< or >
Dashed boundary
Points on the line are not solutions.
y > 2x + 3 uses a dashed line.
Graph linear inequalities step-by-step — see exactly which side to shade and why.
Why this page is different
Calculator Input
Single mode graphs the first linear two-variable inequality and explains the test point method.
Math Keyboard
Tap symbols if you do not want to use the keyboard.
Single Inequality Result
The boundary line is dashed, and the shaded half-plane excludes the origin side.
Boundary: dashed line
Test point: (0, 0)
Test Point Shading Demonstrator
Drag to pan, scroll to zoom, or tap the grid to test a point.
The tapped point is outside the full solution.
(0, 0) is checked against 1 constraint.
Step-by-Step Breakdown
1. Move all terms to one side
Graphing starts from the standard-form boundary line and the sign of the half-plane around it.
2. Solve for y to read the boundary line
Writing the inequality in y-form makes the slope and intercept easy to see.
3. Choose the shaded side
The test point (0, 0) does not satisfy the inequality, so shade the other side.
Constraint Details
Constraint 1
dashed boundaryTest point (0, 0) makes -3 > 0 false.
Core Method
After the boundary line is drawn, the line splits the coordinate plane into two sides. A test point tells you which side contains the solution.
Use (0,0) when it is not on the boundary. If the boundary passes through the origin, switch to a nearby simple point like (1,0) or (0,1), then substitute that point into the original inequality.
If the statement is true, shade the side containing the test point. If the statement is false, shade the opposite side.
False test point
Shade away from the origin.
True test point
Shade the side containing the origin.
Rewrite the inequality as a boundary equation.
Use a dashed line for < or > and a solid line for <= or >=.
Pick a test point that is not on the boundary line.
Substitute the test point and shade the side that makes the inequality true.
For systems, keep only the overlapping shaded region.
Boundary Rules
< or >
Points on the line are not solutions.
y > 2x + 3 uses a dashed line.
<= or >=
Points on the line are included in the solution.
y <= -x + 4 uses a solid line.
AND logic
A point must satisfy every inequality in the system.
The final answer is the shared shaded region.
Graphing Checklist
Convert standard form such as 2x + 3y >= 6 into a boundary line.
Decide dashed versus solid before shading.
Choose a test point that is not on the line.
Substitute the point into the original inequality, not a rearranged version with sign mistakes.
For systems, verify a sample point in the overlap against every inequality.
Dashed line + test point false
Test point (0,0): 0 > 3 is false, so shade away from the origin.
Solid line + test point true
Test point (0,0): 0 <= 4 is true, so shade the side containing the origin.
Two shaded regions, one shared solution
The darker overlap satisfies both inequalities at once, just like AND/intersection logic.
Use the test point method. Pick a point not on the boundary line, often (0,0), substitute it into the original inequality, and shade that point's side if the statement is true. If it is false, shade the other side.
A solid boundary line is used for <= or >= because points on the line are included. A dashed boundary line is used for < or > because points on the line are excluded.
The origin is usually easiest to substitute because both coordinates are zero. The only time to avoid it is when the boundary line passes through (0,0).
Choose another simple point that is not on the boundary, such as (1,0) or (0,1). The calculator automatically switches to a safe backup test point.
Graph each inequality separately, shade each correct half-plane, and keep only the region where all shaded areas overlap. That overlap is the solution to the full system.
A one-variable linear inequality shades a ray or interval on a number line. A two-variable linear inequality shades a region of the coordinate plane because each solution is an ordered pair (x,y).
Only when the inequality uses <= or >=. Strict inequalities use dashed boundaries because equality is excluded.
Substitute the point's x and y values into the original inequality. For systems, the point must make every inequality true.