Graphing Linear Inequalities: Step by Step

To graph a linear inequality in two variables, graph its boundary line, then shade the side that holds the solutions. Use a solid line for ≤ or ≥ and a dashed line for < or >. To choose a side, test a point such as (0, 0). If it makes the inequality true, shade the side with that point.

Before this lesson: You should know how to graph a line in slope-intercept form and how to solve multi-step inequalities, including when to flip the inequality sign.

Extra practice: The Algebra 1 Unit 5 Practice Pack has printable problems, answer keys and a cheat sheet for this unit.

What Does the Graph of an Inequality Show?

A linear equation like $y = x + 1$ has solutions that make a line. A linear inequality like $y \gt x + 1$ has many more solutions. They fill a whole region on one side of the line.

Try two points in $y \gt x + 1$ :

(0, 3): $3 \gt 0 + 1$ is true. So (0, 3) is a solution.
(0, 0): $0 \gt 0 + 1$ is false. So (0, 0) is not a solution.

The line $y = x + 1$ is called the boundary line. It splits the grid into two halves. Every point on one side is a solution, and every point on the other side is not.

Solid or Dashed Line?

The inequality sign tells you whether points on the boundary line are solutions.

Symbol	Meaning	Boundary line
$\lt$	less than	Dashed (points on the line don’t count)
$\gt$	greater than	Dashed (points on the line don’t count)
$\le$	less than or equal to	Solid (points on the line count)
$\ge$	greater than or equal to	Solid (points on the line count)

How to Graph a Linear Inequality

Get y alone if you can. If you multiply or divide by a negative number, flip the inequality sign.
Graph the boundary line. Replace the inequality sign with = and graph that line. Make it dashed for < or >, and solid for ≤ or ≥.
Test a point that is not on the line. (0, 0) is easiest unless the line goes through it.
Shade. If the test point makes the inequality true, shade its side. If not, shade the other side.

Shortcut: When y is alone on the left, $y \gt$ or $y \ge$ means shade above the line. $y \lt$ or $y \le$ means shade below. Still test a point to check.

Worked Examples

Example 1: Check a point

Is (2, 1) a solution of $y \lt 3x - 4$ ?

1 \lt 3(2) - 4

Step 1. Substitute $x = 2$ and $y = 1$ . (See evaluating algebraic expressions.)

1 \lt 6 - 4

Step 2. Multiply on the right side.

1 \lt 2

Step 3. Subtract on the right side.

Step 4. $1 \lt 2$ is true.

Answer: Yes, (2, 1) is a solution.

Example 2: Graph a “greater than or equal to” inequality

Graph $y \ge -x + 3$ .

Step 1. Check the form. $y$ is already alone, so the boundary line is $y = -x + 3$ in slope-intercept form.

Step 2. Plot the y-intercept (0, 3). The slope is −1, so go down 1 and right 1 to (1, 2).

Step 3. The sign is ≥, so draw a solid line through the points.

0 \ge -0 + 3

Step 4. Test the point (0, 0) in the inequality.

0 \ge 3

Step 5. Simplify the right side.

Step 6. This is false. (0, 0) is below the line and is not a solution, so shade above the line.

A solid line y = −x + 3 through (0, 3) and (3, 0), with the region above the line shaded. The point (0, 0) is below the line and is not shaded. — The graph of y ≥ −x + 3 has a solid boundary line and is shaded above.

Answer: The graph is a solid line $y = -x + 3$ with the region above it shaded.

Example 3: Flip the sign

Graph $2x - 3y \gt 6$ .

-3y \gt -2x + 6

Step 1. Get $y$ alone. Subtract $2x$ from both sides.

y \lt \frac{2}{3}x - 2

Step 2. Divide both sides by $-3$ . Dividing by a negative flips $\gt$ to $\lt$ . (See one-step inequalities.)

