashed. The point (-1, 1) is where the two boundaries meet, and it is not a solution because 1 < -1 + 2 is false.
Differentiation Strategies
For Struggling Students
Start with inequalities already in slope-intercept form, and add standard form only after students can graph the boundary and shade reliably
Give a reference card: ≤ and ≥ mean solid line, < and > mean dashed line, and a true test point means shade its side
Use two colored pencils for systems, one per inequality, so the overlap is easy to see
Have students check shading by substituting one point from each side of the line
For Advanced Students
Ask for a system of two inequalities whose solution set is empty, and a system whose solution set is a strip between parallel lines
Give a shaded triangular region and ask students to write the system of three inequalities that produces it, including which boundaries are solid
Ask which corner points of a region with a mix of solid and dashed boundaries belong to the solution set, and why
Assessment Guidance
What to Look For
Check three things on every graph: the boundary line is correct, its style (solid or dashed) matches the symbol, and the shading is on the correct side. Students who shade "above for greater than" without a test point often shade the wrong side after rewriting an inequality with a negative y-coefficient. For systems, ask students to name one point that is a solution and one that is not, and to prove each by substitution. A correct picture with no verification is incomplete evidence of understanding.
02
Classroom Activities
3 Activities
1
Human Coordinate Plane
15 minWhole class
Tape a large coordinate grid on the floor or use the classroom floor tiles, with axes from -5 to 5. Students stand on lattice points and physically form the half-plane for a given inequality, which makes the idea of "all points that satisfy it" concrete.
Procedure
Lay a rope along the boundary line of y ≥ x - 1. Ask students to stand on any point they think is a solution, then call out their coordinates and check them aloud.
Ask whether a student standing on the rope is a solution. Change the inequality to y > x - 1 and ask those students to step off.
Add a second rope for x + y ≤ 3. Only students who satisfy both inequalities stay standing. The class sees the system's solution set as the overlap.
Debrief Questions
Why did some students have to step off when ≥ became >?
How could you tell from where you stood, without substituting, whether you were a solution?
The corner of the two ropes is (2, 1). Is it a solution of the system y ≥ x - 1 and x + y ≤ 3? What changes if one symbol becomes strict?
Modification for Distance Learning
Use a shared graphing app. Each student drags a point to a location they believe is a solution, and the class checks the points on screen before shading the region.
2
Match the Graph
20 minPairs
Pairs receive 9 graph cards and 9 inequality or system cards. Several cards differ only in the boundary style or the shaded side, so students must check each detail.
Card Set
y < 2x + 1, y ≤ 2x + 1, y > 2x + 1 and y ≥ 2x + 1 (same boundary, different style and side)
x - 2y > 4, which students must rewrite as y < 0.5x - 2 before comparing
x ≤ 3 and y > -2 (vertical and horizontal boundaries)
The system y ≥ x and y ≤ -x + 4, whose corner (2, 2) is included
The system y > x + 2 and y < x - 1, which has no solution
Procedure
Pairs match all cards and write the test point they used to confirm each match.
Each pair picks the match they found hardest and explains it to another pair.
Close by asking why the last system has no solution: the lines are parallel and the half-planes point away from each other.
3
Budget Region
25 minGroups of 3
Groups graph a system in context and use the graph to answer questions. This keeps the focus on graphing and reading a solution region, with the inequalities provided.
Scenario
A club sells x bags of popcorn at $3 each and y bottles of water at $2 each. They have room to carry at most 60 items, and they want to raise at least $120. The constraints are x + y ≤ 60, 3x + 2y ≥ 120, x ≥ 0 and y ≥ 0.
Tasks
Graph all four inequalities on one grid with x and y from 0 to 70, and shade the solution region.
Find the corners of the region: (40, 0), (60, 0) and (0, 60), where x + y = 60 meets 3x + 2y = 120.
Decide whether each plan works, using the graph and then substitution: 30 popcorn and 20 water (it works: 50 items and $130); 10 popcorn and 40 water (it does not: $110); 45 popcorn and 20 water (it does not: 65 items).
Explain why a point such as (25.5, 30) lies in the region but is not a realistic plan.
Gallery Walk Variation
Each group posts its graph with one plan marked. Other groups visit, decide whether the plan is in the region, and leave a sticky note with their check.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Graphing a Single Linear Inequality
Panel A uses a solid boundary because ≤ includes the line, and the test point (0, 0) is a solution, so its side is shaded. Panel B uses a dashed boundary because < excludes the line, and the test point (0, 0) is not a solution (open circle), so the opposite side is shaded. Both graphs are drawn to scale, one grid square per unit.
