AI-Math - Graph Inequalities Calculator

A Solution Set You Can See

Solving an inequality gives a set of answers, and a graph is the honest way to display a set. Which picture you draw depends on how many variables are present.

One variable ( $-1 < x \le 3$ ) lives on a number line: a segment with an open circle at an excluded endpoint and a filled circle at an included one.
Two variables ( $y \le 2x - 1$ ) lives on the coordinate plane: a boundary line with one entire side shaded. Every point in the shaded region — thousands of them — satisfies the inequality.

The boundary itself carries information. A solid line means the boundary points are solutions, which happens with $\le$ and $\ge$ . A dashed line means they are not, which happens with the strict $<$ and $>$ . Drawing the wrong style is the same error as writing a bracket where a parenthesis belongs.

Boundary, Style, Test Point

Three steps, always in this order.

1. Draw the boundary

Replace the inequality sign with $=$ and graph that line. Use the intercepts, or slope-intercept form if the equation is already solved for $y$ .

2. Choose solid or dashed

Sign	Boundary
$<$ or $>$	Dashed — endpoints excluded
$\le$ or $\ge$	Solid — endpoints included

3. Shade using a test point

Pick any point not on the boundary — $(0,0)$ whenever the line misses the origin, because the arithmetic is trivial. Substitute it into the original inequality.

True → shade the side containing that point.
False → shade the other side.

This test is why you never need to remember rules like " $\le$ means shade below". Those rules only hold once the inequality has been solved for $y$ , and they fail the moment the coefficient of $y$ is negative — the very case where a division flipped the sign.

Common Mistakes to Avoid

Forgetting the flip. Dividing $-2y > 4$ by $-2$ gives $y < -2$ . Miss this and you shade the wrong half of the plane.
Testing a point on the boundary. $(0,0)$ is useless for $3x + 2y > 0$ ; it makes the inequality an equality. Choose $(1,1)$ instead.
Mixing up circle types. Open circle for $<$ and $>$ , filled circle for $\le$ and $\ge$ .
Shading both sides of a system. With two inequalities the answer is only the overlap; shade lightly and outline the common region.
Assuming the shading is always below for $\le$ . True only after isolating $y$ with a positive coefficient.

示例题目

Step 1: Boundary:

y = 2x - 1

, a line through

(0, -1)

with slope

2

; it also passes through

(1, 1)

Step 2: The sign is

\le

, so draw the line solid

Step 3: Test

(0, 0)

: is

0 \le 2(0) - 1 = -1

? No, that is false

Step 4: Because the test failed, shade the side that does not contain the origin — the region below and right of the line

Step 5: Confirm with

(2, 0)

:

0 \le 2(2) - 1 = 3

✓, and

(2,0)

does lie in the shaded region

Answer: Solid line

y = 2x - 1

with the half-plane below it shaded

Step 1: Boundary:

3x + 2y = 6

. Intercepts are

(2, 0)

and

(0, 3)

Step 2: The sign is strict

>

, so draw the line dashed

Step 3: Test

(0, 0)

: is

3(0) + 2(0) = 0 > 6

? No

Step 4: Shade the side away from the origin

Step 5: Confirm with

(4, 0)

:

3(4) + 2(0) = 12 > 6

✓

Answer: Dashed line through

(2,0)

and

(0,3)

, shaded on the far side from the origin

Step 1: Subtract

1

from all three parts:

-3 < 3x \le 9

Step 2: Divide all three parts by

3

(positive, so no flip):

-1 < x \le 3

Step 3: Draw an open circle at

-1

and a filled circle at

3

, shading between them

Step 4: Check the endpoints: at

x = 3

,

3(3) + 1 = 10 \le 10

✓ included; at

x = -1

,

3(-1) + 1 = -2

, which is not greater than

-2

, so it is correctly excluded

Answer:

-1 < x \le 3

, or

(-1, 3]

常见问题

Dashed for the strict signs < and >, because points on the line do not satisfy the inequality. Solid for ≤ and ≥, because they do. It is the graphical version of choosing parentheses versus brackets in interval notation.

Pick any test point that is not on the boundary — (0,0) if the line does not pass through it — and substitute it into the original inequality. If the statement is true, shade the side containing that point; if false, shade the other side.

Graph each one separately with its own boundary style and shading, then keep only the region where the shadings overlap. That intersection is the solution set; points in just one shaded region satisfy only one of the constraints.

Almost always because the inequality sign was not flipped when both sides were multiplied or divided by a negative number. Using a test point on the original, unmanipulated inequality avoids the problem entirely.

Graph Inequalities Calculator

Boundary, dashed or solid, and the shaded half-plane — decided by a test point, not a guess

A Solution Set You Can See

Boundary, Style, Test Point

1. Draw the boundary

2. Choose solid or dashed

3. Shade using a test point

Common Mistakes to Avoid

示例题目

常见问题

When is the boundary line dashed instead of solid?

Which side of the line do I shade?

How do I graph a system of two inequalities?

Why did my shading end up on the wrong side?

相关求解器

相关学习指南

免费试用 AI-Math

Graph Inequalities Calculator

Boundary, dashed or solid, and the shaded half-plane — decided by a test point, not a guess

A Solution Set You Can See

Boundary, Style, Test Point

1. Draw the boundary

2. Choose solid or dashed

3. Shade using a test point

Common Mistakes to Avoid

示例题目

Problem: Graphy≤2x−1Graph y ≤ 2x - 1Graphy≤2x−1

Problem: Graph3x+2y>6Graph 3x + 2y > 6Graph3x+2y>6

Problem: Graphthecompoundinequality−2<3x+1≤10onanumberlineGraph the compound inequality -2 < 3x + 1 ≤ 10 on a number lineGraphthecompoundinequality−2<3x+1≤10onanumberline

常见问题

When is the boundary line dashed instead of solid?

When is the boundary line dashed instead of solid?

Which side of the line do I shade?

Which side of the line do I shade?

How do I graph a system of two inequalities?

How do I graph a system of two inequalities?

Why did my shading end up on the wrong side?

Why did my shading end up on the wrong side?

相关求解器

相关学习指南

免费试用 AI-Math