Step 3. Graph the boundary line $y = \frac{2}{3}x - 2$ . Plot (0, −2), then go up 2 and right 3 to (3, 0).

Step 4. The sign is $\lt$ , so draw a dashed line.

2(0) - 3(0) \gt 6

Step 5. Test (0, 0) in the original inequality.

0 \gt 6

Step 6. Simplify the left side.

Step 7. This is false. (0, 0) is above the line, so shade below the line. This matches the $\lt$ sign.

A dashed line y = (2/3)x − 2 through (0, −2) and (3, 0), with the region below the line shaded. The point (0, 0) is above the line and is not shaded. — The graph of y < (2/3)x − 2 has a dashed boundary line and is shaded below.

Answer: The graph is a dashed line $y = \frac{2}{3}x - 2$ with the region below it shaded.

Example 4: Horizontal and vertical boundary lines

Graph $x \lt 2$ and, separately, $y \ge -1$ .

x = 2

Step 1. For $x \lt 2$ , the sign is $\lt$ . Draw a dashed vertical line here.

Step 2. Points with x-values less than 2 are to the left of the line, so shade left.

y = -1

Step 3. For $y \ge -1$ , the sign is ≥. Draw a solid horizontal line here.

Step 4. Points with y-values of −1 or more are on or above the line, so shade above.

Answer: $x \lt 2$ : dashed vertical line at 2, shaded left. $y \ge -1$ : solid horizontal line at −1, shaded above.

Example 5: Word problem

A school club sells student tickets for $5 and adult tickets for $10. It needs to raise at least $200. Write and graph an inequality. Will 20 student tickets and 12 adult tickets be enough?

5s + 10a \ge 200

Step 1. Let $s$ be the number of student tickets and $a$ the number of adult tickets. “At least” means ≥.

5s = 200

Step 2. Graph the boundary line $5s + 10a = 200$ using its x- and y-intercepts. Let $a = 0$ .

s = 40

Step 3. Divide both sides by 5.

10a = 200

Step 4. Let $s = 0$ .

a = 20

Step 5. Divide both sides by 10.

Step 6. The sign is ≥, so draw a solid line through (40, 0) and (0, 20).

5(0) + 10(0) \ge 200

Step 7. Test (0, 0).

Step 8. That gives $0 \ge 200$ , which is false. Shade the side away from (0, 0), above the line.

5(20) + 10(12) = 100 + 120 = 220

Step 9. Check (20, 12).

Step 10. $220 \ge 200$ is true, so (20, 12) is in the shaded region.

Answer: $5s + 10a \ge 200$ . Yes, 20 student tickets and 12 adult tickets raise $220, which is enough. Only whole-number points in the shaded region make sense here, because you can’t sell part of a ticket.

Common Mistakes

Mistake	What it looks like	How to fix it
Wrong type of line	A solid line for $y \lt x + 2$	Use a dashed line for < and >, and a solid line for ≤ and ≥.
Not flipping the sign	$-3y \gt -2x + 6$ becomes $y \gt \frac{2}{3}x - 2$	Dividing by −3 flips the sign: $y \lt \frac{2}{3}x - 2$ .
Using the shortcut too early	Shading above for $-y \gt x$	The shortcut only works once y is alone with a positive coefficient. Test a point.
Testing a point on the line	Testing (0, 0) for $y \lt 2x$	(0, 0) is on $y = 2x$ . Pick another point, like (0, 1).

Graphing More Than One Inequality

When you graph two or more inequalities on the same grid, the solutions are the points in the overlap of the shaded regions. You’ll learn this in the next lesson, Systems of Inequalities.

Practice Problems

For problems 1 and 2, decide if the point is a solution. For problems 3–9, describe the graph: the boundary line, solid or dashed, and which side to shade.