Diagram 2: A System as the Intersection of Half-Planes
Each inequality shades one half-plane. The darker overlap is the solution set of the system. The corner (4, 2) satisfies y ≥ x - 2 but lies on the dashed line x + 2y = 8, so it is not a solution. Drawn to scale, one grid square per unit.
04
Homework Assignment
~30 min
HSA.REI.D.12 Homework: Graphing Inequalities and Systems
Directions: Use graph paper and a ruler. For each graph, (a) write the equation of each boundary line and say whether it is solid or dashed, (b) show the test point you used, (c) shade the solution set, and (d) answer the question asked. Label corner points with a closed dot if they are solutions and an open dot if they are not.
Part 1: Single Inequalities (Problems 1-3)
Graph y ≤ -½x + 3. Name one point on the boundary that is a solution.
Graph 4x - y > 2. First solve for y and explain what happens to the inequality symbol. Then decide whether (1, 2) is a solution.
Graph -2x + 5y ≥ -10 using the x- and y-intercepts of the boundary. Is (0, 0) a solution? Is (5, 3)?
Part 2: Systems (Problems 4-6)
Graph the system y > x + 3 and y ≤ -2x + 9. Find the point where the boundaries meet and explain whether it is a solution. Then decide whether (1, 7) is a solution.
Graph the system x + y ≤ 6, x ≥ 0 and y ≥ 1. Describe the shape of the solution region and list its corner points.
Maya can work at most 12 hours a week, split between tutoring (x hours at $15 per hour) and a café job (y hours at $12 per hour). She wants to earn at least $150. The constraints are x + y ≤ 12, 15x + 12y ≥ 150, x ≥ 0 and y ≥ 0. Graph the system and find the corners of the region. Is 4 hours of tutoring and 6 hours at the café a solution? Is 8 hours of tutoring and 3 hours at the café?
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Boundary Line
Correct line with correct solid or dashed style
Correct line but wrong style, or small plotting error
Wrong or missing line
Shading
Correct half-plane or overlap, supported by a test point
Correct shading with no test point shown
Wrong side or no shading
Points and Corners
Corner points found and marked open or closed correctly
Corners found but inclusion not marked or incorrect
Corners not found
Verification and Interpretation
Points checked by substitution and answers explained in context
Checks shown but explanation missing
No checks
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which description matches the graph of y < 3x + 1?
Answer: B
The symbol < is strict, so the boundary y = 3x + 1 is dashed. The test point (0, 0) gives 0 < 1, which is true, and (0, 0) is below the line (the line crosses the y-axis at 1), so shade below. Choice C has the right line style but shades the side where y > 3x + 1. Choices A and D use a solid line, which would wrongly include boundary points.
Question 2 of 20 · Multiple Choice
Which point is a solution of 2x - y ≥ 4?
Answer: C
Substitute: 2(3) - 1 = 5, and 5 ≥ 4 is true. For (2, 1), 2(2) - 1 = 3, which is less than 4; students who pick D often compare with the wrong boundary value. (0, 0) gives 0 and (1, 3) gives -1, both less than 4.
Question 3 of 20 · Multiple Choice
Which statement about the graph of x + 4y > 8 is correct?
Answer: D
The symbol > is strict, so the boundary through (8, 0) and (0, 2) is dashed. Substituting (0, 0) gives 0 > 8, which is false, so the origin is not a solution and the shading is on the other side. Choice B gets the line style right but skips the test-point check.
Question 4 of 20 · Multiple Choice
A student solves -3y ≤ 6x - 9 for y. Which result is correct?
Answer: D
Divide every term by -3 and reverse the symbol: y ≥ -2x + 3. Choice A divides correctly but forgets to reverse the symbol, which leads to shading the wrong half-plane. Choice B reverses the symbol but divides by 3 instead of -3.
Question 5 of 20 · Multiple Choice
Which describes the graph of x ≥ -1 in the coordinate plane?
Answer: A
In two variables, x ≥ -1 is true for every point whose x-coordinate is -1 or greater, whatever y is. The boundary x = -1 is vertical and solid (≥ includes it), and the solutions are to the right. Choice B graphs y ≥ -1. Choice D treats the inequality as a one-variable number-line problem.
Question 6 of 20 · Multiple Choice
Which point is in the solution set of the system y > x and y < 4?
Answer: B
(2, 3): 3 > 2 is true and 3 < 4 is true, so it is in both half-planes. (3, 3) lies on the dashed line y = x, so 3 > 3 is false. (1, 5) fails y < 4. (5, 4) fails both inequalities.
Question 7 of 20 · Multiple Choice
The boundaries of y ≥ 2x - 1 and y < -x + 5 meet at one point. Which statement is correct?
Answer: C
Set 2x - 1 = -x + 5, so 3x = 6 and x = 2, y = 3. The point satisfies y ≥ 2x - 1 (3 ≥ 3) but lies on the dashed line y = -x + 5, so 3 < 3 is false. It is not a solution. Choice A forgets that one boundary is dashed. (3, 5) lies only on the first line, and (3, 2) reverses the coordinates.
Question 8 of 20 · Multiple Choice
Which system has no solution?
Answer: A
The boundaries in A are parallel (slope 2). Points must be above the higher line and below the lower line at the same time, which is impossible. Choice C is the strip between the same two lines, which contains (0, 0). Choice B contains (0, 5). Choice D is the wedge to the left of the origin between the two lines, which contains (0, 0) and (-3, 1).
Question 9 of 20 · Multiple Choice
Why is (0, 0) a poor test point for y < 2x?
Answer: B
The line y = 2x passes through the origin, so (0, 0) is on the boundary and cannot tell you which side to shade. Use a point such as (1, 0): 0 < 2 is true, so shade the side containing (1, 0). Choice A is false: (0, 0) is often a solution, as in y < 2x + 1.
Question 10 of 20 · Multiple Choice
A graph shows a dashed line through (0, -2) and (3, 0), and the side containing (0, 0) is shaded. Which inequality is graphed?
Answer: D
Both points satisfy 2x - 3y = 6, so that is the boundary. The line is dashed, so the symbol is strict. The origin gives 0 < 6, which is true, matching the shading. Choice A shades the side without the origin. Choice B would have a solid line. Choice C has intercepts (2, 0) and (0, -3), a different line.
Question 11 of 20 · Multiple Choice
A student buys x items at $5 and y items at $10 with at most $100, so 5x + 10y ≤ 100, x ≥ 0, y ≥ 0. Which purchase is in the solution region?
Answer: D
5(6) + 10(7) = 100, and 100 ≤ 100 is true, so (6, 7) is on the solid boundary and is included. The others cost $110 (A and B) and $105 (C), which are outside the region. Students who reject D often forget that a solid boundary includes its points.
Question 12 of 20 · Multiple Choice
Which statement about the graph of y ≤ 5 is correct?
Answer: C
Every point with y-coordinate 5 or less is a solution, whatever x is. The boundary y = 5 is horizontal and solid, and the half-plane below it is shaded. Choice B graphs x ≤ 5. Choice D confuses the inequality with the equation y = 5.
Question 13 of 20 · Multiple Choice
Which describes the solution set of the system x ≥ 0, y ≥ 0 and x + y ≤ 4?
Answer: A
Each inequality uses ≥ or ≤, so all three boundaries are solid and the edges are included. The overlap of the three half-planes is the triangle with those corners. Choice B would be correct only if all three symbols were strict. Choice D lists the corners but forgets the points inside.
Question 14 of 20 · Multiple Choice
Which point lies on the boundary of y > -x + 2 but is not a solution?
Answer: A
For (1, 1), -1 + 2 = 1, so it is on the line y = -x + 2, and 1 > 1 is false. (0, 3) and (2, 2) are solutions that are not on the line, and (0, 0) is neither on the line nor a solution.
Question 15 of 20 · Short Answer
Describe how to graph y ≥ 3x - 4: the boundary, its style, the test point and the side to shade.
Boundary: y = 3x - 4, with y-intercept -4 and slope 3 (it also passes through (2, 2)). Style: solid, because ≥ includes the boundary. Test point: (0, 0) gives 0 ≥ -4, which is true. Shade: the side containing (0, 0), which is above the line.
Question 16 of 20 · Short Answer
Graph 2x + 5y < 10. Give the intercepts of the boundary, its style and which side you shade.
Boundary 2x + 5y = 10 has intercepts (5, 0) and (0, 2). It is dashed because < is strict. The test point (0, 0) gives 0 < 10, which is true, so shade the side containing the origin, which is below the line.
Question 17 of 20 · Short Answer
For the system y ≤ x + 3 and y > -2x, find where the boundaries meet, decide whether that point is a solution, and give one point that is a solution.
Set x + 3 = -2x, so x = -1 and y = 2. The corner (-1, 2) is on the dashed line y = -2x (2 > 2 is false), so it is not a solution. One solution is (1, 1): 1 ≤ 4 is true and 1 > -2 is true. Other points in the overlap also work.
Question 18 of 20 · Short Answer
Is (3, -1) a solution of the system x - y > 3 and x + y ≤ 2? Explain where it lies on the graph.
Yes. x - y = 3 - (-1) = 4, and 4 > 3 is true. x + y = 3 + (-1) = 2, and 2 ≤ 2 is true. The point lies on the solid boundary x + y = 2 and inside the half-plane x - y > 3, so it belongs to the solution set.