(1, 4); $y \gt 2x + 1$
(−2, 0); $y \le -x - 3$
$y \lt x + 2$
$y \ge -2x + 1$
$y \gt \frac{3}{4}x - 3$
$3x + y \le 4$
$x - 2y \lt 6$
$x \ge -3$
$y \lt 4$
Pens cost $2 and notebooks cost $4. You can spend at most $20. Write an inequality. Can you buy 4 pens and 3 notebooks?

Show answers

4 \gt 2(1) + 1

Substitute x = 1 and y = 4.

4 \gt 2 + 1

Multiply.

4 \gt 3

Add.

This is true. Yes, (1, 4) is a solution.

0 \le -(-2) - 3

Substitute x = −2 and y = 0.

0 \le 2 - 3

Simplify the opposite of −2.

0 \le -1

Subtract.

This is false. No, (−2, 0) is not a solution.

y = x + 2

The sign is $\lt$ , so draw a dashed line.

0 \lt 0 + 2

Test (0, 0)

0 \lt 2

Simplify.

This is true, so shade the side with (0, 0), below the line.

y = -2x + 1

The sign is ≥, so draw a solid line.

0 \ge -2(0) + 1

Test (0, 0)

0 \ge 1

Simplify.

This is false, so shade the side without (0, 0), above the line.

y = \frac{3}{4}x - 3

The sign is $\gt$ , so draw a dashed line.

0 \gt \frac{3}{4}(0) - 3

Test (0, 0)

0 \gt -3

Simplify.

This is true, so shade the side with (0, 0), above the line.

y \le -3x + 4

Subtract 3x from both sides.

y = -3x + 4

The sign is ≤, so draw a solid line.

0 \le -3(0) + 4

Test (0, 0)

0 \le 4

Simplify.

This is true, so shade the side with (0, 0), below the line.

-2y \lt -x + 6

Subtract x from both sides.

y \gt \frac{1}{2}x - 3

Divide both sides by −2 and flip the sign (see one-step inequalities)

y = \frac{1}{2}x - 3

The sign is $\gt$ , so draw a dashed line.

0 \gt \frac{1}{2}(0) - 3

Test (0, 0)

0 \gt -3

Simplify.

This is true, so shade the side with (0, 0), above the line.

x = -3

The sign is ≥, so draw a solid vertical line.

x-values greater than or equal to −3 are on or to the right of the line
Shade to the right.

y = 4

The sign is $\lt$ , so draw a dashed horizontal line.

y-values less than 4 are below the line
Shade below.

10.

2p + 4n \le 20

Let p = pens and n = notebooks. “At most” means ≤.

Check (4, 3): $2(4) + 4(3) = 8 + 12 = 20$

20 \le 20

Compare.

This is true. Yes, you spend exactly $20, so the point is on the solid boundary line.

Want more practice? The Algebra 1 Unit 5 Practice Pack has 55 pages of problems on this and the other lessons in the unit, with step-by-step answers and a cheat sheet.

Graphing Linear Inequalities FAQ

How do you know whether to shade above or below the line?

Test a point that isn’t on the line, such as (0, 0). If it makes the inequality true, shade its side. If it makes it false, shade the other side.

When do you use a dashed line?

Use a dashed line for < and >, because points on the line are not solutions. Use a solid line for ≤ and ≥.

Why do you flip the inequality sign?

Multiplying or dividing both sides by a negative number reverses the order of numbers. For example, $2 \lt 5$ , but $-2 \gt -5$ . So the sign must flip to stay true.

What is the solution of a linear inequality in two variables?

The solution is every point in the shaded region. If the line is solid, the points on the line are solutions too.

Summary

Graph the boundary line first, using = in place of the inequality sign.
Dashed line for < or >. Solid line for ≤ or ≥.
Test a point not on the line, then shade the side that makes the inequality true.
Flip the sign when you multiply or divide by a negative number.
When y is alone, > and ≥ shade above, and < and ≤ shade below.

Previous lesson: Systems of Equations Word Problems

Next lesson: Systems of Inequalities

Back to the Algebra 1 Index