Question 19 of 20 · Short Answer
A graph shows a dashed line along y = x with the region above it shaded, and a solid vertical line x = 3 with the region to its left shaded. Write the system that the overlap represents, and say whether (3, 3) is a solution.
y > x and x ≤ 3. The dashed line gives a strict symbol, and the solid line gives ≤. The point (3, 3) is on both boundaries: it satisfies x ≤ 3, but 3 > 3 is false, so it is not a solution.
Question 20 of 20 · Short Answer
A theater sells x adult tickets at $12 and y student tickets at $8 and needs at least $480, so 12x + 8y ≥ 480 with x ≥ 0 and y ≥ 0. Is 20 adult and 25 student tickets a solution? Is 30 adult and 15 student tickets? Why is the point (22.5, 26.5) in the shaded region but not a realistic answer?
20 adult and 25 student tickets bring in 12(20) + 8(25) = $440, which is less than $480, so no. 30 adult and 15 student tickets bring in 12(30) + 8(15) = $480, which is on the solid boundary, so yes. The point (22.5, 26.5) satisfies the inequality (12(22.5) + 8(26.5) = 482), but tickets come in whole numbers, so only points with whole-number coordinates in the region are realistic plans.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does it mean that the solutions form a half-plane?
A line divides the coordinate plane into two regions called half-planes. For a linear inequality such as y > 2x - 3, every point on one side of the line y = 2x - 3 makes the inequality true and every point on the other side makes it false. So the solution set is not a few points or a line, it is an entire half-plane, possibly with its boundary line included.
How do I know whether to draw a solid or a dashed boundary line?
Look at the symbol. For ≤ or ≥, points on the line make the inequality true, so draw a solid line. For < or >, points on the line make it false, so draw a dashed line. The standard itself says to exclude the boundary in the case of a strict inequality.
Can I just shade above the line for greater than and below for less than?
Only when the inequality is solved for y, as in y > mx + b. In standard form, such as 3x - 2y ≤ 12, the direction depends on the sign of the y-coefficient, and students who skip that step often shade the wrong side. A test point works in every form, including vertical boundaries like x < 4, where "above" and "below" do not apply. Teach the test point as the reliable method and "above or below" as a shortcut to check it.
What if the boundary line goes through the origin?
Then (0, 0) is on the line and cannot tell you which side to shade. Choose any point that is clearly off the line, such as (1, 0) or (0, 1), and substitute it into the original inequality.
Is the corner point where two boundary lines meet part of the solution set?
Only if it satisfies every inequality in the system. If both boundaries are solid, the corner is included. If either boundary is dashed, the corner makes that strict inequality false, so it is not included. Mark it with an open dot. Substituting the corner into each inequality is a quick way to check.
Can a system of linear inequalities have no solution?
Yes. If the half-planes do not overlap, the solution set is empty. This happens with parallel boundaries whose half-planes face away from each other, such as y > x + 2 and y < x - 1. Systems with three or more inequalities can also have no overlap even when no two boundaries are parallel.
What are the common mistakes students make on this standard?
Forgetting to reverse the inequality symbol when dividing by a negative number while solving for y
Drawing a solid line for a strict inequality, or a dashed line for ≤ or ≥
Using (0, 0) as the test point when it lies on the boundary
Shading each inequality in a system but not identifying the overlap as the solution set
Marking a corner point as a solution when one of its boundaries is dashed
How is HSA.REI.D.12 usually tested?
Students are asked to graph an inequality or system, to choose the graph that matches an inequality, to decide whether given points are solutions, or to write the inequality shown by a graph. On the digital SAT, linear inequalities in one or two variables and systems of linear inequalities are part of the Algebra domain, often in context, where a question asks which point satisfies a set of constraints.
Why do word problems add x ≥ 0 and y ≥ 0?
Quantities like hours worked or tickets sold cannot be negative, so these two inequalities limit the region to the first quadrant, including the axes. Even then, some points in the region may not make sense, such as half a ticket. Writing the constraints from a situation is the focus of HSA.CED.A.3, and this standard focuses on graphing them and reading the result.
What comes after HSA.REI.D.12?
Students use these graphs when they write and interpret constraints in context (HSA.CED.A.3) and when they study how graphs of equations and their intersections represent solutions (HSA.REI.D.11). In later courses, graphing a feasible region is the first step of linear programming, where a quantity such as profit is maximized over the region.
07
Related Standards
6 standards
These standards connect to HSA.REI.D.12: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.EE.B.4Prerequisite
Use variables to write and solve simple equations and inequalities from word problems
